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## Notes

### Scalar vs Vector quantities

• Scalar quantities:described by a magnitude only.
• eg. distance, mass, length, temperature
• Vector quantities: quantities described by a magnitude and direction
• eg. displacement, weight, acceleration, force, momentum

### Some terms

• Displacement: The distance measured along a straight line in a stated direction w.r.t. the
• original point (vector).
• Velocity: Rate of change of displacement
• Acceleration: Rate of change of velocity
• Note: Negative Acceleration = Retardation

### Acceleration of free-fall

• ﻿The acceleration of free-fall near the surface of the Earth is constant and is approximately 10m/s2. It is derived from the gravitational force felt by objects near the Earth surface and independent of the mass of any object.
• Speed of a free-falling body (experiencing no other forces other than gravity) increases by 10m/s every second or when the body is thrown up, it decreases by 10m/s every second.
• The higher the speed of an object, the greater the air resistance.
• Terminal Velocity: When an object is moving at constant velocity, acceleration is 0.
• As an object falls, it picks up speed, increasing air resistance. Eventually, air resistance becomes large enough to balance the force of gravity where the acceleration of the object is 0, reaching constant velocity.

### Displacement-Time Graphs

• Used to show displacement over time.
• Horizontal line: Body at rest.
• Straight line with positive gradient: Uniform velocity.
• Straight line with negative gradient: Uniform velocity in the opposite direction.
• Curve: Non – uniform velocity.
• The gradient of the tangent of this graph gives the instantaneous velocity of the object.

### Velocity-Time Graphs

• Used to show velocity over time.
• Such a graph can be used to find:
• Velocity
• Distance travelled: Area under the graph

### The Equations

They are called the 'suvat' equations because the quantities s, u, v, a and t are used in the equations, with four of the symbols used in each equation.

= displacement (measured in metres)
= initial velocity (measured in metres per second, ms-1)
= final velocity (also measured in ms-1)
= acceleration (measured in metres per second per second, ms-2)
= time (measured in seconds, s)

Below are the equations:

#### Note

• It is important to bear in mind that these equations can only be used for CONSTANT ACCELERATION ONLY. When acceleration is not constant, these equations do not work. For variable acceleration, either graphical methods or calculus would be needed.
• Furthermore, these equations can only be used for motion in a straight line or one-dimensional motion.
• Thus these equations are known as the equations of rectilinear motion.
• Rectilinear motion is one-dimensional motion with uniform acceleration.

## MCQ Questions

1. Which of the following is a vector?
a. area
b. volume
c. density
d. force

2. The displacement of an object from a fixed point is the distance moved by the object
a. in a particular interval of time
b. in a particular direction
c. at a constant speed
d. at a constant velocity

3. A car accelerates from rest at 5ms-2 for 0.5 minute. The final velocity of the car is
a. 150ms-1
b. 5.5ms-1
c. 10ms-1
d. 2.5ms-1

4. When the brakes of a bicycle were applied, the bicycle was brought to rest from 4ms-1 in 2 minutes. What is the acceleration of the bicycle?
a. -1/30 ms-2
b. 1/30 ms-2
c. -2ms-2
d. 2ms-2

5. A free falling object is said to be in linear motion. This is because the object is falling
a. due to its weight
b. at constant velocity
c. at constant acceleration
d. in one direction

6. An object has been falling freely from rest for 3 s. The maximum velocity of the object is
a. 30ms-1
b. 3.3ms-1
c. 10ms-1
d. 13ms-1

7. A body is moving in a circle at a constant speed. Which of the following statements about the body is true?
a. There is no acceleration
b. There is no force acting on it
c. There is a force acting at a tangent to the circle
d. There is a force acting towards the centre of the circle
e. There is a force acting away from the centre of the circle

8. What must change when a body is accelerating?
a. the force acting on the body
b. mass of the body
c. speed of the body
d. velocity of the body

9. The velocity of a car which is decelerating uniformly changes from 30m/s to 15m/s in 3.3s and covers a distance of 75m. After what further distance will the car finally come to rest?
a. 25m
b. 37.5m
c. 50m
d. 75m

10. A ball is thrown vertically upwards into the air with an initial velocity of 5.0m/s. It takes 2.0s to reach the maximum height where it comes to rest momentarily. What is the acceleration of the ball at the maximum height of its path?
a. 0m/s2
b. -2.5m/s2
c. -10m/s2
d. 12.5m/s2

11. A tow truck pulls a broken down car of mass 1000kg and moves at a constant velocity of 9m/s along a horizontal road. It is known that the frictional force acting on the car is 500N. Find the tension in the cable connecting the truck and the car.
a. 0N
b. 500N
c. 9000N
d. 10500N

12. A trolley runs down a slope with constant acceleration a due to a forward force F acting on it in a direction along the slope. This force F is proportional to its mass. The mass of the trolley is now doubled and the trolley is allowed to run down the same slope. In both cases, effects of friction and air resistance are negligible. Which statement is correct for the second experiment?
a. The acceleration is 10 a
b. The acceleration is -0.5 a
c. The acceleration is a
d. The acceleration is 2 a

13. A motorcycle starts from rest on a straight road and reaches a velocity of 25ms-1 after 5.0s. What is its average acceleration in that time?
a. 5.0ms-2
b. 25ms-2
c. 125ms-2
d. 250ms-2

1. d
2. b
3. a
4. a
5. c
6. a
7. d
8. d
9. a
10. c
11. b
12. c
13. a

## Structured Questions Worked Solutions

1. A car accelerates uniformly from a speed of 20ms-1 to a speed of 25ms-1 in 2s. Calculate
a. the average speed for this period of 2s
b. the distance travelled during this period
c. the acceleration

Solution

1a. 22.5 m/s
1b. 45 m
1c. 2.5 m/s2

2.
Describe the motion of the lorry over the following sections of graph along...
a. PQ
b. QR
c. RS

Solution

2a. moving with decreasing speed
2b. stationary/zero speed
2c. moving with increasing speed

3.
a. How far has the object travelled during the first 5 seconds?
b. What is the acceleration of the object
c. For how long does the object move at uniform velocity?
d. What is the average speed of the object during the first 15 seconds?

Solution

3a. 25 m
3b. 2 m/s2
3c. 0
3d. 11.7 m/s2

4. In an experiment, a student measured the accelerations of a steel ball and a feather falling freely. Will the two accelerations be the same or different? Give a reason for your answer.

Solution

The two accelerations are different. The air resistance slowed the downward motion of the falling feather much more than that of the falling steel ball.

5. The figure below shows the velocity of a bus moving along a straight road over a period of time.
a. What does the portion of the graph between O and A indicate?
b. What can you say about the motion of the bus between B and C?
c. What is the deceleration of the bus between C and D?
d. What is the total distance travelled by the bus in 100 s?
e. What is the average velocity of the bus?

Solution

5c. 1 m/s2
5d. 2000 m
5e. 20 m/s

6. Two cyclists, A and B, start a race. A accelerates for the first 5 s, until his velocity reaches 12ms-1, after which he travels with constant velocity. B accelerates for the first 10 s, until his velocity reaches 15ms-1, after which he travels with constant velocity.
a. Sketch the velocity-time graphs for the two cyclists
b. Calculate the distance travelled by both cyclists in the first 10 s
c. Who is in the lead after 10 s?

Solution

6b. 90 m; 75 m

7. A car travels along a straight road. The speedometer reading after every 5 s is tabulated below
 Time/s 0 5 10 15 20 25 30 35 40 Velocity m/s 0 10 20 30 30 30 30 15 0

a. Draw a velocity-time graph to show the variation of velocity with time
b. Describe the motion of the car
c. How far from the starting point is the car after 20 s
d. What is the total distance travelled by the car
e. What is the average velocity of the car for the whole journey

Solution

7c. 373 m
7d. 825 m
7e. 20.6 m/s

8. Find the average velocity of a car which travels 360km in 6 hours in
a. km/h
b. m/s

Solution

8a. 60 km/h
8b. 16.7 m/s

9. Find the average velocity of an athlete who runs 1500m in 4 minutes in
a. m/s
b. km/h

Solution

9a. 6.25 m/s
9b. 22.5 km/h

10. The graph below shows the speed-time graph for a child on a swing
a. Write down
i. the maximum speed
ii. the time at which the maximum speed occurs

bi. On the graph, mark with 'Z' the point where the magnitude of acceleration of the child is maximum.
bii. Mark with 'M' one point at which the acceleration is zero

c. Estimate the distance travelled by the child in 1.2s

d. Describe briefly the changes in acceleration during the period shown on the graph

Solution

10ai. 6.0 m/s

10aii. 0.60 s

10b.
10c. 3.6 m

10d. The acceleration increases from zero initially to a maximum and decreases to zero. The body then decelerates with deceleration increasing to a maximum and decreases to zero.

11. A body is accelerated uniformly from rest and in the first 8.0s of its motion it travels 20m. Calculate
a. the average speed of this period of 8 s
b. the speed at the end of this period
c. the acceleration

Solution

11a. 2.5 m/s
11b. 5 m/s
11c. 5/8 m/s2

12. A car of length 6.0m accelerates from rest along a straight level road as shown.
The car takes 2.0s to pass the point P.
10.0s later the car has just passed point Q.
The car takes 0.40s to pass point Q.

Calculate
ai. the average speed of the car as it passes P
aii. the average speed of the car as it passes Q
aiii. the average acceleration of the car between P and Q
bi. Estimate the distance between P and Q
bii. What assumption did you make when you estimated the distance between P and Q?

Solution

12ai. 3 m/s
12aii. 15 m/s
12aiii. 1.2 m/s2
12bi. 90 m
12bii. The car accelerates uniformly between P and Q. The acceleration remains constant.

13a. What is meant by the period of a simple pendulum?

13b. The period of a simple pendulum 1 m long is approximately 2 s. State accurately how you would determine the period of such a pendulum as accurately as possible, using a stopclock accurate to within 0.1 s.

Solution

13a. The period of a simple pendulum is the time taken for one complete oscillation made by the pendulum.

13b.
The pendulum as shown above is slowly brought to point X and then released. Simultaneously, the stopclock is started. For every time the bob returns to X, it is considered one oscillation. Count to 30 oscillations and read the time taken. The whole experiment is repeated three times. Then average the three readings. The period is found by taking average time over 30.

14. Students, investigating motion down an inclined plane, measure the speed of a steel ball at one second intervals after the ball starts to roll from rest down one such plane:

 time in s 0 1 2 3 speed in m/s 0 0.6 1.2 1.8

a. calculate the average acceleration over the first 3.00 s

b. calculate the average speed over the first 3.00 s

c. what was the distance travelled by the ball in the first 3.00 s?

d. how do the numbers in the table show that the acceleration was constant?

Solution

14a. average acceleration = (final speed - initial speed) / time = (1.80 - 0.00) / 3.00 = 0.60 m/s2

14b. average speed = 1/2 x (u + v) = 1/2 x (0.00 + 1.80) = 0.90 m/s

14c. distance travelled = average speed x time = 0.90 x 3 = 2.7 m

14d. The speed increases by a constant value of 0.60 m/s every one second, hence the acceleration can be seen to be constant.

15. A metal box, attached to a small parachute, is dropped from a helicopter.

a. Explain in terms of the forces acting, why
i. its velocity increased immediately after being dropped
ii. it reached a uniform velocity after a short time

b. The total force opposing the motion of the box and parachute at a particular instant during its fall is 30N. The combined mass of the box and parachute is 5.0kg. Calculate the resultant downward force on the box and parachute. (g = 10 N/Kg)
Briefly describe the motion of the box and parachute at this time

c. At the end of this fall the parachute is caught on a tall tree. The box is then cut loose and falls from rest to the ground. The time of fall is 2.4 s. Calculate
i. the velocity with which the box strikes the ground
ii. the average velocity during its fall
iii. the distance fallen (g = 10 m/s2)

Solutions

15ai. This is because at the moment the box left the helicopter, the force of gravity pulled it downwards, causing an increase in velocity

15aii. The presence of the parachute increased the air resistance which acted as a constant retarding force on the box. Once this force was equal to the weight of the box, velocity became constant.

15b. resultant downward force = (5 x 10) - 30 = 20 N

From Newton's Second Law of Motion, the box and parachute at this time will accelerate at a rate of 4 m/s2

15ci. given t = 2.4 s, u = 0, g = 10 m/s2,

v = u + at
v = 10(2.4) = 24 m/s

15cii. average velocity = (u + v)/2 = (0 + 24)/2 = 12 m/s

15ciii. distance = 12 x 2.4 = 28.8m

16. The figure below shows part of the route of a rollercoaster in which the passenger cars are pulled up to point A and released.
During one run, a car and passengers of total mass 800kg are released from rest at point A, a height of 30m above the terminal platform. The car travels a distance of 120m along the track to reach the highest point B of the vertical loop which is 20m above the terminal platform. The track itself is frictionless but at the terminal platform, brakes are applied to stop the car.

a. In moving from point A to point B, calculate
i. the loss in gravitational potential energy of the car
ii. the speed of the car at point B
b. Once the car reaches the terminal platform, a constant braking force of 7500N is applied to the car. Find the braking distance of the car.
c. In the design of the rollercoaster, do you think summit C can be higher than point A? Explain your answer.