NCERT Solutions Class 9 Science Chapter 8: Motion Intext Question

Intext Question Chapter 8 Class 9 Science Motion Ncert Solutions

In text Questions of page no 100

Q. No 1: An object has moved through a distance. Can it have zero displacement? If yes, support your answer with an example.
Ans:
Yes. Displacement can be zero even when an object moved through a certain distance.
Example: If a person is moving on a circular track after completing one round he will reach at the same position. His displacement will be zero because the actual distance from the initial and final position is zero even though he has covered the distance equal to the circumference of the circle.

In text Questions of page no 102

Q. No. 1 Distinguish between speed and velocity.
Ans:

Speed	Velocity
Speed has only magnitude and no direction and hence it is a scalarquantity.	Velocity has magnitude as well as direction and hence it is avector quantity.
Speed of a moving object cannot be zero.	Velocity of a moving object can be zero.
The rate of change of distance is known as speed.	The rate of change of displacement is known as velocity.

Q. No 2: Under what condition(s) is the magnitude of average velocity of an object equal to its average speed?
Ans: The condition where the total distance covered by an object is the same as its displacement, then its average speed would be equal to its average velocity.

Q. No 3: What does the odometer of an automobile measure?
Ans:The odometer of an automobile measures the total distance covered by an automobile.

Q. No 4: What does the path of an object look like when it is in uniform motion?
Ans: When an object is in uniform motion then its path will be a straight line straight line.

Q. No 5: During an experiment, a signal from a spaceship reached the ground station in five minutes. What was the distance of the spaceship from the ground station? The signal travels at the speed of light, that is, 3 × 10⁸ m s⁻¹.

Ans: Time taken by the signal = 5 minute = 5 × 60 = 300 s
Speed of the signal = 3 × 10⁸ m/s
We know that
Speed= Distance/Time
Therefore Distance= Speed × Time
= 3 × 10⁸ × 300
= 9 × 10¹⁰ m
So the distance of the spaceship from the ground station is 9 × 10¹⁰ m.

In text Questions of page no 103

Q. No 1: When will you say a body is in (i) uniform acceleration? (ii) non-uniform acceleration?
Ans:
(i) A body is said to have uniform acceleration if it travels in a straight path and the velocity of a body increases or decreases by equal amounts in an equal interval of time.
(ii) A body is said to have non-uniform acceleration if it travels in a straight and the velocity of a body increases or decreases in unequal amounts in an equal interval of time.

Q. No 2: A bus decreases its speed from 80 km h^-1 to 60 km h^-1 in 5 s. Find the acceleration of the bus.
Ans:

Initial velocity of the bus, u = 80 km/h
$ = 80 \times {5 \over {18}} = 22.22m/s$

Final velocity of the bus, v = 60 km/h
$ = 60 \times {5 \over {18}} = 16.67m/s$

Time take (t) = 5 s
Now acceleration(a)

$\eqalign{& = {{v - u} \over t} \cr& = {{16.67 - 22.22} \over 5} \cr& = {{ - 5.55} \over 5} \cr& = - 1.11m{s^{ - 2}} \cr} $

when acceleration is negative then it is called retardation.

Q. No 3: A train starting from a railway station and moving with uniform acceleration attains a speed 40 km h^-1 in 10 minutes. Find its acceleration.

Ans:

Initial velocity of the train (u) = 0 m/s

Final velocity of the train, v = 40 km/h
= 40×5/18=100/9 m/s
Time taken, t = 10 min = 10 × 60 = 600 s
Now, acceleration

$\eqalign{& a = {{v - u} \over t} \cr& = {{{{100} \over 9} - 0} \over {600}} \cr& = {{100} \over {9 \times 600}} \cr& = {1 \over {54}}m/{s^2} \cr} $

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