Kinematics: Class 9 Physics Notes, Short Questions and Numericals

Rest and motion, scalars and vectors, distance and displacement, speed and velocity, acceleration, graphs of motion, the three equations of motion and free fall under gravity.

Key points

Kinematics is the study of motion without asking what causes it. A body is at rest if its position does not change with time relative to its surroundings. It is in motion if its position changes. Rest and motion are relative: a passenger sitting in a moving bus is at rest relative to the bus but in motion relative to the road.

Types of motion are:

Translatory motion: the whole body moves along a path. It can be linear (a car on a straight road), circular (a stone whirled on a string) or random (a bee flying).
Rotatory motion: the body spins about an axis, such as a fan or a wheel.
Vibratory motion: the body moves to and fro about a mean position, such as a pendulum or a guitar string.

A scalar has magnitude only, for example mass, time, distance, speed and energy. A vector has magnitude and direction, for example displacement, velocity, acceleration and force. A vector is drawn as an arrow. The length of the arrow shows the magnitude and the arrowhead shows the direction.

Distance is the total length of the path travelled. It is a scalar and never negative. Displacement is the shortest straight line from the starting point to the end point, in the direction from start to end. It is a vector. If you walk 3 m east and then 4 m north, the distance is 7 m but the displacement is 5 m. A body that returns to its start has zero displacement.

Speed is the distance covered per unit time. Its SI unit is m/s. If the speed stays the same, it is uniform speed. Average speed is total distance divided by total time. Velocity is the rate of change of displacement, or displacement per unit time. It is a vector, with the same unit m/s. A car moving round a circle at constant speed has changing velocity because its direction keeps changing.

Acceleration is the rate of change of velocity: a = (vf − vi)/t. Its SI unit is m/s². Positive acceleration means velocity is increasing in the chosen direction. Negative acceleration, also called retardation or deceleration, means the body is slowing down. Acceleration is uniform when velocity changes by equal amounts in equal intervals of time.

Graphs of motion are read as follows:

Distance-time graph: the slope gives speed. A straight sloping line means uniform speed, a horizontal line means the body is at rest, and a curve means the speed is changing.
Speed-time graph: the slope gives acceleration. A horizontal line means uniform speed, a straight sloping line means uniform acceleration, and a line along the time axis means rest.
The area under a speed-time graph gives the distance travelled. For a straight line from rest, the area is a triangle, ½ × base × height.
In a velocity-time graph, the area gives displacement.

The equations of motion apply to uniform acceleration in a straight line. Here vi is initial velocity, vf is final velocity, a is acceleration, t is time and S is displacement:

First equation: vf = vi + at
Second equation: S = vi t + ½ a t²
Third equation: 2aS = vf² − vi²

These come from the velocity-time graph. The slope of the line is a, which gives the first equation. The area under it is a rectangle plus a triangle, which gives the second equation. Eliminating t between the first two gives the third. Use signs carefully: take the direction of motion as positive, and give a negative value to a when the body slows down.

Free fall is motion under gravity alone, with air resistance ignored. All bodies fall with the same acceleration g near the Earth's surface. In calculations we use g = 10 m/s² (the more exact value is 9.8 m/s²). A body dropped from rest has vi = 0. A body thrown upwards loses speed at 10 m/s every second, stops at the top and then falls back. When air is present, a feather falls more slowly than a coin, but in a vacuum they fall together.

Use signs consistently: choose a positive direction before substituting into equations. A negative acceleration means acceleration points opposite that direction; it does not always mean that speed is falling.

Study tip: Write the given data with units first, then choose the equation that contains the unknown and has no unknown you do not need. Fix a positive direction and keep to it. For graph questions, label the axes and say clearly whether you use slope or area. Always check the answer with a second equation if time allows.

Definitions and flashcards

Say the meaning first, then open the card to check.

Rest

A body whose position does not change with time relative to its surroundings.

Scalar

A quantity with magnitude only, such as speed or time.

Vector

A quantity with both magnitude and direction, such as velocity.

Displacement

The shortest distance from the start point to the end point, with direction.

Velocity

Displacement covered per unit time.

Acceleration

Rate of change of velocity, measured in m/s².

Uniform acceleration

Equal change of velocity in equal intervals of time.

Free fall

Motion of a body under the pull of gravity alone.

Retardation

Negative acceleration, when a body slows down.

Formula sheet

Name	Formula	What it means
Average speed	v = S/t	Distance divided by time
Acceleration	a = (vf − vi)/t	Change of velocity per unit time
First equation	vf = vi + at	Final velocity after time t
Second equation	S = vi t + ½ a t²	Displacement in time t
Third equation	2aS = vf² − vi²	Relates velocity and displacement without time
Free fall from rest	h = ½ g t², v = g t	Height fallen and speed gained

Short questions with model answers

These are the kind of questions asked for 2, 3 and 5 marks. Try each one before opening the answer.

Differentiate between distance and displacement. (2 marks)

Distance is the total length of the path travelled. It is a scalar and is always positive. Displacement is the shortest straight line from start to end, with direction. It is a vector and can be zero.

What does the slope of a speed-time graph tell us? (2 marks)

The slope of a speed-time graph gives the acceleration of the body. A steeper line means a larger acceleration. A horizontal line has zero slope, so the acceleration is zero.

Why is velocity changing when a car moves at constant speed round a bend? (2 marks)

Velocity is a vector, so it depends on direction as well as speed. On a bend the direction keeps changing even though the speed is constant. Hence the velocity changes and the car is accelerating.

Derive the equation vf = vi + at. (3 marks)

Acceleration is the change of velocity divided by time, so a = (vf − vi)/t. Multiply both sides by t to get at = vf − vi. Rearranging, vf = vi + at. Here vi is the initial velocity, vf is the final velocity and t is the time taken. The same result is found from the slope of a velocity-time graph, which is a straight line for uniform acceleration.

Describe how the distance travelled can be found from a speed-time graph. (3 marks)

The area between the graph line and the time axis equals the distance travelled. For a body starting from rest with uniform acceleration the graph is a straight line and the area is a triangle. Its area is ½ × base × height, where the base is the time t and the height is the final speed vf. This gives S = ½ vf t. For a trapezium, add the rectangle and the triangle separately.

Explain free fall and state the value of g used in calculations. (5 marks)

Free fall is the motion of a body when gravity is the only force acting on it. Air resistance is ignored. Every body then falls with the same acceleration, called the acceleration due to gravity g. Near the Earth's surface g is about 9.8 m/s², and we use 10 m/s² in numericals. A body dropped from rest gains 10 m/s of speed every second. A body thrown upwards loses speed at the same rate, so it stops at the highest point and then falls back. In a vacuum a stone and a feather reach the ground together.

A delivery rider travels 120 m east and then 50 m west. Find distance and displacement. (2 marks)

Distance = 120 + 50 = 170 m. Displacement = 120 - 50 = 70 m east. Distance counts the whole route; displacement compares the end positions.

How can a velocity-time graph show both acceleration and displacement? (3 marks)

Its gradient gives acceleration. Its signed area gives displacement. An area below the time axis is negative displacement, so it must not automatically be added as positive distance.

Explain a uniformly accelerated journey from rest to 12 m/s in 6 s, including its graph and distance. (5 marks)

Acceleration = (12 - 0)/6 = 2 m/s². The velocity-time graph is a straight rising line from the origin. Its triangular area is 1/2 x 6 x 12 = 36 m. This area gives displacement; because the direction does not reverse, the distance is also 36 m.

Numericals with worked steps

A car starts from rest and reaches a speed of 20 m/s in 8 s. Find its acceleration and the distance covered.

vi = 0, vf = 20 m/s, t = 8 s
a = (vf − vi)/t = (20 − 0)/8 = 2.5 m/s²
S = vi t + ½ a t² = 0 + ½ × 2.5 × 8² = ½ × 2.5 × 64
S = 80 m
Check with 2aS = vf²: 2 × 2.5 × 80 = 400 = 20², so it agrees

Answer: a = 2.5 m/s², S = 80 m

A train moving at 30 m/s applies brakes and stops in 150 m. Find the retardation and the time taken to stop.

vi = 30 m/s, vf = 0, S = 150 m
2aS = vf² − vi², so 2 × a × 150 = 0 − 900
300 a = −900, so a = −3 m/s²
t = (vf − vi)/a = (0 − 30)/(−3) = 10 s
Check: S = vi t + ½ a t² = 300 − ½ × 3 × 100 = 300 − 150 = 150 m

Answer: Retardation 3 m/s² (a = −3 m/s²), time 10 s

A stone is dropped from the top of a tower and hits the ground after 3 s. Take g = 10 m/s². Find the height of the tower and the speed on landing.

vi = 0, a = g = 10 m/s², t = 3 s
h = ½ g t² = ½ × 10 × 9 = 45 m
vf = vi + g t = 0 + 10 × 3 = 30 m/s
Check: vf² = 2gh = 2 × 10 × 45 = 900, so vf = 30 m/s

Answer: Height 45 m, landing speed 30 m/s

A ball is thrown vertically upwards at 25 m/s. Take g = 10 m/s². Find the time to reach the highest point and the maximum height.

vi = 25 m/s, vf = 0 at the top, a = −10 m/s²
t = (vf − vi)/a = (0 − 25)/(−10) = 2.5 s
h = vi t + ½ a t² = 25 × 2.5 − ½ × 10 × 6.25 = 62.5 − 31.25 = 31.25 m
Check: 2aS = vf² − vi² gives −20 h = −625, so h = 31.25 m

Answer: Time 2.5 s, maximum height 31.25 m

A trolley has initial velocity 5 m/s and uniform acceleration 1.5 m/s² for 4 s. Calculate displacement.

Use s = ut + at²/2 because acceleration is uniform.
Substitute: s = 5 x 4 + 1.5 x 4² / 2.
Add 20 m + 12 m.

Answer: 32 m in the positive direction.

A car brakes uniformly from 16 m/s to rest over 32 m. Find its acceleration.

Use v² = u² + 2as.
0 = 16² + 2a x 32.
a = -256/64 = -4 m/s².

Answer: -4 m/s² relative to the initial direction of motion.

Common mistakes

Using distance and displacement as if they were the same. They differ whenever the path is not a straight line or the body returns to its start.
Forgetting the negative sign of acceleration when a body slows down, which gives a wrong answer in the equations of motion.
Reading the slope of a distance-time graph as acceleration. It gives speed. Only the slope of a speed-time graph gives acceleration.
Mixing units, for example using km/h with seconds. Convert 1 km/h to 1000/3600 = 5/18 m/s first.
Using vi = 10 m/s for a body dropped from rest. A dropped body starts with vi = 0.

Practise this chapter

Practice questions and timed test Printable worksheet