CBSE Important Questions Class 9 Science Chapter 7 – Motion

Class 9 Science Chapter 7 Motion is part of the CBSE Class 9 Science syllabus. It explains what motion is, why a reference point is needed, and the difference between distance and displacement, speed and velocity. Students learn about uniform and non-uniform motion, acceleration and retardation, the three equations of motion, distance-time and velocity-time graphs, and uniform circular motion.

These CBSE important questions for Class 9 Science Chapter 7 cover the definition of motion and the origin as a reference point, numerical problems on acceleration and deceleration, reading a speed-time graph to find distance, uniform circular motion, relative speed of two trains and the three equations of motion. Every question has a clear step-by-step answer in simple English with figures, so students can practise on their own, parents can check the working, and teachers can use them for class tests.

Q1. Define motion. State one common characteristic of all objects in motion.

Answer: Motion is the change in the position of a body with respect to its surroundings and with time.
The common characteristic of all moving objects is that their position keeps changing with time with respect to a fixed reference point. If the position does not change, the object is said to be at rest.

Q2. A reference point used to specify the position of a moving object is also known as
(a) origin
(b) distance
(c) final position
(d) initial position

Answer: (a) origin
To describe where an object is, we first choose a fixed point from which all distances are measured. This fixed reference point is called the origin. Distance and position are measured from it, so they are not the reference point itself.

Q3. A motorcyclist decreases his speed from 50 m/s to 20 m/s in 10 seconds. Find the acceleration of the motorcyclist.

Answer: Given: initial speed u = 50 m/s, final speed v = 20 m/s, time t = 10 s.
a = (v − u) / t
= (20 − 50) / 10
= −30 / 10 = −3 m/s²
The negative sign shows that the motorcyclist is slowing down. This is called retardation or deceleration.

Q4. On the way to the zoo a speed breaker comes, and Mohan has to apply the brakes to slow his car from 70 km/h to 20 km/h in 10 seconds. The approximate acceleration of the car will be
(a) −1.4 m/s²
(b) 1.4 m/s²
(c) −5 m/s²
(d) 5 m/s²

Answer: (a) −1.4 m/s²
First change the speeds into m/s by multiplying by 5/18:
u = 70 × 5/18 = 19.44 m/s and v = 20 × 5/18 = 5.56 m/s
a = (v − u) / t = (5.56 − 19.44) / 10 = −13.88 / 10 = −1.4 m/s²
The negative sign shows the car is slowing down.

Q5. Starting from rest, Roma pedals her bicycle to reach a velocity of 6 m/s in 30 s. Then she applies the brakes so that the velocity comes down to 4 m/s in the next 5 s. The acceleration of the bicycle in the first and second cases would be
(a) 1.0 m/s² and 0.2 m/s² respectively
(b) 0.1 m/s² and −0.2 m/s² respectively
(c) 0.2 m/s² and −0.4 m/s² respectively
(d) 0.4 m/s² and −0.2 m/s² respectively

Answer: (c) 0.2 m/s² and −0.4 m/s² respectively
First case: u = 0, v = 6 m/s, t = 30 s
a = (6 − 0)/30 = 0.2 m/s²
Second case: u = 6 m/s, v = 4 m/s, t = 5 s
a = (4 − 6)/5 = −2/5 = −0.4 m/s²

Q6. If a bus moving along a straight line with a velocity of 90 km/h decreases its velocity to 54 km/h in 10 s, then the deceleration of the bus is
(a) 5 m/s²
(b) 4 m/s²
(c) 3 m/s²
(d) 1 m/s²

Answer: (d) 1 m/s²
u = 90 × 5/18 = 25 m/s and v = 54 × 5/18 = 15 m/s
a = (15 − 25)/10 = −10/10 = −1 m/s²
So the deceleration (retardation) is 1 m/s².

Q7. A driver of a car travelling at 52 km/h applies the brakes and the car stops in 5 s. Another driver going at 3 km/h in another car applies his brakes slowly and stops in 10 s. On the same graph paper, plot the speed-time graphs for the two cars. Which of the two cars travelled farther after the brakes were applied?

Speed-time graphs of two cars after the brakes are applied

Answer: On a speed-time graph, the distance travelled is the area under the graph line.
First car: u = 52 km/h = 14.44 m/s, stops in 5 s.
Distance = ½ × base × height = ½ × 5 × 14.44 = 36.1 m
Second car: u = 3 km/h = 0.83 m/s, stops in 10 s.
Distance = ½ × 10 × 0.83 = 4.15 m
So the first car travelled farther after the brakes were applied.

Q8. If a body is revolving along the circumference of a circle with constant speed, then there is a change in
(a) the mass of the body
(b) the magnitude of velocity
(c) the direction of velocity
(d) both the magnitude and direction of velocity

Answer: (c) the direction of velocity
In uniform circular motion the speed stays the same, so the magnitude of velocity does not change. But the direction of motion keeps changing at every point of the circle, and velocity depends on direction too. So the velocity changes, which means the body is accelerating even at constant speed.

Q9. A driver changes the speed of a car from 25 m/s to 10 m/s in 5 seconds. Find the acceleration of the car.

Answer: Given: u = 25 m/s, v = 10 m/s, t = 5 s.
From the first equation of motion, v = u + at, so a = (v − u)/t
= (10 − 25)/5
= −15/5 = −3 m/s²
The negative sign means the car has a retardation of 3 m/s².

Q10. A bus starts from rest and attains a velocity of 20 m/s in 300 s. Assuming that the acceleration is uniform, the acceleration is
(a) 1/25 m/s²
(b) 1/21 m/s²
(c) 1/19 m/s²
(d) 1/15 m/s²

Answer: (d) 1/15 m/s²
u = 0, v = 20 m/s, t = 300 s
a = (v − u)/t = 20/300 = 1/15 m/s² (about 0.067 m/s²)

Q11. Two trains A and B, moving in opposite directions, are about to pass each other. The speeds of A and B are 30 m/s and 40 m/s respectively. The trains completely pass each other in 10 seconds. Calculate the combined length of the two trains.

Two trains moving in opposite directions, initially and finally

Answer: When two bodies move in opposite directions, their relative speed is the sum of their speeds.
Relative speed = 30 + 40 = 70 m/s
In the time the trains take to pass each other completely, the distance covered at this relative speed is equal to the combined length of the two trains.
Distance = speed × time = 70 × 10 = 700 m
So the combined length of the two trains is 700 m.

Q12. Write the three equations of uniformly accelerated motion and state what each symbol means.

Answer: The three equations of motion for uniform acceleration are:
1. v = u + at
2. s = ut + ½at²
3. v² = u² + 2as
Here:
u = initial velocity, v = final velocity, a = acceleration, t = time taken, and s = distance travelled in that time.

Related Study Material on Extramarks

Q.1 Define the term displacement.
Marks:1
Ans

The change in the position of a moving body in a particular direction is called its displacement. It is the shortest distance measured from the initial position to the final position of a body.

Q.2 A bus covers a distance of 250 km from Delhi to Jaipur towards West in 5 hours in the morning and returns to Delhi in the evening covering the same distance of 250km in the same time of 5 hours.
Find
(a) Average speed
(b) Average velocity of the bus for the whole journey.
Marks:2
Ans

(i) Average speed = Total distance travelled /Total time taken

= [250 km + 250 km] / [5hr + 5hr]

= 50 km/ hr

(ii) Bus travels from Delhi to Jaipur = +250km & then come back it means Jaipur to Delhi = -250km

Average velocity =Total displacement /Total time taken

= [250 km – 200 km] / [5hr+5hr]

= 0km / 10hr

Hence, average velocity = 0 km/h

Q.3 When do we say the acceleration of a body is zero?
Marks:1
Ans

Acceleration = Change in velocity/ time interval

When the velocity of a body does not change i.e. final velocity is equal to the initial velocity, then we say that it’s acceleration is zero.

Mathematically, a = (v – u) / t
= (u – u) / t
= 0 / t
or, a = 0

Q.4 Two cars come to rest with the same constant retardation, a. If the first car travelling with initial speed 3u, covers a distance d1 and the second car travelling with an initial speed of u, covers a distance d2, then

A. d1 = 9d2

B. d2 = 9d1

C. d1 = 16d2

D. d2 = 16d1

Marks:1
Ans

d1 = 9d2

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