Motion Class 9 Important Questions

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Very Short Answer Questions (1 Mark Questions)

Ques. What is an odometer?

Ans. An odometer is a device that shows the distance traveled by an automobile. 

Ques. When a body is said to be positively accelerated?

Ans. A body is said to be positively accelerated when it starts from rest and its velocity increases continuously.

Ques. What would be the effect on velocity and speed, when a body moves uniformly along the circle?

Ans. When a body moves uniformly along a circle, its speed remains the same but its velocity changes due to continuous change in direction of motion.

Ques. Can the displacement be zero?

Ans. Yes, when the initial and final position of the object is the same, the displacement is said to be zero.

Ques. What is the origin?

Ans. Origin is a reference point to describe the position of an object.

Ques. Name the two scientists who have given the phenomenon of motion.

Ans. Galileo Galilei and Isaac Newton are the two scientists who have given the phenomenon of motion.

Ques. What is Motion?

Ans. Motion is the change in position of an object with respect to its surroundings.

Ques. What is the SI unit of Speed?

Ans. The SI unit of speed is meters per second (m/s).

Ques. Differentiate between Speed and Velocity.

Ans. The main difference between speed and velocity is

• Speed is a scalar quantity, representing the rate of motion, while
• Velocity is a vector quantity, indicating the rate of motion along with direction.

Ques. State the First Equation of Motion.

Ans. The first equation of motion is given by

v = u + at

where

• v is the final velocity
• u is the initial velocity
• a is acceleration, and
• t is time.

Ques. What is the relationship between Distance, Speed, and Time?

Ans. The relationship is given by the formula

s = v/t

Where

• s is distance
• v is speed, and
• t is time.

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Short Answer Questions (2 Marks Questions)

Ques. Can a body have variable velocity but constant speed?

Ans. Yes, when a body moves in a uniform circular motion, it has constant speed but due to the change in the direction of motion, its velocity changes at every point.

Ques. What is the difference between scalar and vector quantity?

Ans. The differences between scalar and vector quantities are as follows:

Ques. Can the magnitude of the displacement and distance be equal?

Ans. Yes, the magnitude of displacement can be equal to the distance traveled by the object, if the body or object moves along a straight line.

Ques. State the statement as true or false.

(a) Displacement can be zero.

(a) The magnitude of displacement is greater than the distance traveled by the object.

Ans. (a) True

Explanation: When the initial position coincides with the final position, the displacement is zero.

(b) False

Explanation: Displacement is the shortest distance covered by an object, so it can never be greater than the distance covered by the object.

Ques. List the type of acceleration if a car is travelling along a straight road, and increases its speed by unequal amounts in equal intervals of time.

Ans. A car changes its speed in equal intervals of time. Such a type of acceleration is known as non-uniform acceleration. The acceleration of the object is said to be non-uniform when its velocity increases or decreases in an equal interval of time.

Ques. Define uniform acceleration.

Ans. In uniform acceleration, an object travels in a straight line, and its velocity increases or decreases by equal amounts in equal intervals of time. The motion of a freely falling body is an example of uniform acceleration.

Ques. Define non-uniform acceleration.

Ans. An object is said to be in non-uniform motion when it increases its speed by unequal amounts in equal intervals of time. For example, A car traveling along a straight road increases its speed by unequal amounts in equal intervals of time.

Ques. Explain the concept of relative states of rest and motion. 

Ans. The states of rest and motion are relative, meaning they depend on the observer's frame of reference. An object considered at rest by one observer might be in motion from the perspective of another observer in a different frame. For example, a person sitting on a moving train sees a book on the seat as at rest. However, an observer on the platform sees both the person and the book in motion. This illustrates that the determination of rest or motion is relative to the observer's point of view.

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Long Answer Questions (3 Marks Questions)

Ques. A bus starting from rest moves with a uniform acceleration of 0.3 m s^-2 for 2 minutes. Find the distance travelled by bus.

Ans. Given,

• The initial velocity of the bus, u= 0
• Acceleration of bus, a = 0.3 m s^-2
• Time, t = 2 minutes = 120 seconds

From the equation of motion, we have

v = u + at

On substituting the values, we get

v = 0 + (0.3 x 120)

⇒ v = 36 ms^-1

Let S be the distance travelled by bus, then we have

S = ut + 1/2 at²

On substituting the values, we get

⇒ S = (0 x 120) + (1/2 x 0.3 x 120²)

⇒ S = 0 + (1/2 x 0.3 x 120²)

⇒ S = 2160 m

Ques. Mohan moves from point A to point B in a circle with O as the center. Given AO = OB = 10 cm. Calculate

(a) the distance covered

(b) the displacement, when it reaches B.

Ans. Given

• diameter of the circle, d = AO = OB = 10 cm

(a) Distance covered = Circumference of circle

= 2πr = 2π(d/2) ( Radius = d/2)

= 2 x 3.14 x  5

= 31.4 cm

(b) Displacement = 2 × OB

= 2 × 10 = 20 cm along AB

Ques. Find the acceleration of the bus, when it decreases its speed from 80 km h^–1 to 60 km h^–1 in 5 s.

Ans. Given

• Initial speed of bus (u) = 80 km h^–1 = 22.22 ms^–1
• Final speed of bus (v) = 60 km h^–1 = 16.67 ms^–1
• Time (t) = 5 s

We have

Acceleration (a) = (v − u)/t

⇒ a = (16.67 − 22.22)/5

⇒ a = −1.11m/s²

Therefore, the acceleration of the bus will be -1.11 m/s²

Read More: Average Velocity Formula

Ques. Find the acceleration of the train when it starts from a railway station and moves with uniform acceleration attaining a speed of 40 km h^-1 in 10 minutes.

Ans. Given

• The initial velocity of the train, u = 0
• Final velocity, v = 40 km h^-1 = 11.11 ms^-1
• Time, t = 10 min = 600 seconds

Put values of u, v, and t in the acceleration formula

Acceleration (a) = (v − u)/t

⇒ a = (11.11 – 0)/600

⇒ a = 0.018 m/s²

Therefore, the acceleration of the train will be 0.018 m/s².

Ques. Sara travels from her home to the grocery store. Her initial odometer reading is 1520 km, and the final reading is 1535 km. If the time taken for the trip is 30 minutes, calculate Sara's average speed and velocity.

Ans. Given:

• Initial odometer reading = 1520 km
• Final odometer reading = 1535 km
• Time taken = 30 minutes = 0.5 hour

Average Speed = Total Distance/Total Time

Average Speed = (1535 km − 1520 km)/0.5 hour

⇒ Average Speed= 30 km/h​

Average Velocity = Total Displacement/Total Time = (1535 km − 1520 km)/0.5 hour

Average Velocity ​= 30km/h

Now, convert to m/s: 30 km/h= 30 × 1000/3600 m/s

Average Velocity ≈ 8.33 m/s

Therefore, Sara's average speed is 30 km/h and her average velocity is approximately 8.33 m/s.

Ques. An electron moving with a velocity of 5 × 10⁴ ms^-1 enters into a uniform electric field and acquires a uniform acceleration of 10⁴ ms^-2 in the direction of its initial motion.
(i) Calculate the time in which the electron would acquire a velocity double its initial velocity.
(ii) How much distance the electron would cover in this time?

Ans. Given

• The initial velocity of the electron, u = 5 × 10⁴ ms^-1
• Acceleration of the electron, a = 10⁴ ms^-2

(i) To calculate the time in which the electron would acquire a velocity double its initial velocity.

The equation of motion for uniformly accelerated motion is given by:

v = u + at

Where

• v is the final velocity,
• u is the initial velocity,
• a is the acceleration,
• t is the time.

Here, the final velocity v is double the initial velocity u, so v = 2u. The acceleration a is given as 10⁴ ms², and the initial velocity u is 5 x 10⁴ ms^-1

Substituting the values into the equation, we get:

2u = u + at

⇒ 2 x 5 x 10⁴ ms^-1 = 5 x 10⁴ ms^-1 + 10⁴ t

Solving for t

10⁵ = 5 x 10⁴ ms^-1 + 10⁴ t

⇒ t = 5 s

So, the time in which the electron would acquire a velocity double its initial velocity is 5 s.

ii) To calculate the distance the electron would cover in this time

The equation for distance covered s in uniformly accelerated motion is given by:

s = ut + 1/2 at²

Where s is the distance travelled by the electron

Substituting the known values:

s = 5 x 10⁴ ms^-1 x 5 s + 1/2 10⁴ x 5²

⇒ s = 2.5 x 10⁵ + 1/2 x 10⁴ x 25

⇒ s = 2.5 x 10⁵+ 1.25 x 10⁵

⇒ s = 3.75 x 10⁵ m

So, the electron would cover a distance of 3.75 x 10⁵ ​m in this time.

Ques. A car accelerates from rest at a rate of 2m/s² for a time period of 8s. Determine the final velocity of the car using the equation v = u + at.

Ans. Given:

• Initial velocity u = 0 m/s (car starts from rest)
• Acceleration a = 2 m/s²
• Time t = 8 s

Using the equation v = u + at, we can calculate the final velocity v:
v = 0 + (2 x 8)
⇒ v = 0 + 16 
⇒ v = 16 m/s
The final velocity of the car is 16 m/s after accelerating for 8 s.

Ques. What are the uses of a distance-time graph?

Ans. A distance-time graph, also known as a displacement-time graph, is a visual representation of the motion of an object. It depicts the relationship between the distance covered by an object and the time taken to cover that distance. Here are some uses of a distance-time graph:

• The gradient (slope) of the graph represents the object's speed.
• A steeper slope indicates higher speed.
• The direction of motion can be inferred from the slope: an upward slope indicates motion away, a downward slope indicates motion toward, and a horizontal line suggests rest.
• The area under the graph, particularly in the case of a straight-line graph, represents the distance travelled.
• The graph helps in distinguishing between distance and displacement, where displacement is the overall change in position.
• A horizontal line indicates that the object is at rest (zero speed).
• A sloping line, whether upward or downward, indicates motion.
• Multiple objects or different segments of motion can be compared on a single graph.
• Changes in slope represent acceleration or deceleration.
• Distance-time graphs provide a visual aid for solving problems related to motion.

Ques. The walls of your classroom are in motion but appear stationary. Explain.

Ans. The phenomenon described is an example of relative motion.

• In this case, although the walls of the classroom are in motion, they appear stationary to the students inside the classroom.
• This is because the students and the entire classroom are in the same frame of reference and are moving with the walls.
• When all objects in a given frame of reference move together with the same velocity, there is no relative motion perceived among them.
• Therefore, from the perspective of the students inside the classroom, the walls seem stationary, creating an illusion of rest despite the actual motion.

Ques. Displacement for a course of motion may be zero but the corresponding distance covered is not zero. Explain

Ans. Suppose the object moves from O to A and then back to B, distance = 60 km + 25 km = 85 km

Displacement from O to B = 35 Km

When the object travels from O and reverts back to O, then the final position coincides with the initial position and makes the displacement equal to zero.

So, displacement for a course of motion may be zero but the corresponding distance covered is not zero.

Ques. What is the nature of the distance-time graphs for uniform and non-uniform motion of an object?

Ans. For uniform motion, the nature of the distance-time graph will be linear that is it would be in a straight line.

Distance-Time Graph for Uniform Motion

For non-uniform motion, the nature of the distance-time graph will be a curved line.

Distance-Time Graph for Non-Uniform Motion

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Very Long Answer Questions (5 Marks Questions)

Ques. The brakes applied to a car produce an acceleration of 4 ms^-2 in the opposite direction to the motion. If the car takes a time of 3s to stop after the application of brakes, calculate the distance it travels during this time.

Ans. Given

• The acceleration of the car, a = – 4 ms^-2
• Time taken, t = 3 s
• Final velocity of the car, v = 0

Let u be the initial velocity of the car, then we have

v = u + at

⇒ 0 = u + (- 4 x 3)

⇒ u = 12 ms^-1

The distance travelled by the car is given by

s = ut + 1/2 at²

⇒ s = (12 x 3) + [1/2 x (-4) x (3)²]

⇒ s = (12 x 3) + [(-2) x 9]

⇒ s = 36 + (-18)

⇒ s = 18 m

Therefore, the car will move 18 m before it stops after the application of brakes.

Ques. A body begins to slide over a horizontal surface with an initial velocity of 0.3 m/s. Due to friction, the velocity decreases at the rate of 0.05 m/s². Calculate how much time will it take for the body to stop.

Ans. Given

• Initial velocity, u = 0.3 m/s
• Final velocity, v = 0
• Acceleration, a = -0.05 m/s²

Now, according to the first equation of the motion,

v = u + at

⇒ 0 = 0.3 + (- 0.05t)

⇒ 0.05 t = 0.3

⇒ t = 0.3/0.05

⇒ t = 6 s

Thus, the body will take 6 s to stop.

Ques. Write the differences between distance and displacement.

Ans. The differences between distance and displacement are as follows-

Ques. Ramesh throws a stone vertically upwardly with a velocity of 5 ms^-1. If the acceleration of the stone during its motion is 12 m s^-2 in the downward direction, what will be the height attained by the stone and how much time will it take to reach there?

Ans. Given:

• The initial velocity of stone, u = 5 ms^-1
• Acceleration of stone, a = -12 ms^-2 (-ve sign for downward acceleration)
• Final velocity, v = 0 ( at maximum height velocity will be zero)

We have, v = u + at

On substituting values, we get

0 = 5 + (-12)t

⇒ t = 5/12 = 0.41 s

Time taken by stone to reach earth, t = 0.41 s

The height attained by the stone is given by

S = ut + 1/2at²

On substituting the values, we get

S = 5 x 0.41 + [1/2 x -12 x (0.41)²]

⇒ S = 2.05 + 1.00

⇒ S = 3.05 m

Ques. An athlete completes one round of a circular track of diameter 200 m in 40 seconds. What will be the distance covered and the displacement at the end of 2 minutes 20 s?

Ans. Given

• The diameter of the circular track, d = 200m
• Time taken to complete the track, t = 40s

Distance covered in one round = Circumference of a circle = 2πr

= 2 × π × 100  ( r = d/2)

The time that an athlete takes to complete one round = 40 seconds.

Number of rounds completed in 140s = 140/40 = 3.5

Distance covered = 3.5 × circumference

= 3.5 × 2πr

= 3.5 × 2 × 3.14 × 100

Distance covered = 2198 ≈ 2200 m= 22km

Displacement is calculated as 200 m as it returns to the initial position.

Read More: Differences between Acceleration and Velocity

Ques. Abdul, while driving to school, computes the average speed for his trip to be 20 km h^-1. On his return trip along the same route, there is less traffic and the average speed is 40 km h^-1. Calculate what is the average speed of Abdul’s trip.

Ans. Given

• Average speed while driving to school = 20 km h^-1
• Average speed while returning from school = 40 km h^-1

Let, the distance from Abdul’s home to school = x km

Time taken while driving to school is given by

t = x/20 hr

Time taken while returning from school is given by

t = x/40 hr

Total distance travelled = x + x = 2x

total time = t + t

= x/20 + x/40

= 3x/40

The average speed for Abdul’s trip is given by

Average speed = Total distance traveled / total time taken

= 2x / (3x/40)

= 26.67 Km/hr

Ques. A ball is dropped from a height of 20 m. if its velocity increases uniformly at the rate of 10 m/s, with what velocity will it strike the ground?

Ans. Given:

• Initial velocity (u) = 0 m/s (as the ball is dropped)
• Acceleration (a) = 10 m/s²
• Displacement (s) = 20 m

Using the second equation of motion v² = u² + 2as

v² = 400 m²/s²

Taking the square root of both sides to find v = 20m/s

The ball will strike the ground with a velocity of 20m/s. 

Ques. (a) Is it possible that an object with a constant acceleration but with zero velocity?

(b) Is it possible that an object moving in a certain direction with acceleration in the perpendicular direction?

Ans. (a) Yes, it is possible. When an object is thrown vertically up, then at the peak of its trajectory its velocity is zero, but a downward acceleration will act on the body.

(b) Yes, it is possible. For example- when an athlete moves in a circular path in the stadium, then its direction is along the tangent of the circle but the centripetal acceleration is towards the radius of the circle. The athlete moving along a circular path is an example of an accelerated motion where acceleration is always perpendicular to the direction of motion of an object at a given instance.

Ques. What are the differences between uniform and non-uniform motion?

Ans. The differences between uniform and non-uniform motion are

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