Class 9 Physics Notes – Chapter 2: Kinematics

Punjab Board | 2026

A. Multiple Choice Questions (MCQs)

2.1. The numerical ratio of displacement to distance is:

(a) Always less than one
(b) Always equal to one
(c) Always greater than one
(d) Equal to or less than one

Correct Answer: (d) Equal to or less than one

Explanation: The magnitude of displacement is always less than or equal to the distance travelled. Therefore, the ratio is always less than or equal to 1.

2.2. If a body does not change its position with respect to some fixed point, then it will be in a state of:

(a) Rest
(b) Motion
(c) Uniform motion
(d) Variable motion

Correct Answer: (a) Rest

Explanation: Rest is the state of a body when its position does not change with respect to its surroundings.

2.3. A ball is dropped from the top of a tower. The distance covered by it in the first second is:

(a) 5 m
(b) 10 m
(c) 50 m
(d) 100 m

Correct Answer: (a) 5 m

Explanation:

Using the equation:

S = vᵢt + ½gt²

Given:

Initial velocity, vᵢ = 0
Acceleration due to gravity, g = 10 m/s²
Time, t = 1 s

S = 0 + ½ × 10 × 1²

S = 5 m

2.4. A body accelerates from rest to a velocity of 144 km h⁻¹ in 20 seconds. The distance covered by it is:

(a) 100 m
(b) 400 m
(c) 1400 m
(d) 1440 m

Correct Answer: (b) 400 m

Explanation:

Convert the final velocity into metres per second:

vᶠ = 144 × 5/18 = 40 m/s

Given:

vᵢ = 0, t = 20 s

Acceleration:

a = (vᶠ − vᵢ)/t

a = (40 − 0)/20 = 2 m/s²

Distance:

S = vᵢt + ½at²

S = 0 + ½ × 2 × 20²

S = 400 m

2.5. A body is moving with constant acceleration starting from rest. It covers a distance S in 4 seconds. How much time does it take to cover one-fourth of this distance?

(a) 1 s
(b) 2 s
(c) 4 s
(d) 16 s

Correct Answer: (b) 2 s

Explanation:

For a body starting from rest and moving with constant acceleration:

S ∝ t²

If the distance becomes S/4, then:

S/4 ∝ t²

The time becomes half of the original time:

t = 4/2 = 2 s

2.6. The displacement-time graphs of two objects A and B are shown in the figure. Point out the true statement.

(a) The velocity of A is greater than B.
(b) The velocity of A is less than B.
(c) The velocity of A is equal to that of B.
(d) The graph gives no information in this regard.

Correct Answer: (a) The velocity of A is greater than B.

Explanation: The slope of a displacement-time graph represents velocity. According to the supplied solution, graph A has a steeper slope than graph B.

Note: The referenced graph was not included in the supplied text.

2.7. The area under the speed-time graph is numerically equal to:

(a) Velocity
(b) Uniform velocity
(c) Acceleration
(d) Distance covered

Correct Answer: (d) Distance covered

Explanation:

Area under a speed-time graph = Speed × Time

For uniform speed, this gives the distance covered.

2.8. The gradient of the speed-time graph is equal to:

(a) Speed
(b) Velocity
(c) Acceleration
(d) Distance covered

Correct Answer: (c) Acceleration

Explanation:

Gradient = Change in speed / Change in time

The change in speed per unit time gives acceleration.

2.9. The gradient of the distance-time graph is equal to:

(a) Speed
(b) Velocity
(c) Distance covered
(d) Acceleration

Correct Answer: (a) Speed

Explanation:

Gradient = Change in distance / Change in time

The change in distance per unit time gives speed.

2.10. A car accelerates uniformly from 80.5 km h⁻¹ at t = 0 to 113 km h⁻¹ at t = 9 s. Which graph best describes the motion of the car?

(a) Curved upward speed-time graph
(b) Straight inclined speed-time line starting from a non-zero intercept
(c) Straight line starting from zero
(d) Downward-sloping line

Correct Answer: (b) Straight inclined speed-time line starting from a non-zero intercept

Explanation: Uniform acceleration is represented by a straight inclined line on a speed-time graph. Since the initial speed is not zero, the line starts above the origin.

B. Short Answer Questions

2.1. Define scalar and vector quantities.

Answer:

Scalar quantity: A physical quantity described completely by its magnitude and unit. Examples include mass, time and speed.

Vector quantity: A physical quantity that requires both magnitude and direction for its complete description. Examples include displacement, velocity and force.

2.2. Give five examples each of scalar and vector quantities.

Answer:

Scalar quantities:

Mass
Length
Time
Temperature
Speed

Vector quantities:

Displacement
Velocity
Acceleration
Force
Momentum

2.3. State the head-to-tail rule for addition of vectors.

Answer: The head-to-tail rule states that vectors are added by placing the tail of the second vector at the head of the first vector. The resultant vector is drawn from the tail of the first vector to the head of the last vector.

2.4. What are distance-time and speed-time graphs?

Answer:

Distance-time graph: A graph with time on the x-axis and distance on the y-axis. It is used to determine the speed of an object.

Speed-time graph: A graph with time on the x-axis and speed on the y-axis. It is used to determine acceleration and distance covered.

2.5. Falling objects near the Earth have the same constant acceleration. Does this imply that a heavier object will fall faster than a lighter object?

Answer: No. The acceleration due to gravity, g, is the same for all freely falling objects regardless of their mass, provided air resistance is neglected. Therefore, heavy and light objects fall at the same rate under these conditions.

2.6. Vector quantities are sometimes written in scalar notation (not boldface). How is the direction indicated?

Answer: In ordinary notation, the direction of a vector may be indicated by an arrow over its symbol, such as v⃗, or by explicitly stating its direction or sign alongside the magnitude.

2.7. A body is moving with uniform speed. Will its velocity be uniform? Give a reason.

Answer: Not necessarily. Uniform speed means that the magnitude of velocity remains constant. If the direction changes, as in circular motion, the velocity also changes because velocity depends on both magnitude and direction.

2.8. Is it possible for a body to have acceleration when moving with (i) constant velocity and (ii) constant speed?

Answer:

(i) Constant velocity: No. Acceleration is the rate of change of velocity. If velocity is constant, acceleration is zero.

(ii) Constant speed: Yes. If the direction of motion changes, the velocity changes and the body can have acceleration. For example, a body moving in a circle at constant speed has centripetal acceleration.

C. Constructed Response Questions

2.1. Distance and displacement may or may not be equal in magnitude. Explain this statement.

Answer: Distance is the total length of the actual path covered, while displacement is the straight-line distance from the initial position to the final position, with direction.

When a body moves along a straight line in one direction without turning back, distance and the magnitude of displacement are equal. When it moves along a curved or zigzag path, or returns to its starting point, distance is greater than or equal to the magnitude of displacement.

2.2. When a bullet is fired, its velocity as it leaves the barrel is called the muzzle velocity of the gun. The muzzle velocity of one gun with a longer barrel is less than that of another gun with a shorter barrel. In which gun is the acceleration of the bullet larger? Explain your answer.

Answer: The supplied solution states that acceleration is larger in the gun with the shorter barrel. It uses the equation:

a = (vᶠ² − vᵢ²)/(2s)

where vᶠ is final velocity, vᵢ is initial velocity and s is the distance travelled.

For the same initial velocity, a larger final velocity reached over a shorter distance can indicate greater acceleration.

Note: The information provided does not specify the initial velocities or the actual muzzle velocities of the two guns. Therefore, the acceleration comparison cannot be established conclusively from barrel length alone.

2.3. For a complete trip, average velocity was calculated and its value came out to be positive. Is it possible that its instantaneous velocity at any time during the trip had a negative value? Give justification.

Answer: Yes. Average velocity depends on total displacement divided by total time:

Average velocity = Total displacement / Total time

During the trip, an object may move backward for part of the journey, giving it a negative instantaneous velocity. The average velocity can still be positive if the net displacement in the positive direction is greater.

D. Comprehensive Questions

2.1. How can a vector be represented graphically? Explain.

Answer: A vector is represented graphically by a directed line segment with an arrowhead. The length of the line, drawn according to a suitable scale, represents the magnitude of the vector, while the arrowhead indicates its direction.

2.2. Differentiate between (i) rest and motion, and (ii) speed and velocity.

Answer:

(i) Rest and motion

Rest: A body is at rest if its position does not change with respect to its surroundings.

Motion: A body is in motion if its position changes with time relative to its surroundings.

(ii) Speed and velocity

Speed: Distance covered per unit time. It is a scalar quantity.

Velocity: Displacement per unit time. It is a vector quantity.

2.3. Describe different types of motion. Also give examples.

Answer:

Translatory motion: Motion in which a body moves from one place to another. Example: a car moving along a road.
Rotatory motion: Motion in which a body rotates about a fixed axis. Example: a ceiling fan.
Vibratory or oscillatory motion: To-and-fro motion about a mean position. Example: a swinging pendulum.

2.4. Explain the difference between distance and displacement.

Answer:

Distance: A scalar quantity representing the total length of the actual path travelled by a body.

Displacement: A vector quantity representing the change in position from the starting point to the ending point.

Distance depends on the actual path travelled, whereas displacement depends only on the initial and final positions.

2.5. What do the gradients of distance-time and speed-time graphs represent? Explain with diagrams.

Answer:

Gradient of a distance-time graph = Speed

Speed = Change in distance / Change in time

Gradient of a speed-time graph = Acceleration

Acceleration = Change in speed / Change in time

The gradient is calculated by dividing the change in the vertical-axis quantity by the change in time.

2.6. Prove that the area under a speed-time graph is equal to the distance covered by an object.

Answer: For an object moving with uniform speed v for time t:

Distance = Speed × Time

S = vt

The area under its speed-time graph is the area of a rectangle:

Area = Height × Width

Area = v × t

Therefore:

Distance covered = Area under the speed-time graph

For variable speed, the area under the graph can be divided into small strips. The sum of these areas gives the total distance covered.

2.7. How can the equations of motion be applied to bodies moving under the action of gravity?

Answer: For freely falling bodies under gravity, acceleration a is replaced by acceleration due to gravity g, approximately 9.8 m/s² or 10 m/s². For vertical motion, displacement may be represented by height h, with signs chosen according to the direction taken as positive.

Using downward as the positive direction for a falling body, the equations are:

vᶠ = vᵢ + gt
h = vᵢt + ½gt²
vᶠ² − vᵢ² = 2gh

For upward motion, the sign of gravity must be reversed if upward is taken as positive.

E. Numerical Problems

2.1. Draw representative lines of the following vectors:

(a) A velocity of 400 m s⁻¹ making an angle of 60° with the x-axis.

(b) A force of 50 N making an angle of 120° with the x-axis.

Answer:

(a) Velocity vector

Scale: 100 m/s = 1 cm

Line length = 400/100 = 4 cm

Draw a 4 cm line at an angle of 60° counter-clockwise from the positive x-axis. Add an arrowhead to show the direction.

(b) Force vector

Scale: 10 N = 1 cm

Line length = 50/10 = 5 cm

Draw a 5 cm line at an angle of 120° from the positive x-axis. Add an arrowhead to show the direction.

2.2. A car is moving with an average speed of 72 km h⁻¹. How much time will it take to cover a distance of 360 km?

Answer:

Given:

Speed = 72 km/h

Distance = 360 km

Using:

Time = Distance / Speed

t = 360/72

t = 5 hours

2.3. A truck starts from rest. It reaches a velocity of 90 km h⁻¹ in 50 seconds. Find its average acceleration.

Answer:

Given:

vᵢ = 0, t = 50 s

Convert final velocity:

vᶠ = 90 × 5/18 = 25 m/s

Using:

a = (vᶠ − vᵢ)/t

a = (25 − 0)/50

a = 0.5 m/s²

2.4. A car passes a green traffic signal while moving with a velocity of 5 m s⁻¹. It then accelerates at 1.5 m s⁻². What is its velocity after 5 seconds?

Answer:

Given:

vᵢ = 5 m/s, a = 1.5 m/s², t = 5 s

Using:

vᶠ = vᵢ + at

vᶠ = 5 + (1.5 × 5)

vᶠ = 5 + 7.5

vᶠ = 12.5 m/s

2.5. A motorcycle initially travelling at 18 km h⁻¹ accelerates at a constant rate of 2 m s⁻². How far will the motorcycle travel in 10 seconds?

Answer:

Convert initial velocity:

vᵢ = 18 × 5/18 = 5 m/s

Given:

a = 2 m/s², t = 10 s

Using:

S = vᵢt + ½at²

S = (5 × 10) + ½ × 2 × 10²

S = 50 + 100

S = 150 m

2.6. A wagon is moving on the road with a velocity of 54 km h⁻¹. Brakes are applied suddenly. The wagon covers a distance of 25 m before stopping. Determine its acceleration.

Answer:

Convert initial velocity:

vᵢ = 54 × 5/18 = 15 m/s

Given:

vᶠ = 0, S = 25 m

Using:

vᶠ² − vᵢ² = 2aS

0² − 15² = 2 × a × 25

−225 = 50a

a = −4.5 m/s²

The negative sign indicates that the acceleration is opposite to the direction of motion.

2.7. A stone is dropped from a height of 45 m. How long will it take to reach the ground? What will be its velocity just before hitting the ground?

Answer:

Given:

h = 45 m, vᵢ = 0, g = 10 m/s²

(a) Time taken

Using:

h = vᵢt + ½gt²

45 = 0 + ½ × 10 × t²

45 = 5t²

t² = 9

t = 3 s

(b) Final velocity

Using:

vᶠ = vᵢ + gt

vᶠ = 0 + (10 × 3)

vᶠ = 30 m/s downward

2.8. A car travels 10 km with an average velocity of 20 m s⁻¹. Then it travels in the same direction through a diversion at an average velocity of 4 m s⁻¹ for the next 0.8 km. Determine the average velocity of the car for the total journey.

Answer:

Part 1

d₁ = 10 km = 10,000 m

v₁ = 20 m/s

t₁ = d₁/v₁ = 10,000/20 = 500 s

Part 2

d₂ = 0.8 km = 800 m

v₂ = 4 m/s

t₂ = d₂/v₂ = 800/4 = 200 s

Total distance:

d = 10,000 + 800 = 10,800 m

Total time:

t = 500 + 200 = 700 s

Since both parts are described as travelling in the same direction, the supplied solution calculates average velocity as:

Average velocity = 10,800/700

Average velocity ≈ 15.4 m/s

2.9. A ball is dropped from the top of a tower. The ball reaches the ground in 5 seconds. Find the height of the tower and the velocity with which the ball strikes the ground.

Answer:

Given:

t = 5 s, vᵢ = 0, g = 10 m/s²

(a) Height of the tower

h = vᵢt + ½gt²

h = 0 + ½ × 10 × 5²

h = 5 × 25

h = 125 m

(b) Final velocity

vᶠ = vᵢ + gt

vᶠ = 0 + (10 × 5)

vᶠ = 50 m/s downward

2.10. A cricket ball is hit straight up into the air. An observer notes that it takes 3 seconds to reach the highest point. What was the initial velocity of the ball? If the ball was hit 1 m above the ground, how high did it rise above the ground?

Answer:

Given:

t = 3 s, vᶠ = 0, g = 10 m/s²

(a) Initial velocity

Taking upward as the positive direction:

vᶠ = vᵢ − gt

0 = vᵢ − (10 × 3)

vᵢ = 30 m/s

(b) Maximum height above the ground

Height gained above the release point:

h = vᵢt − ½gt²

h = (30 × 3) − ½ × 10 × 3²

h = 90 − 45

h = 45 m

The ball was initially 1 m above the ground:

H = 45 + 1

H = 46 m

Notes by Sir Dilawar

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