Question

a body of mass 1kg moving with Speed 100 cm/ s. It is brought to rest in 10 sec by applying a force against its motion . What distance will it move before coming at rest?​

Answer 1

Answer :-

The body will move a distance of 5 metres before coming at rest.

Explanation :-

We have :-

→ Mass (m) = 1 kg

→ Initial velocity (u) = 100 cm/s = 1 m/s

→ Time taken (t) = 10 sec

→ Final velocity (v) = 0

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Firstly let's calculate acceleration of the body by using the 1st equation of motion.

v = u + at

⇒ 0 = 1 + a(10)

⇒ 0 - 1 = 10a

⇒ -1 = 10a

⇒ a = -1/10

⇒ a = -0.1 m/s²

Now, we can calculate the distance covered by the body either by using the 2nd or 3rd equation of motion.

v² - u² = 2as

⇒ 0 - (1)² = 2(-0.1)s

⇒ 0 - 1 = -0.2s

⇒ -1 = -0.2s

⇒ s = -1/-0.2

⇒ s = 5 m

Answer 2

Given :-

A body of mass 1kg moving with Speed 100 cm/ s. It is brought to rest in 10 sec by applying a force against its motion .

To Find :-

Distance covered

Solution :-

We know that

100 cm/s = 1 m/s

Now

We know that

a = v - u/t

a = 0 - 1/10

a = -1/10

a = -0.1 m/s²

Now

s = ut + 1/2 × at²

s = 1 × 10 + 1/2 × (-0.1)(10)²

s = 10 + 1/2 × (-0.1) × 100

s = 10 - 1/2 × 0.1 × 100

s = 10 - 0.1 × 50

s = 10 - 5

s = 5 m

Hence

Distance travelled is 5 m

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