Page No 113: - Chapter 8 - Motion - Ncert Solutions class 9 - Science

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Question 7:

A ball is gently dropped from a height of 20 m. If its velocity increases uniformly at the rate of 10 m s⁻², with what velocity will it strike the ground? After what time will it strike the ground?

Answer:

Distance covered by the ball, s = 20 m

Acceleration, a = 10 m/s²

Initially, velocity, u = 0 (since the ball was initially at rest)

Final velocity of the ball with which it strikes the ground, v

According to the third equation of motion:

v² = u² + 2 as

v² = 0 + 2 (10) (20)

v = 20 m/s

According to the first equation of motion:

v = u + at

Where,

Time, t taken by the ball to strike the ground is,

20 = 0 + 10 (t)

t = 2 s

Hence, the ball strikes the ground after 2 s with a velocity of 20 m/s.

Question 8:

The speed-time graph for a car is shown is Fig. 8.12.

Fig. 8.12

(a) Find out how far the car travels in the first 4 seconds. Shade the area on the graph that represents the distance travelled by the car during the period.

(b) Which part of the graph represents uniform motion of the car?

Answer:

(a)

The shaded area which is equal to  represents the distance travelled by the car in the first 4 s.

(b)

The part of the graph in red colour between time 6 s to 10 s represents uniform motion of the car.

Question 9:

State which of the following situations are possible and give an example for each of these:

(a) an object with a constant acceleration but with zero velocity.

(b) an object moving in a certain direction with an acceleration in the perpendicular direction.

Answer:

(a) Possible

When a ball is thrown up at maximum height, it has zero velocity, although it will have constant acceleration due to gravity, which is equal to 9.8 m/s².

(b) Possible

When a car is moving in a circular track, its acceleration is perpendicular to its direction.

Question 10:

An artificial satellite is moving in a circular orbit of radius 42250 km. Calculate its speed if it takes 24 hours to revolve around the earth.

Answer:

Speed = DistanceTimeDistance = 2πr = 2×3.14×42250 

= 265330 km Time = 24 hSpeed = 26533024 

= 11055.4 km/h