You need to understand velocity, speed, and acceleratifion

School Science Lessons
Physics
2025-10-24
Speed and acceleration
Contents
14.1.0 Acceleration
14.2.0 Acceleration due to gravity
14.3.0 Catapults
14.4.0 Falling
14.5.0 Scalars and vectors
14.6.0 Speed and road safety
14.7.0 Velocity and speed

14.2.0 Acceleration due to gravity
The acceleration caused by the gravitational attraction of the Earth.
14.2.1 Acceleration down an incline
14.2.2 Acceleration of marbles down an incline
14.2.3 Galileo's acceleration experiment
14.2.4 Galileo's inclined plane
14.2.5 Galileo's pendulum experiment
14.2.6 Monkey and hunter
14.2.7 Parabolic trajectory of thrown chalk
14.2.8 Path of a projectile
14.2.9 Swing bucket in vertical circle
14.2.10 Trajectory model

14.3.0 Catapults
Catapults, slingshots, trebuchet, mangonel, onager, ballista
14.3.1 Catapults
14.3.2 Thrust and recoil

14.4.0 Falling
Falling, measuring standard gravity, g
See diagram 14.5.0: Falling joke
14.4.1 Free fall
14.4.2 Equations of uniformly accelerated motion
14.4.3 Balance with a metre stick
14.4.4 Ball bearings fall together
14.4.6 Catch a coin
14.4.8 Coin and feather
14.4.9 Coins falling at the same time
14.4.10 Drop coin and paper
14.4.11 Drop paper and ball
14.4.12 Drop strings with attached weights
14.4.13 Dropping balls timer
14.3.14 Falling chimney paradox
14.4.15 Falling washers on a string
14.4.16 Falling water from a tap
14.4.17 Forces on a falling object
14.4.18 Free fall timer
14.4.19 Ink jet marker
14.4.20 Man on a springboard
14.4.21 Paths of projectilesl
14.4.22 Pendulum-timed free fall
14.4.23 Rotating turntable
14.4.24 Three-holes bottle
14.4.25 Time a falling body with a stopwatch
14.4.26 Ticker timer

14.5.0 Scalars and vectors
14.5.1 Scalars and vectors
14.5.2 Aeroplane ground speed
14.5.3 Boat crossing river
14.5.4 Addition of vectors when P and Q at right angles
14.5.5 Polygon of vectors
14.5.6 Vector addition
14.5.7 Vector resolution
14.5.8 Vectors addition problem
14.5.9 Verify the parallelogram law

14.7.0 Velocity and speed
14.4.2 Velocity
14.4.2 Equations of uniformly accelerated motiony
Experiments
14.7.3 Muzzle velocity
14.7.4 Terminal velocity
14.7.5 Time of flight
14.7.6 Toy car running on moving paper

14.1.0 Acceleration
Average acceleration = change in velocity / time taken, a = (v-u) / t
Instantaneous acceleration is the rate of change of velocity with respect to time as measured by an accelerometer in an aircraft.
You can read velocity and acceleration from displacement-time graphs and velocity-time graphs.
Deceleration occurs when the speed of a vehicle decreases, because the vectors for velocity and acceleration are in opposite directions.

14.2.1 Acceleration down an incline
See diagram 8.235: Acceleration down an incline
1. Use a grooved three metre plank.
Incline it so that marbles will roll down the groove.
Arrange small tin flags hung from wires so that the marbles hit them and make "clinks" sounds.
Put the flags at regular intervals, e.g. 25, 50, 75, 100 cm, from the beginning of the plank.
Roll a marble down the groove and listen to the intervals between "clink" sounds.
The time between the "clinks" will reduce as the ball rolls down the incline.
Arrange the flags so that the clinks occur at equal intervals of time.
Measure the distance between the flags.
The distance between the flags increases down the incline.
2. Use an inclined air track.
For timing use a stop clock and metre stick.
Put lights that flash every second along an incline and horizontal track such that they are flashing at the moment the ball passes.
3. Observe a car on an inclined wire.
Stretch a long wire diagonally across the chalkboard with chalk marks at every metre.
Time a car as it accelerates past the marks.
4. Observe a ball on an incline.
Roll a ball bearing down the groove of a plastic metre stick.
Use a slow roller solid wheel turning on a small axis to roll down an incline.
5. Use a Duff's plane, chalk ball on incline.
Use a ball that leaves a trail while rolling down a chalk covered trough.

14.2.2 Acceleration of marbles down an incline
Use a 3 m plank of wood with a groove down the centre.
Incline the plank so that marbles can roll down the groove.
Arrange small tin flags hung from wires so that the marbles hit them and make "clinks" sounds.
Put the flags at regular intervals, e.g. 25, 50, 75, 100 cm, from the end of the plank.
Roll a marble down the groove and listen to the time intervals between "clink" sounds.
The time intervals between the "clinks" reduce as the ball rolls down the incline.
Arrange the flags so that the clinks occur at equal intervals of time.
Measure the distance between the flags.
The distance between the flags increases down the incline in the ratio 1: 3: 5: 7: 9.

14.2.3 Galileo's acceleration experiment
See diagram 14.2.9 Galileo's acceleration experiment
When Galileo was trying to measure the acceleration due to gravity with only a human pulse as a time keeper he had the idea of slowing down the acceleration so that it could be measured more accurately.
So, while keeping in mind the previous pendulum experiment, he devised a system of two ramps with inclinations that could be varied and he collected much data on the movement of a ball down one ramp
and up the other.
1. Similarly to the previous experiment, if a round bronze ball rolls down a ramp that is smoothly connected to another steeper upward ramp, the ball will roll up the second ramp to a level almost equal to the level it started at, even if the two ramps have different slopes.
Then it will then continue to roll backwards and forwards between the two ramps, coming to rest, because of friction and air resistance.
So the ball must be going the same speed coming off one ramp as it does coming off the other.
If we imagine the second ramp steeper and steeper the ball as just falling.
So for a ball rolling down a ramp, the speed at various heights is the same as the speed the ball it would have attained by just falling vertically from its starting point to that height ignoring any difference
between a rolling ball and a smoothly sliding or falling ball.
Having placed this board in a sloping position, by raising one end roll the ball, along the channel, and note the time required to make the descent.
Now roll the ball only one quarter the length of the channel and having time of its descent is one half of the former.
Try other distances and compare the time for the whole length, the spaces traversed were to each other as the squares of the times, and for all inclinations of the channel, along which the ball was rolled.
2. Use a large vessel of water placed in an elevated position with a small diameter pipe attached to the bottom that allows water to be collected time of each descent, whether for the whole length of the channel or for part of its length.
Weigh the water collected after each descent.

14.2.4 Galileo's inclined plane
See diagram 14.2.11 Galileo's inclined plane
Galileo's experiment, said to be done by himself at the Leaning Tower of Pisa, showed that different weight objects fall with the same acceleration and reach the ground simultaneously.
To study how gravity affects acceleration, he did not use falling objects, because they accelerate too quickly for accurate measurement.
So he used an inclined plane.
He found that the acceleration of a billiard ball at an angle θ = 60o down the inclined plane is not much slower than acceleration of a free falling object.
However, at an angle θ = 30o he could measure time accurately enough using a water clock, a clepsydra, and musical intervals.
He used the distance down the inclined plane during the first time interval as his unit of measure.
He discovered that the distances down the inclined plane increased by odd numbers of intervals of 1, 3, 5, 7, 9, at all angles.
Table 14.4.10 Time units

Time in units	Distance down inclined plane	Time squared
after 1 time unit	1 unit	12 = 1
after 2 time units	1 + 3 = 4 units	22 = 4
after 3 time units	1 + 3 + 5 = 9 units	32 = 9
after 4 time units	1 + 3 + 5 + 7 = 16 units	42 = 16

So the distance covered is directly proportional to the square of the time, d ∝ t2.

14.2.5 Galileo's pendulum experiment
See diagram 14.2.9 Galileo's pendulum experiment
Attach an iron ball to a thread, pull it back to height h1 and let it swing as a pendulum from a nail, n1, in a board to h2, then swings almost back to h1 where it started.
Note that h1 and h2 are almost at the same height above the ground level.
Stop the swinging and insert a nail, n2 in the board directly below n1, so the n1 h1 > n1 n2.
Pull the thread back to h1 and let the iron ball swing.
As the pendulum swings through its lowest point, the thread hits the nail, n2 to shorten the pendulum so that the iron ball swings up more steeply, to G with the nail at n2.
However, the pendulum swings almost back to h1 where it started.
Note that the heights of h1 and G are almost at the same height above the level ground.
So the speed of the iron ball passes through the lowest point in the swing is the same in both directions as before n2 was inserted.
The iron ball must move at the same speed when the thread hit n2 as when the nail wasn't there.
So how far it swings out from the vertical depends on how fast it is moving at the lowest point.

14.2.6 Monkey and hunter
See diagram 8.239: Monkey and hunter 1 and 2
See diagram 14.2.14: Monkey and hunter 3
1. Aim the gun at the monkey when the monkey is held up high.
When the ball leaves the gun, the monkey should drop.
The ball will hit the monkey since they fall at the same rate.
Lower velocity means each falls a greater distance before hitting.
A hunter shoots a ball from compressed air gun at a target released when the gun is fired.
The ball hits the target in mid air.
2. Monkey and hunter, monkey and bullet, mid-air target, path of a projectile.
The apparatus consists of a "gun" aimed at a toy monkey suspended by an electromagnet.
The bullet, on leaving the barrel, activates a light sensitive switch that releases the monkey.
The gun is aimed to hit the monkey.
The bullet can be aimed at different angles and distances so that collision occurs at different points of the monkey's path.
3. Acceleration of a freely falling body is equal to the acceleration of the vertical component of a projectile.
This experiment shows that the vertical and the horizontal velocities of a projectile are independent of each other and that the acceleration of a freely falling body is the same as the acceleration of the vertical component of a body in projectile motion.
A "gun" is aimed at a "monkey" suspended by an electromagnet.
When the "bullet" leaves the barrel of the "gun" it activates a light sensitive switch to release the "monkey".
This experiment is called "The monkey and hunter", but nowadays use a drink-can instead of a poor monkey!
4. The projectile is a ball bearing and the target "monkey" is a metal drink-can hanging from an electromagnet.
The circuit of the electromagnet includes two bared wires fixed parallel to and each side of the axis of a cardboard tube.
They project about 2.5 cm beyond the end of the tube.
Complete the electrical circuit with a short length of copper wire resting on the projecting wires as a switch.
Fix the cardboard tube in a stand so that it points towards the drink-can.
Note the angle of the cardboard tube above the horizontal.
Blow the ball bearing up the cardboard tube.
When the ball bearing passes the end of the cardboard tube, it displaces the piece of copper wire, opens the switch and no electric current flows through the electromagnet.
So the metal can is released to fall.
The ball bearing can hit the metal drink can in mid-air if the angle of the cardboard tube above the horizontal is correct.
The ball bearing will hit the drink-can at different angles of the cardboard tube (between 0o and 90o), different distances, and different initial speeds of the ball bearing up the cardboard tube.

14.2.7 Parabolic trajectory of thrown chalk
See 2.0.5: Conic sections, parabola
See 2.0.6: Parabola equation
Throw a piece of chalk so it follows a parabolic path drawn on the chalkboard.
Roll ink dipped balls down an incline onto a tilted stage on an overhead projector.
Roll a tennis ball covered with chalk dust across a tilted blackboard.
To demonstrate a model that shows the various parabolic trajectories corresponding to different initial conditions.

14.2.8 Path of a projectile, water from a hose
1. Water from a hose
Observe the path of water from a hose held at different angles and with different water pressure.
Note the angle that results in the greatest distance reached by the water, 45o.
2. Flight of a shuttlecock
Hit a shuttlecock in front of a blackboard marked with chalked squares.
Observers can draw similar curves on graph paper to show the flight path of the shuttlecock.

14.2.9 Swing bucket in vertical circle, swinging pail
This experiment is best done outside the classroom on the grass.
1. Fill a bucket with water and attach a heavy rope or chain to the handle.
Hold on to the end of the rope or chain and swing the bucket to and fro, until you can swing it quickly in a vertical circle without the water spilling out.
Ask an observer to time the circular swings.
Start swinging the bucket slower and slower until water starts to fall out.
Ask the observer to tell you the slowest swing speed before the water started falling out.
2. Repeat the experiment with a bucket full of nails.
Observe when the nails start to drop down or hit the bucket making a noise at the top of the swing.

14.2.10 Trajectory model
See 14.2.15: Trajectory model
Construct a model that shows the various parabolic trajectories corresponding to different initial conditions.
Attach pendulums to a rod at equal intervals with the length of each pendulum proportional to the square of its distance from the pivot point.
Vary the angles of inclination of the rod to show different parabolic trajectories corresponding to different initial conditions.

14.3.1 Catapults
Catapults, trebuchet, mangonel, onager, ballista
"Da Vinci Catapult" (siege weapon), soft clay balls reach 4 m, (toy product)
See diagram 4.140: Trebuchet
Safety
Projectiles may be considered as "weapons" or "firearms" under the local "Weapons Act", so before making such a device consult the categories of "weapons" provided by the "Police Service".
Even if not classified as a weapon, a teacher may consider it too dangerous for a school activity.
For example, a potato cannon may be classified as a "Muzzle Loading Firearm", if a "firearm" is defined as a "weapon that on being aimed a a target can cause death or injury".
Catapult, trebuchet, bow & arrow, baseball bat, kitchen knife, are usually not considered weapons, because bodily harm is not intended, but if they are used for a "behavioural offence" they then become a "weapon".
Catapult
A catapult is any kind of device that shoots or launches a projectile by mechanical means.
Catapults use stored energy (potential energy) to hurl a projectile (kinetic energy).
A simple catapult used to shoot small stones is made from a forked stick and an elastic band.
Catapults can be use to launch gliders from the side of a hill or aircraft from ships.
Before the invention of guns, catapults were military machines.
Trebuchet
A trebuchet was a siege engine used to smash walls or to throw projectiles over them.
It uses leverage to propel a projectile accurately.
The sling and the arm swing up to the vertical position, where usually assisted by a hook, one end of the sling releases, propelling the projectile towards the target with great force.
The trebuchet used a falling, hinged, counterweight to launch a projectile.
The counterweight was much heavier than the projectile.
It was very accurate and consistent in operation.
The counterweight pivots around a much shorter distance than the projectile, so the projectile reaches a much higher velocity than the counterweight end of the beam.
The limiting speed of the falling counterweight is the speed of free fall.
A teacher will limit on the size of the throwing arm, e.g. use paddle pop ice cream sticks and limit the size of the counterweight or the projectile, e.g. golf ball, or the projectile should be soft, e.g. a softball.
Mangonel
The mangonel had an arm with a bowl-shaped bucket or sling containing a projectile attached to the end.
The arm is pulled back from the vertical the it rotates at a high speed and throws the projectile out of the bucket, towards the target.
The launch velocity of the projectile is equal to the velocity of the arm at the bucket end.
The launch angle was controlled by a crossbar.
The mangonel was used for launching projectiles at lower angles than the trebuchet.
The mangonel used an energy storage mechanism to rotate the arm consisting of a tension device, a bent cantilever, or a torsion device, e.g. a twisted rope, connected to the arm.
It relied on deformable materials that would wear out and lose elasticity during use.
Elastic potential energy is stored in the tension of the rope and the arm.
It does not release its energy in a linear fashion, because the arm makes an arc with radius equal to the arm's length.
So potential energy is transferred into rotational kinetic energy and to translational kinetic energy of the projectile.
Onager
The onager was like a small trebuchet, but using a torsion bundle to rotate the arm, similar to the mangonel.
Ballista
The ballista was like a big crossbow using twisted rope as the energy source, like a mangonel.
It was used to shoot bolts, spears and other projectiles.

14.3.2 Thrust and recoil
A force applied on a surface in a direction perpendicular or normal to the surface is called thrust, a reaction force of Newton's laws of motion.
A thrust force tends to increase velocity or momentum.
A drag force tends to decrease velocity or momentum.
The propellor of a ship or plane and a jet or rocket engine exerts this propulsive force.
The recoil of a gun describes how it springs back with the force of discharge.
Some guns have a hydraulic cylinder that compresses and so absorbs the shock of recoil and decelerates the rearward thrust of the firearm.
Experiments
1. Use a bow and arrow or catapult.
The force exerted by one arm to pull back equals the force used to hold steady the bow or the fork of the catapult.
2. Hold a hose in your hand then turn of the tap.
Feel the backwards force on your hand.
Drop the hose on the ground and see it move backwards in a snake-like motion.
3. Fire a rifle.
As the bullet leaves the rifle, you feel the recoil force on your shoulder.
When large guns are fired, they tend to move backwards.

14.4.1 Free fall
Free fall, measuring g
An object, mass m, falling freely, without friction or air resistance, where the gravitational field is g newton / kg, experiences a force on it of mg, Newton, called its weight.
F = ma, Weight = mg
So acceleration due to gravity during free fall = g, where g = the size of the earth's gravitational field at that place.
Near the surface of the earth, g = 9.8 Newton per Kg, so free fall acceleration = 9.8 metres per second2, 9.8 m/sec2.

14.4.2 Equations of uniformly accelerated motion, gravity
Displacement, velocity and acceleration, linear kinematics
1. v = u + at
a = (v-u) / t
2. s = (u + v) t / 2
3. s = ut + at2 / 2
s = ut + 0.5 at2
1.4 v2 = u2 + 2as
Where u = initial velocity, v = velocity after time t, s = distance travelled in time t, a = constant acceleration (v, s, and a are positive in the direction of u).

14.4.3 Balance with a metre stick
Balance with a metre stick, stationary meeting point, centre of mass, centre of gravity
See diagram 8.146: Stationary meeting point
A body acts as if its mass is concentrated at a single point, the centre of mass.
Gravity acts through the same point, the centre of gravity.
If a vertical line through the centre of gravity of an object does not pass through its base, the object falls over.
An object, e.g. a motor car, will not roll over easily if it has a low centre of gravity and a wide base.
The centre of gravity of a metre stick or uniform rod is in the centre.
If two fingers support the rod and one finger moves towards the centre of gravity the rod begins to tip towards that finger to increase the weight and increase the force of friction.
The other finger feels less wight and has less friction so the rod easily slides above it.
Experiments 1. Support a metre stick or uniform rod over your two index fingers so that each finger is exactly 1 cm from the end.
The weight on the fingers feels exactly the same.
Keep the left finger in place, but slowly move the right finger towards the centre until it is half way between the centre and the end.
The metre stick feels heavier on the right finger than on the left finger.
Move the fingers together while keeping the metre stick balanced.
As your left finger moves towards the right finger, the metre stick feels heavier on it.
The weight on each finger feels about the same when the two fingers move together to be just each side of the centre of gravity.
2. Repeat the experiment by moving one finger quickly and the other finger slowly.
Maintain the ruler in balance while moving the fingers.
If the metre stick remains horizontal, the two fingers always meet at the centre of the metre stick.
3. Repeat the experiment using two round smooth pencils on a level table instead of fingers.
Move the right pencil towards the middle of the rod while holding the left pencil in place.
As the right pencil approaches the middle of the rod the pencils have the same distance to the ends of the rod.
4. Repeat the experiment by hanging your hat on one end of the metre stick.
Note the new position of the centre of gravity.
5. Repeat the experiment with a broom to find its centre of gravity.
6. Slide two kitchen scales under a loaded beam.
Note the scale readings of the moving and stationary scales change in the same way that your fingers feel change in weight under the metre stick.
7. Put an empty drink-can on a rough wooden board.
Raise one end of the board until the drink-can falls over.
At that angle, a vertical line through the centre of gravity of the drink-can passes outside its base.
8. Stand still then raise your right arm sideways.
Nothing happens.
Raise your right leg sideways.
If your upper body moves to the left, your centre of gravity remains over your left foot so you remain stable.
If you keep your upper body rigid, your centre of gravity moves to the right and is no longer over your left foot, so you fall over.

14.4.4 Ball bearings fall together
See diagram 8.234: Simultaneous fall
See diagram 14.2.4: Spring-loaded device
1. A spring loaded device drops one ball and projects the other horizontally.
2. Two balls simultaneously dropped and projected horizontally hit the floor together.
Drop one billiard ball and shoot another out simultaneous.
One ball is projected horizontally as another is dropped simultaneously.
Instructor rolls a ball off the hand while walking at a constant velocity.
3. Use two clothes-pegs, a pair of ball bearings and a wide rubber band.
Fix the band lengthways around one peg.
Then open the peg and force a ball against the tension of the rubber band between the prongs of the peg.
Grip the other ball with the second peg.
Hold the pegs side-by-side, pointing away horizontally above the floor.
Squeeze both pegs at once.
At the same moment, one ball begins to fall vertically, and the other is shot forwards.
Note what happens by looking and listening very carefully.
Repeat the experiment from different heights and with a tighter rubber band.
If the experiment is done correctly, while the ball bearings land in different places they strike the ground simultaneously.
2. Use two clothes pegs, a pair of ball bearings A and B, a wide rubber band about 8 cm long.
Fix the rubber band lengthways around one clothes peg.
Then open the clothes peg and force ball bearing A against the tension of the rubber band between the prongs of the clothes peg.
Grip ball bearing B with the second peg.
Hold the two pegs side by side, pointing away horizontally above a sheet of tin on the floor.
Squeeze the ends of the arms of both clothes pegs.
At the same moment, ball bearing A begins to fall vertically, and ball bearing B is shot forwards.
Observe what happens by looking and listening when the ball bearings hit the sheet of tin.
Repeat the experiment at heights 0.5 m, 1.0 m, 1.5 m, 2.0 m and with a tighter rubber band.
If the experiment is done correctly, ball bearings A and B land in different places on the sheet of tin, but they strike the ground simultaneously.
3. Two balls simultaneously dropped and projected horizontally hit the floor together.
Drop one billiard ball and shoot another out simultaneous.
One ball is projected horizontally as another is dropped simultaneously.
Instructor rolls a ball off the hand while walking at a constant velocity.

14.4.6 Catch a coin
Catch a coin starting with the fingers at the midpoint of the coin.

14.4.7 Catch a metre stick
Catch a metre stick, reaction time, drop the ruler
See diagram 9.249: Reaction time
1. Drop a metre stick and use the distance it drops before catching it to find the reaction time of the catcher.
If a body falls from a height s, the distance it falls after t seconds = gt2 / 2.
So if you measure s, you can obtain t, t = 2S / g.
Hold metre ruler vertically with the zero on the scale down and the 100 on the zero on the scale up.
With your arm stretched horizontally, hold the ruler vertically between the thumb and first finger with the lower edge of the first finger at the zero
on the scale.
Open your fingers then close them again as quickly as possible to catch the ruler again.
Record the distance to the downward edge of your first finger.
Repeat the experiment and calculate the average distance down the metre stick.
Use t = 2S / g to calculate the time of the falling ruler, i.e. your reaction time.
For example, if distance dropped = 9.0 cm, t = 2S / g, = 2 x .09 / 9.81 = 0.135 seconds
2. Repeat the experiment under the following pairs of conditions:
1. Hold the ruler first with your left hand.
Hold the ruler first with your right hand.
2. Talk to others while doing the experiment.
Do not talk to others while doing the experiment.
3. Allow loud background music.
Do not allow loud background music.
4. Try other contrasting conditions to see whether your reaction time is affected.
5. Fast money.
Use the thumb and first finger to hold the top of a vertical new bank note.
Hold it above the hand of another person.
Tell that person to catch the bank note.

14.4.8 Coin and feather
Coin and feather, penny and feather, guinea and feather experiment, free fall tube
See diagram 14.2.6: Guinea and feather experiment
(A guinea was a coin originally worth one pound sterling.) This famous experiment was repeated by an astronaut on the moon in 1971, without the need of an air pump!
1. Shake a coin and feather to one end of the 150 cm sealed glass tube full of air.
Stand the glass tube on its end.
Quickly invert the glass tube.
The coin falls much faster, because air resistance affects the feather more.
Connect the glass tube to an air pump.
Turn on the pump to evacuate the glass tube.
Stand the glass tube on its end.
Invert the glass tube.
The coin and feather fall at same the uniform acceleration in the absence of air resistance.
2. Repeat the experiment with same size cork and lead ball.
3. Use a one metre long glass tube containing a lead sinker and a feather.
Observer the accelerations of the objects when you invert the glass tube.
Attach the glass tube to an evacuating pump and evacuate the tube for about two minutes.
Observe the accelerations of the objects after evacuation.

14.4.9 Coins falling at the same time
See diagram 16.159: Flicking
Observe the falling object and horizontal projectile object start off at the same time and fall down to the ground at the same time too.
Fold a paper card in half, then fold each side one third from the end outwards to form a convex.
Place coins on each side of the centre ridge of the card and hold one end on the table edge.
Flick the ridge of the card to the side with your middle finger of right hand.
One of the coins will be thrown several metre s away, the other will fall straight to the floor at the exact same moment.
First observe if the two coins start motion at the same time.
Then repeat the steps above, note which coin hits the floor first.
Flick the card lightly, hear the click.
If you hear only one click, that means the direction of the falling object at first has not any influence on the falling times.
This is, because the accelerations of the two falling coins are the same if the air resistance is disregarded.

14.4.10 Drop coin and paper
1. Drop a large coin, e.g. half dollar, with a very small piece of paper on it.
Hold the coin horizontally and give it a slight spin to it horizontal as it falls.
The coin and paper fall together.
2. Show that currents of air flowing around the falling coin does not cause the paper to stick to it.
Hold the coin high and horizontally with the paper on it.
Carry the coin straight down with a speed greater than the falling coin.
The piece of paper is left behind the coin.

14.4.11 Drop paper and ball
1. Practice dropping heavy and light objects simultaneously with one hand.
Use two identical sheets of paper.
Crumple one sheet of paper into a tight ball.
Simultaneously drop the ball of paper and sheet of paper from the same height.
2. Practice scrunching a standard sheet of paper to form a sphere.
Select a rubber ball with a similar diameter.
Simultaneously drop the rubber ball and the standard sheet of paper.
Repeat the experiment by simultaneously dropping the rubber ball and the sheet of paper scrunched to form a sphere.

14.4.12 Drop strings with attached weights
Drop two strings with attached weights, one with equal distance intervals and the other with equal time intervals (1, 4, 9, 16).
Let strings hit a hard floor and listen to the sounds of weights hitting the floor.
Drop a string with wood blocks or lead balls tied at unit intervals and equal time intervals.

14.4.13 Dropping balls timer
Use a latching relay system for turning a standard timer on and off so that an electromagnet releases the ball and starts the clock and a catcher stops the clock.
Drop light and heavy balls through a multiple pass light beam and show the output on an oscilloscope.

14.3.14 Falling chimney paradox
1. Falling chimney
See photo, Falling chimney
See diagram 14.3.17: Falling chimney paradox
When a tall brick chimney falls, because of a demolition charge, the upper part tends to spin slower than the bottom part as the length increases.
The initial angular velocity = sqrt 3g/L, g is the acceleration due to gravity and L is the height of the chimney.
The falling chimney tends to break at L/3.
2. Falling hinged metre stick The vertical component of the acceleration of the end of a hinged stick is greater than g where the vertical angle < 35
See diagram 14.3.17: Falling chimney paradox
Connect two metre sticks at one end.
Place the two sticks flat on the bench.
Raise one end of the upper metre stick to make an angle of 35o with the lower metre stick, then release it.
The raised end falls to meet the end of the lower metre stick.
Repeat the experiment, but place a coin at the end of the raised metre stick.
Place a plastic cup 5 cm from the end of the lower meter stick.
Let the raised end fall.
The coin falls into the cup.

14.4.15 Falling washers on a string
See diagram 8.154: Falling washers on a string
1. Tie an iron washer to one end of a 2 m string.
Tie another six washers every 30 cm along the string.
Stand on a chair, hold the end of the string with no washer attached and let the string hang down.
Let the bottom washer be 30 cm above the floor.
Release the string and listen to the sound of the washers hitting the floor.
The sound intervals are not the same, they get smaller as the washers hit the floor.
2. To repeat the experiment, tie an iron washer to one end of a 4 m string, then tie on other washers 15 cm, 45 cm, 75 cm, 105 cm and 135 cm apart.
So the distances of the washers from the end of the string are 15 cm, 60 cm, 135 cm, 240 cm and 375 cm.
Stand on a chair, hold the end of the string with no washer attached and let the string hang down.
Let the bottom washer be 30 cm above the floor.
Release the string and listen to the sound of the washers hitting the floor.
The sound intervals are the same - (0.175 seconds).
The acceleration, because of gravity (acceleration of free fall) g = 9.8 m / sec2 .
For a free falling body, h = gt2 / 2, so as time increases, the speed of a free falling body is faster and faster.
Using the formula h = gt2 / 2, the falling distances of a free falling body at 1, 2, 3, 4, 5 seconds are 0.5g, 2g, 4.5g, 8g, 14.5g.
The falling distances every second are (0.5g) (2g - 0.5 g = 1.5g) (4.5g - 2 g = 2.5g) (8 g - 4.5 g = 3.5g) (14.5 g - 8 g = 4.5),
i.e. the ratio is 0.5: 1.5: 4.5: 3.5: 4 = 1: 3: 5: 7: 9.
For similar falling objects separated by the ratio of distances 1: 3: 5: 7: 9, the time intervals of falling are equal.

14.4.16 Falling water from a tap
The velocity of water in free fall from a tap increases with the distance of fall.
However, the volume of water coming out of the tap remains constant as long as the tap is not turned further on or off.
If the volume remains constant during the fall the cross-sectional area must decrease during the fall, i.e. cross sectional area X velocity = a constant.
So the steam of water gets thinner, starting when it leaves the tap.

14.4.17 Forces on a falling object
The two forces on an object falling through the air are 1. the gravitational force, expresses as weight of the object, and 2. the aerodynamic drag of the object.
Weight, W = mass of the object, m X gravitational acceleration, g, where g = 9.8 m sec-2, on the earth's surface.
Drag
F_D = 1/2 rho x u2 x C_D x A
where F_D, is the drag force, i.e. the force component in the direction of the flow velocity, rho is the mass density of the fluid, u is the flow velocity relative to the object, C_D is the drag coefficient, and A is the reference area.
So Drag depends on u2.
Force, F = mass, m X acceleration, a, F = m a, so a = F / m
Net force on the falling object = W- D, so a = (W - D) / m
Falling body accelerates until D = W, then (W - D) = 0, so no net force and acceleration = 0, so body now falls with constant velocity, called terminal velocity.
Equations for free falling objects
Acceleration due to gravity, g = 9.81 m / s2
Position as a function of time, s = gt2
Velocity as a function of time, v = gt
Velocity as a function of position, v = 2gx

Table 14.2.0 Velocity and position

t	v = gt	s = gt2
0	0	0
1 second	9.81 m / s	4.91 m
2 seconds	19.62 m /s	19.62 m
3 seconds	29.43 m /s	44.15 m
4 seconds	39.24 m / s	78.48 m
5 seconds	49.05 m /s	122.63

1. An iron ball is dropped from a height, followed second later by an identical iron ball dropped from the same height.
While both iron balls are still falling, does the vertical separation between them stay the same, increase or decrease?
After 1 second the first iron ball has fallen 4.91 m and the second iron ball has fallen 0 m.
The separation is 4.91 - 0 = 4.91 m.
After 2 seconds the first iron ball has fallen 19.62 m and the second iron ball has fallen 4.91 m.
The separation is 19.62 - 4.91 = 14.7 m.
So the separation increases.

14.4.18 Free fall timer
See diagram 14.3.1m: Measuring g
1. The ball can be released from four different heights, 0.5m, 1m, 1.5m and 2m.
Place the ball between the contacts.
Slide the rod to the left and tighten set screw to hold ball in place.
Tap the contact pad and reset the timer, in that order.
Loosen set screw to release ball.
The timer will start.
The timer stops when the ball strikes the pad.
Move ball release to a new position and repeat.
The experiment gives g to within 5%.
2. Time a ball as it drops 0.5 m, 1.0 m, 1.5 m, or 2.0 m.
Drop a magnet through several equally spaced coils of wire and examine the induced voltage on an oscilloscope.

14.4.19 Ink jet marker
Drop a length of wood near an ink marker that leaves lines at equal time intervals.
Use a rotating ink jet to spray a paper sleeve on a falling metre stick.

14.4.20 Man on a springboard
Is the second lower diagram correct?
See diagram: Springboard
Drag force
The drag equation is a formula used to calculate the force of drag experienced by an object due to movement through a fully enclosing fluid.
F_D = 1/2 rho x u2 x C_D x A
D is the drag force, the force component in the direction of the flow velocity, rho is the mass density of the air, u is the flow velocity relative to the man,
A is the reference area, (area of the man), CD is the drag coefficient (depending on the actual situation).
So Drag depends on u2.
However, once the man jumps, the net force on him is just the drag force backwards so the man will move backwards relative to the car.
The second diagram is correct.
Comments: Assuming that the car was travelling at 100 KPH, the drag force would be significant and if there was a head wind it would increase further.
(Try sticking your head out of the window at 100 KPH!)
It all then depends how high he jumps as his time in the air will be the time that the drag force applies.
I doubt that the diver would end up behind the car, but they could be in danger of hitting the diving board on the way down.
If you want to include the full formulae as part of your answer you could include s= ut + 1/2 at2 etc to get the time in the air.
In fact a fully worked out answer would be a good addition to your text book on line.
Cycling professionals can probably give you some data on the actual drag factor.
They go to extraordinary lengths to streamline riders and bikes for time trial events.

14.4.21 Paths of projectiles
See diagram 8.239: Path of a projectile
See 2.0.5: Conic sections
See 2.0.6: Parabola equation
See diagram 4.152: Path of a projectile: A Copper gate, B Ball
An object is in free fall if its motion is only affected by the force of gravity.
So, ignoring air resistance, an object is in free fall the moment it is released when dropped or projected.
1. The apparatus is used to show that the vertical and horizontal velocities of a projectile are independent.
The projectile is a metal ball, e.g. a ball bearing.
The target is a metal drink-can, suspended by an electromagnet.
The circuit includes two wires attached to a copper gate at the entrance to the cardboard tube.
When the circuit is closed, the drink-can is kept in position by the electromagnet.
Sight along path p1 then blows the ball up the cardboard tube.
The ball hits the copper gate to open the circuit and let the drink-can fall.
The ball travels through path p2 and hits the drink-can target in mid-air.
2. The following experiment can be applied to different projectiles, e.g. golf balls, cannon balls, darts, discus, shot put, slingshot, catapult and long jumper.
If a projectile has initial horizontal velocity before it starts falling, its trajectory is a parabola, e.g. ball rolling off a table.
The only force on a projectile is gravity.
Throw a ball vertically as high as you can.
Note the time between when the ball leaves your hand and the ball stops rising.
Also, note the time between when the ball stops rising and the ball descends to the height of your hand.
The times are the same.
Throw the ball up at different angles to the horizontal and note the times taken for rising and descent.
3. Hold ball 1 just over the edge of the table so that it can hit the floor when you drop it.
Put an identical ball 2 in the middle of the table and use a stick to push it steadily towards the edge of the table.
When ball 2 passes the edge of the table immediately drop ball 1.
Ball 2 has original horizontal velocity, but ball 1 has no horizontal velocity.
However, both balls fall and land on the floor simultaneously, because the acceleration due to gravity is the same, whatever their state of motion.
Both balls are projectiles.
4. A body in free fall descends height, h = 1/2 gt2, where t = time and g = acceleration, because of gravity, 9.8 m / sec2.
The time interval for both balls to reach the floor, t = (2h / g).
If the table is 1 m high and velocity of ball B is 5 m / sec, the time of fall, t = (2 X 1) / 98 = 0.45 sec.
The distance ball B travelled before reaching the floor, d = vt = 5 X 0.45 = 4.25 m.
Find your "hang time", i.e. the time you are off the ground when you make a vertical jump.
Hold a piece of adhesive tape between your thumb and finger, jump up next to a wall, and leave the sticky tape on the wall at the top of your jump.
Use the formula: t = (2h / g), where h is the distance from the sticky tape to the floor.
How high can you jump?
An American basketball player can do a 1.25 metres vertical jump!
5. Use a bow and arrow by pulling the bowstring back a certain distance and pointing the arrow at a certain angle to the horizontal before releasing the bowstring.
Note the maximum height of the arrow and the distance it travels before hitting the ground.
Repeat the experiment by pulling the bowstring back the same distance, but change the angle to the horizontal.
Find the angle to the horizontal where the arrow attains the greatest height and longest distance before hitting the ground.

14.4.22 Pendulum-timed free fall
Let a pendulum released from the side hit a ball dropped from the height that gives a fall time equal to a quarter period of the pendulum.

14.4.23 Rotating turntable
Drop a ball on a turning phonograph turntable.
Time a bouncing ball for many bounces and find g using the coefficient of restitution.

14.4.24 Three-holes bottle
Three-holes can, water-jets from a bottle, water bottle with three holes, spouting cylinder, 3-hole can, vase with three holes, Torricelli tank
See diagram 14.3.15: Three-holes bottle
See diagram 4.153: Three-holes can, right and wrong
1. Punch three identical holes in the side of a plastic drink bottle at 1/4, 1/2 and 3/44 of the height, but offset so that the streams of water do not interfere with each other.
Plug the holes then fill the bottle with water.
Put the bottle on a table with a sink draining top.
Attach a tube to a tap to keep a constant head of water when you remove the plugs.
Remove the plugs.
Note the speed of the water through the three holes.
Feel the water with your finger as it comes out of the hole.
The fastest water stream is through the lowest hole.
Note how much water passes through each hole in the same period.
Note the path of the water streams.
Draw a diagram of the three water streams showing the distances travelled by each stream to the table top.
Diagram 4.153.2 is incorrect, although it occurs in some textbooks.
Diagram 4.153.1 is correct.
Water from the middle hole hits the table at the greatest distance from the bottle, d2.
Water from the bottom and top holes both hit the table at the same lesser distance from the bottle, d1.
The greater the depth, the greater the pressure.
Liquid pressure increases with depth.

2. Use a 1.25-L or 2-L plastic drink bottle or a vertical acrylic tube, or a 1.5 m length and 12 cm diameter water down pipe.
* Use a nail held in pliers, or heated knitting needle, or punch, to make three identical round holes in the wall of the bottle:
Three holes equidistant by 6 cm vertically apart, or, three holes at 1/4 and 1/2 and 3/4 of the height, but offset so that the streams of water do not
interfere with each other, or, one hole 2 cm up from the bottom, then another hole 2 cm up, but 2 cm to the side, then another hole 2 cm up, but 2 cm to the side.
So the three holes will not be directly above each other.
* Plug the three holes by covering them with adhesive tape (electrician's tape).
* Put the bottle on a table with a sink drain top and attach a plastic tube to a tap to keep a constant head of water when you remove the plugs.
Remove the plugs from the holes, one by one
* Note the speed of the water streams through the three holes.
The water streams follow parabolic paths towards the sink.
Feel the water with your finger as it comes out of each hole.
The fastest water stream is through the lowest hole.
* Measure how much water passes through each hole in the same period.
* Observe the paths of the water streams and draw a diagram of the three water streams showing the distance, d, travelled by each water stream until it reaches the sink drain top.
The water stream from the middle hole hits the table at the greatest distance from the bottle.
The water streams from the bottom and top holes both hit the table at the same lesser distance from the bottle.
* Diagram 14.4.24 show a right diagram, and a wrong diagram that occurs in some textbooks.
* Diagram 14.4.24a shows the distances travelled by the water streams, d, with respect to vertical distances below the horizontal surface, a.
Eventually, the ranks for distance, d, are 1. the water stream from the bottom hole, 2. the water stream from the middle
hole, 3. the water stream from the top hole.
* Does this experiment show that in a fluid, the greater the depth, the greater the pressure or liquid pressure increases with depth?
Does it show the direct relationship of pressure and depth?
It is true that in a fluid at rest, hydrostatic pressure increases with depth.
However, the water streams are not at rest, so Bernoulli theorem may be used to calculate the projected trajectory of the water stream, as a projectile, assuming that water is an almost non-viscous liquid.
The Bernoulli principle states that in a non-viscous fluid, increase in the speed of the fluid occurs with a decrease in pressure or potential energy of the fluid.
Using the Bernoulli's equation, H = h + v2/ 2g, so v2 = 2g X (H-h), the same as free fall
Where v = speed of water stream at the exit hole, g = acceleration due to gravity, and (H-h) = the difference between the height of the water surface and the height of the hole.
Let time of free fall = tf
tf2 = 2h / g
At the holes the water streams are horizontal, so range, d = velocity at the hole X time of free fall.
So d = (Vh X tf) = 2 X [h (H-h)]½.
When two holes have heights adding up to H, i.e. hole 1, and hole 3, their ranges are identical.
The distance travelled by the water streams depends on the exit velocity at the holes and the time of flight until the water stream reaches the horizontal surface.
The maximum range of a water stream occurs from the centre hole at half the height of the water.
The combined ranges from the two holes equidistant from the centre hole equals the maximum range.

3. Apply projectile motion to water streaming out of a hole in a 3-hole can
For the time for a water drop to fall vertically, use s = ut + 1/2gt2,
so t = sqrt (s / 0.5 g), where g = the acceleration, because of gravity, 9.8 m / sec2, and s = vertical distance from the hole to table.
For the top hole, s = 3, so t = sqrt (3 / 0.5 g), for the middle hole, s = 2, so t = sqrt (2 / 0.5 g), and for the bottom hole, s = 1,
so t = sqrt (1 / 0.5 g).
For the horizontal velocity of the drop, v, use Torricelli Theorem, v = sqrt (2gh), where h is the height from the hole to the TOP of the can.
For the top hole, h =1, so sqrt (2gh) = sqrt (2g), for the middle hole, h = 2, so sqrt (2gh) = sqrt (4g), and for the bottom hole, h =3, so sqrt (2gh) = sqrt (6g).
To find the horizontal distance the drop travels, use distance = vt.
For the top hole, vt = sqrt (3 / 0.5 X sqrt (2g) = 3.464.
For the middle hole, vt = sqrt (2 / 0.5) X sqrt (4g) = 4.
For the bottom hole, vt = sqrt (1 / 0.5) X sqrt (6g) = 3.464.
So water from the middle hole hits the table at the greatest distance, 4, from the bottle, and water from the top and bottom holes, both hit the table at the same distance, 3.464, from the bottle.

4. If a bottle has five equidistant holes, water passing through hole 3 and hits the table top at distance 3, d3.
Water passing through hole 2 and hole 4 and hits the table together at distance 2, d2.
Water passing through holes 1 and 5 and hit the table together at distance 1, d1.
The farthest distance from the bottle is d3, then d2, then d3.
Also (d3 - d2) = (d2 - d1).
The range of water hitting the table is greatest from the middle hole and less the farther the hole is from the middle hole.

5, Torricelli tank allows water to stream from holes at different heights in a vertical glass tube.
The flow from the holes changes as you move the tube up and down.
Determine the velocity of efflux of water by the comparing the parabolic trajectory of each stream, or attach a manometer to the different openings.
Holes of different size at the same height show that the flow is independent of the diameter.

14.4.25 Time a falling body with a stopwatch
See diagram 8.151.1: Throw up and fall down | See diagram 8.151.2: Graph of throw up and fall down
1. Time a falling body.
Throw a ball as high as you can.
Use a stopwatch to measure
1.1 from when it leaves your hand to when the ball reaches the greatest height and stops rising,
1.2 from when the ball starts to fall and reaches the height of your hand.
Time up equals time down.

14.4.26 > Ticker timer, Recording timer
Time a falling body with an electric "ticker timer" (recording timer)
1. Recording timer, AC 5 - 12 V, ticker timer, heavy base with rubber feet, includes disc and small recording tape
Carbon paper disc, suits recording timer
2. Recording tape, large, suits recording timer
| See diagram 8.238: Ticker timer
| See diagram 8.238.1: Ticker timer tape
A ticker timer uses low volt AC power source to make a vibrating spring print a series of points on paper tape using circular carbon paper.
When the paper tape moves with the object, the greater the space between points the further the object has moved.
The time between any two points is equal to the reciprocal of the AC frequency, 50 HZ, so the time interval is 0.02 s.
Experiments
1. Time a falling body with a ticker timer.
Attach a weight to a strip of paper tape.
Pass the tape between the armature of an electric bell and a pad of carbon paper.
Release the paper tape so that the weight falls and drags the paper after it.
The end of the arm of the timer hits the carbon paper against the tape and makes marks on it at equal time intervals.
Measure the distance between the marks.
2. Make a ticker timer from an electric bell mechanism.
Remove the clapper and extend the armature by soldering a strip of metal to it.
At the end of this extension, drill a hole to fit a small round head screw.
Fix the screw head downwards to act as a marking hammer.
Fasten the mechanism to a wooden base.
Fix a 3 cm diameter disc of a carbon paper disc to the base with a drawing pin.
The drawing pin holds the disc loosely at the centre so that the disc can rotate to expose a new surface as the tape passes under it.
Attach staples to the base to guide the path of the ticker tape.
If the extension to the armature strikes the paper too hard, the timing may be uneven.
3. Attach a weight to a strip of paper tape.
Pass the tape between the armature of an electric bell and a pad of carbon paper.
Release the paper tape so that the weight falls and drags the paper after it.
The end of the arm of the timer hits the carbon paper against the tape and makes marks on it at equal time intervals.
Measure the distance between the marks.
4. To study the motion of a long window blind, tape the paper tape to the bottom of the blind and let the paper tape moves up through the ticker timer.
Find a convenient starting point A on the paper tape and record it as 0, even if it is not the first point.
Then mark each point every 5 dots, i.e.
5 X 0.02 = 0.1 second.
Measure the distance between points AB, BC, CD, DE and calculate the average speed in every space, e.g. from A to B, v = distance / time = 5.0 cm / 0.1s.
5. Ticker timer using an electric bell mechanism
See diagram 32.5.4.4: Electric bell circuit
Remove the clapper and extend the armature by soldering a strip of metal to it.
At the end of this extension drill a hole to fit a small round headed screw.
Fix the screw head downwards to act as a marking hammer.
Fasten the mechanism to a wooden base.
Fix a 3 cm diameter disc of a carbon paper disc to the base with a drawing pin.
The drawing pin holds the disc loosely at the centre so that the disc can rotate to expose a new surface as the tape passes under it.
Attach staples to the base to guide the path of the ticker tape.
If the extension to the armature strikes the paper too hard, the timing may be uneven.
6. To draw the (V - t) graph, calculate the instantaneous speed equal to the average speed.
Note that the window blind had begun to move before the selecting point A of t = 0, so there is a part of the graph before t = 0.
The graph shows that the curtain moves in constant acceleration before 0.35 seconds, it moves in constant speed of 2 m / s between 0.35 s and
0.65 s, and then it moves in constant deceleration.
Calculate the acceleration in AD stage from changes in the instantaneous speed in the centre of the time intervals AB and CD: (1.5 - 0.5) / (0.25 - 0.05) = 5.0 (m / s2).
Calculate the acceleration in each stage by this method and draw an acceleration-time graph of motion of the window blind.
Table 14.2.3

Points	Distance cm travelled in 0.1 sec	Average v, m / s
AB	05.0	0.5
BC	10.0	1.0
CD	15.0	1.5
DE	20.0	2.0
EF	20.0	2.0
FG	20.0	2.0
GH	20.0	2.0
HI	17.0	1.7
IJ	08.0	0.8
JK	04.0	0.4

14.5.1 Scalars and vectors
1. A scalar is a quantity with magnitude (size), but no direction, e.g. density, energy, mass, work.
Addition of scalars is an algebraic addition.
So when a 100 g body is added to a 200 g body, its residual mass is 200 g.
Scalar quantities are like the rungs of a ladder.
2. A vector is a quantity with magnitude, size, and direction, e.g. force, momentum, velocity.
A vector can be represented by an arrow with its length proportional to the magnitude of the vector and its direction being parallel to the direction
of the vector and so showing its direction.
More than one vector acting at a point may be represented by a single resultant vector being equivalent to all of them in magnitude and direction.
Vector quantities have both size and direction and you can represent vectors by lines, with length proportional to the size and direction indicated relative to some reference direction, e.g. displacement, velocity, acceleration, force and momentum.
You can add vectors to find the resultant vector, by placing drawn vectors head to tail.
You can resolve vectors into two perpendicular component vectors, rectangular components, so that the two components are equivalent to the single vector.
If all the forces on a body are acting in the same direction the net force is the simple arithmetical sum of the forces.
So a 90 kg force is needed to raise a 90 kg mass, but if three people share the load, each person needs exert only 30 kg force.

14.5.2 Aeroplane ground speed, vectors problem
An aeroplane is travelling north with airspeed 240 km h-1 through a wind blowing 70.0 km h-1 from the west.
Its ground speed is (2402 + 702) = 250 km h-1 and its track is about N 16o E (cos θ = 240 / 250 = 0.96, cos 16o = 0.9613).
Ground speed is the speed of an aircraft relative to the ground.

14.5.3 Boat crossing river, vectors problem
See diagram 14.1.1: 2.1 Boat crossing river problem
1. A boat travelling due east at 4 km / hour is in a river flowing due south at 2.0 km / hour.
The velocity of the boat relative to a point on the river bank is 4.47 km / hour in direction 26o33' south of east.
2. Boat travel displacement, triangle of vectors
See diagram 14.1.3: Triangle of vectors, boat travel
If vectors P and Q may be represented by AB and AC as sides of a triangle, the resultant is represented by the third side AC.
So the resultant of vectors acting on a point or body at different directions may be calculated or found with a vector diagram.
A boat travels 4.5 km north then 6.5 km north east.
The total displacement of the boat is represented by the third side of the triangle of vectors.
3. A man has a small boat which can hold only himself and one animal. How does he take a hyaena, a sheep and a cabbage creoo the river?

14.5.4 Addition of vectors when P and Q are at right angles
See diagram 14.1.1: P and Q are at right angles
1. P is the resolved part of R in the direction OX.
P = R cos α and Q = R sin α.
2. The resolved part of any vector F in a direction making an angle α with the direction of the vector is F cos θ.
The resolved part is the effective part of the given vector in the required direction.
For example, if a force of 20 N be acting in a given direction, the fraction of the force in a direction at an angle of 60o with the direction of the available force = 20 cos 60o = 10 N.
3. The potential energy of a body mass m above the level AB = mgh.
If the mass m was rolled up the slope length l, the work done against the force F down the slope, i.e. the resolved part of the weight of m down the slope, through distance l = mgh.
So Fl = mgh, F = mg X h / l = mg cos (90 - θ) = mg sin θ.

14.5.5 Polygon of vectors
See diagram 14.1.3: Polygon of vectors
Four vectors, acting on one point can be represented in size and direction by the sides AB, BC, CD, DE of a polygon.
The resultant of these four forces is represented in size and direction by the resultant, R, i.e. the side AE that completes the polygon.
If the vectors are forces, a force equal and opposite to R, the equilibrant, would produce equilibrium in the system.
If four coplanar forces 40, 40, 10 and 30 newton are acting at a point at angle 60o in succession to each other, the equilibrant is 50 newton parallel to the second force of 40 newton.

14.5.6 Vector addition, parallelogram law, vector diagram
See diagram 14.1.1: Parallelogram of vectors
Triangle of vectors
1. Addition of vectors using the parallelogram law
If two vectors, AB and AD, acting at a point, may be represented by in magnitude and direction by two adjacent sides of a parallelogram, ABCD, the resultant, R, is the diagonal AC.
R2 = AD2 + AC2 + 2 X AD X AC X cos BAD
The usual practice is to call the two vectors P and Q and call the angle between them θ, theta
So R2 = P2 + Q2 + 2PQ cos θ>
P and Q are called the components of R.
Also, the direction of the resultant relative to AD is the angle α
tan α = Q sin θ / P + Q cos θ
Experiment
See diagram 14.1.02 Cartesian coordinates
The angle between the vector and any axis is not the same as the angle between the axis and the projection of the vector onto a place defined by two axis.

14.5.7 vector resolution, Pull a wagon, canal boat
See diagram 14.1.4: Resolution of vector into components
Often it is convenient to break up a single vector, F, into two vectors, i.e. resolve one force into two component forces.
If P and Q are at right angles, P is called the resolved part of R in the direction of P.
P = R cos α
Q= R sin α
A student uses a rope to pull a wagon along a road with force F.
The angle of the rope to the horizontal is θ.
The force F on the rope can be resolved into a horizontal component Fcos θ that causes the wagon to move along the road and a vertical component F sin θ that tends to lift the wagon.
So some of the force needed to move the wagon is lost, because the rope is not pulled parallel to the road.
Experiments
1. A big box is pulled along a bench by a cord at 30o to the horizontal bench top.
The tension in the cord measured by spring balance is 80 N.
What is (i) the effective part of the pull in moving the box, and (ii) the force tending to lift the box off the bench?
(i) F = 80 cos 30o = 80 X 0.866 = 69.28 N (ii) F = 80 sin 30o = 80 X 0.5 = 40 N.
2. A canal boat is pulled along a canal by a tug boat with a rope at angle 10o horizontally to the direction of movement along the canal.
The captain of the tug boat sets the rudder to ensure that the canal boat moves directly along the middle of the canal.
If the tension of the rope is 150 kN (kilonewton), the resolved part of the force on the canal boat in that direction is 150 cos 10o = 150 X 0.98 = 147 kN.
(1 kilonewton, kN = 1 000 X the amount of force required to accelerate a body with a mass of one kilogram at a rate of one metre per second per second, 1 kN = 101.97132 kg force load).

14.5.8 Vectors addition
See diagram 14.1.1: Addition of two vectors
The magnitude and direction of the resultant may be found by drawing to scale AB, AD and the angle θ, completing the drawing of the parallelogram and carefully measuring the length of R and the size of angle α relative to AD.
In the diagram, to add vector B to vector A, move B parallel to itself until the tail of B is at the end of A.
In the new position of B it has its original length and direction.
The sum A + B is the vector R drawn from the tail of A to the head of B.

14.5.9 Verify the parallelogram law
See diagram 14.1.2: Verify the parallelogram law
Verify the parallelogram law by vertically supporting a 1 kg mass by two spring balances as in the diagram.
Place the apparatus next to a vertical board with graph paper stuck to it.
Trace the pathway of the three strings onto the graph paper and read the forces measured by the two spring balances.
Draw a parallelogram with side P and Q proportional to the values on the spring balances using the scale 1 cm = 1 N (newton).
The diagonal OC should be 9.8 cm.

14.6.0 Speed and road safety
Student discussion topics: "Every K over speed limit K is a killer"
Does the design of a car have any influence on fatality rates due to speed related crashes?
How does braking a car going over the speed limit reduce the chance of a fatality?
What effect does the velocity of a car have on the fatality rate in a frontal collision with a pedestrian?
Does the reaction time and the distance between cars effect the amount of force delivered on a specific vehicle?
How do different types of brakes (disk and drum brakes) affect the performance of a car travelling at 5 km/h over the speed limit?
What force should car bodies be cleared for to survive and protect a passenger from fatality in a high-level crash?
Are the school zone speed limits safe enough for young children to survive a collision or do they need to be lowered?
How does a small increase in speed determine the severity of a car crash?
How does braking during a collision affect survival rate compared to travelling at a constant velocity?
All people travelling over the speed limit have a reduced reaction time leading to more fatalities on the road.
To what extent does driving over the speed limit effect the reaction time of the driver and therefore the impact force and the survival rate of a pedestrian hit by a car?
Does the safety distance between vehicles need to be increased depending on the speed limit?
How do factors such as reaction rate, breaking distance and less control contribute to the amount of speed-related crashes to pedestrian children aged 0-14?
How does in increase of 1km per hour over the speed limit affect the outcome of a head on collision?

14.7.1 Velocity and speed
1. Displacement and distance
If an object changes position from A to B along any path, you measure its displacement by a straight line from starting point A to finishing point B.
This is a vector quantity that has magnitude (size), and direction.
Displacement is the change in position, usually with positive direction to the right.
Displacement is a vector quantity.
It is the distance between the initial point and final point of an object.
Distance = total length of travel, from beginning to end in any direction.
Distance is a scalar quantity representing the interval between two points.
It is the magnitude of the interval.
A person walks in a rectangular path A -> B -> C -> D -> A, starting from A.
At A, distance = 0, displacement = 0.
At B, distance = AB, displacement = AB, in the direction AB.
At C, distance = AB + BC, displacement = AC, in the direction AC.
At D, distance = AB + BC + CD, displacement = AD, in the direction AD.
Back at A, distance = AB + BC + CD + DA, displacement = 0.
If a motor vehicle is driven in a circle, when it returns to the starting point the displacement is zero and the distance travelled is the circumference of the circle.
The motor vehicle can be driven at constant speed, but not constant velocity, because its direction of motion is continually changing.
2. Instantaneous velocity
Instantaneous velocity is the rate of change of displacement on one direction with respect to time as measured by a speedometer.
However, a speedometer in a motor vehicle continues to function whether it moves only in one direction or changes direction, so a speedometer in a motor vehicle is a "speed meter" not a "velocity meter".
The instrument that shows the distance travelled by a vehicle is called an odometer.
It should show the total distance travelled by a motor vehicle unless it has been illegally wound back by an unscrupulous second hand car dealer.
An odometer that can be reset to zero to show the distance travelled during short trips in commonly called a trip meter or tripometer.

14.7.3 Muzzle velocity, bullet timer
Fire a bullet to pass through two aluminium foil strips and show the signal on an oscilloscope.
Fire an air gun through two rotating cardboard discs separated by a known distance.

14.7.4 Terminal velocity
Let a marble roll down a tube of water at a slight incline to reach terminal velocity and allow you to measure slow constant velocity.

14.7.5 Time of flight: Release a projectile fired from a pendulum apparatus by timing signals from two microphones.

14.7.6 Toy car running on moving paper
1. A toy wind-up car moves perpendicular across a pulled sheet of paper that is moving at half the speed of the toy car or at the same speed.
Demonstrate superposition principle of velocities by pulling on the sheet of paper below the toy car.
2. Run a toy car at constant speed in both directions on moving paper to show how velocities add and subtract.
Time a toy car with a stop clock as you pull it across the table at constant velocity in front of a metre stick.
3. Photograph uniform motion.
Take an open shutter photograph of a toy clockwork motor car.