# NCERT Class 9 Science Solutions: Chapter 8 – Motion Part 5

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Question 5:

Q.A stone is thrown in a vertically upward direction with a velocity of 5 m If the acceleration of the stone during its motion is 10 m in the downward direction, what will be the height attained by the stone and how much time will it take to reach there?

**Answer:**

Initially, velocity of the stone, m/s

Final velocity, (since the stone comes to rest when it reaches its maximum height)

Acceleration of the stone, acceleration due to gravity, (in downward direction)

There will be a change in the sign of acceleration because the stone is being thrown upwards.

Acceleration,

Let s be the maximum height attained by the stone in time t.

According to the first equation of motion:

According to the third equation of motion:

Hence, the stone attains a height of 1.25 m in 0.5 s.

Question 1:

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

**Answer:**

Diameter of a circular track,

Radius of the track,

Circumference

In 40 s, the given athlete covers a distance

The athlete runs for 2 minutes 20 s=140 s

The athlete covers one round of the circular track in 40 s. This means that after every 40 s, the athlete comes back to his original position. Hence, in 140s he had completed 3 rounds of the circular track and is taking the fourth round.

He takes 3 rounds in Thus, after 120 s his displacement is zero.

Then, the net displacement of the athlete is in 20 s only. In this interval of time, he moves at the opposite end of the initial position. Since displacement is equal to the shortest distance between the initial and final position of the athlete, displacement of the athlete will be equal to the diameter of the circular track.

Distance covered by the athlete in 2 min 20 s is 2200 m and his displacement is 200 m.

Question 2:

Q. joseph jogs from one end A to the other end B of a straight 300 m road in 2 minutes 50 seconds and then turns around and jogs 100 m back to point C in another 1 minute. What are joseph’s average speeds and velocities in jogging (a) from A to B and (b) from A to C?

**Answer:**

(a) 1.765 m/s,1.765 m/s(b)1.739 m/s,0.87 m/s

(a) From end A to end B

Distance covered by joseph while jogging from A to B

Time taken to cover that distance min 50 seconds

Total distance covered300 m

Total time taken170 s

Displacementshortest distance between A and B 300m

Time interval170 s

The average speed and average velocity of joseph from A to B are the same and equal to 1.765 m/s.

(b) From end A to end C

Total distance covered = Distance from A to B + Distance from B to C=300+100=400 m

Total time taken=Time taken to travel from A to B + Time taken to travel from B to C = 170 + 60 =230 s

Total distance covered=Distance from A to B + Distance from B to C

=300+100=400 m

Total time taken=time taken to travel from A to B + Time taken to travel from B to C=170+60==230s

Displacement from A to

Time interval=time taken to travel from A to B + time taken to travel from B to C = 170+60=230 s

Average velocity

The average speed of joseph from A to C is 1.739 m/s and his average velocity is 0.87 m/s.