1. Rate of change of speed and direction Rate of change of momentum
magnitude
direction
Shortest distance and direction from the start point to the current location
Speed and direction
Rate of change of speed and direction
Rate of change of momentum
Constant speed
Stationary
Acceleration
Constant speed (in reverse)
Deceleration
v = x/t
v = 1.2/5.0
v = 0.24m/s

2. v =x/t
m/s
a = (v-u)/t
m/s2
v =x/t
400m
2.5m/s
a = (v-u)/t
15s
v =400/15
1.5m/s
a = ( )/3
26.7
m/s 3s
0.33
m/s2
v =x/t
26.7m/s
0m/s
a = (v-u)/t
x = vt
17.5s
2.5m/s
t = (v-u)/a
-4m/s2
x = 26.7x17.5
t = (0-2.5)/-4
467 m
0.63 s

3. Increasing rate of acceleration Deceleration
Constant speed
Increasing rate of acceleration
Deceleration
Decreasing rate of acceleration
a = (v-u)/t
a = (7-0)/3.6
a = 1.94 m/s2

4. area2 = base x height
= 3 x 5
= 15m
area4 = base x height
= 5 x 1
= 5m 2 3
area1 = ½ base x height
= ½ x 2 x 5
= 5m 1 4
area3 = ½ base x height
= ½ x 3 x 4
= 6m
Total area =
= 31m
17000N
1500N
1200N
300N
1500
17000N

5. Force of the face acting on the fist
Force of exhaust gases pushing the rocket engine forwards N
800N
Force of water pushing the engine forwards
2000
right
1000
left
left/right
Acceleration
Deceleration
Constant speed

6. Car A would have half the acceleration of car B.
According to Newton’s second law, F=ma (rearranged to a=F/m) a doubling of the mass would lead to a halving of acceleration in a situation where the force is constant.
Car C would have twice the acceleration of car D.
According to Newton’s second law, F=ma (rearranged to a=F/m) a doubling of the thrust (force) would lead to a doubling of acceleration in a situation where the mass is constant.

7. F=ma N w=mg N w=mg F=ma F=12 000 x 0.8 w=70x10 9600 N 700 N F=ma w=mg
70kg kg
0.8m/s2
F= x 0.8
10N/kg
w=70x10
9600 N
700 N
F=ma
w=mg kg
1100N N
a=F/m
m=w/g
10N/kg
a=19 000/12 000
m=1100/10
1.58
m/s2
110 kg

8. Set up a glider (air cart) on an air track with light gates set to calculate acceleration. Hang a mass hanger from the end of a string which is attached to the glider, and runs along a pulley. Place the masses on the glider. Start the glider from a reference point and allow it to accelerate freely. Record the acceleration. Move one of the masses from the glider to the hanger and repeat the process. Do this for all of the masses. Repeat the experiment 3 times and plot a graph of the weight of the masses on the hanger (force) against acceleration. Measure and record the mass of the moving system (masses, hanger, string and glider). Repeat the experiment with gliders of different masses and compare the gradients of the lines drawn on the various graphs. 1/gradient = mass (since m=F/a and the gradient is dy/dx = a/F)
When air is present, and a penny and feather are released from the top of the container at the same time, the hits the bottom of the container first. When the air is removed by a vacuum pump, however, both objects hit the bottom of the container at the same time. This means that they must fall with identical motion, hence with identical acceleration.

9. (Negative acceleration)
Unbalanced
Acceleration
The skydiver has just stepped out of the plane so has zero vertical speed, so no air resistance yet. The downward force is his/her weight.
Unbalanced
Acceleration
Air resistance is proportional to air speed. The skydiver has gained some vertical airspeed and so has generated some air resistance, however it has not yet become equal to his/her weight
Balanced
Constant speed
Due to his/her acceleration, the skydiver has now reached an airspeed where his/her air resistance is the same size has his/her weight. This is called terminal velocity.
Unbalanced
Deceleration
(Negative acceleration)
Air resistance is also proportional to surface area. The skydiver has gained a massive amount of surface area by deploying the parachute, so the very large air resistance, which acts in the opposite direction to his/her motion, is causing deceleration.
Balanced
Constant speed
Due to his/her deceleration, the skydiver has now again reached an airspeed where his/her air resistance is the same size has his/her weight. This is a lower air speed than previously, due to the high surface area. This is a lower terminal velocity.