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Page No 110: - Chapter 8 Motion class 9 ncert solutions Science - SaraNextGen [2024]


Question 2:

A train is travelling at a speed of 90 km h−1. Brakes are applied so as to produce a uniform acceleration of −0.5 m s−2. Find how far the train will go before it is brought to rest.

Answer:

Initial speed of the train, u = 90 km/h = 25 m/s

Final speed of the train, = 0 (finally the train comes to rest)

Acceleration = −0.5 m s−2

According to third equation of motion:

v2 = u2 + 2 as

(0)2 = (25)2 + 2 (−0.5) s

Where, s is the distance covered by the train

https://img-nm.mnimgs.com/img/study_content/curr/1/9/8/120/500/Chapter%208_html_2c081c8f.gif

The train will cover a distance of 625 m before it comes to rest.

 

Question 3:

A trolley, while going down an inclined plane, has an acceleration of 2 cm s−2. What will be its velocity 3 s after the start?

Answer:

Initial velocity of the trolley, u = 0 (since the trolley was initially at rest)

Acceleration, a = 2 cm s−2 = 0.02 m/s2

Time, t = 3 s

According to the first equation of motion:

v = u + at

Where, v is the velocity of the trolley after 3 s from start

v = 0 + 0.02 × 3 = 0.06 m/s

Hence, the velocity of the trolley after 3 s from start is 0.06 m/s.

 

Question 4:

A racing car has a uniform acceleration of 4 m s−2. What distance will it cover in 10 s after start?

Answer:

Initial velocity of the racing car, u = 0 (since the racing car is initially at rest)

Acceleration, a = 4 m/s2

Time taken, t = 10 s

According to the second equation of motion:

https://img-nm.mnimgs.com/img/study_content/curr/1/9/8/120/505/Chapter%208_html_36ac3e10.gif

Where, s is the distance covered by the racing car

https://img-nm.mnimgs.com/img/study_content/curr/1/9/8/120/505/Chapter%208_html_40663799.gif

Hence, the distance covered by the racing car after 10 s from start is 200 m.

 

 

Question 5:

A stone is thrown in a vertically upward direction with a velocity of 5 m s−1. If the acceleration of the stone during its motion is 10 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?

Answer:

Initially, velocity of the stone,u = 5 m/s

Final velocity, v = 0 (since the stone comes to rest when it reaches its maximum height)

Acceleration of the stone, a = acceleration due to gravity, g = 10 m/s2

(in downward direction)

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

Acceleration, a = −10 m/s2

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

According to the first equation of motion:

v = u + at

0 = 5 + (−10) t

https://img-nm.mnimgs.com/img/study_content/curr/1/9/8/120/509/Chapter%208_html_m4137d0cf.gif

According to the third equation of motion:

v2 = u2 + 2 as

(0)2 = (5)2 + 2(−10) s

https://img-nm.mnimgs.com/img/study_content/curr/1/9/8/120/509/Chapter%208_html_m6572d2fd.gif

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

Also Read : Page-No-112:-Chapter-8-Motion-class-9-ncert-solutions-Science

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