Question

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Answer 1

Answer:

A body moving with uniform retardation covers 3 km before its speed is reduced to half of its initial value it comes to rest in another distance of 1 km.

Explanation:

We will employ the use of the kinematic equations to solve for the distance covered:

We know that  v² = u² + 2as

Where v is the final velocity - we can say this is v m/s in this case or u/2

          u is the initial velocity - which we can say is u m/s

          a is the acceleration

           s is the displacement, 3 km

v² = u² + 2as  

(u/2²) = u² + 2×a×3

6a = (u/2²) - u²

6a = u²/4 - u²

a = (u²/4 - u²)/6

a = u²/24 - u²/6

a =- 3u²/24

a = -u²/8

This is the deceleration at which the body is moving with, -u²/8

⇒We can now consider, the retardation of the body from when its velocity is u until it gets to rest, zero velocity v = 0

v² = u² + 2as  

0² = u² + 2 × - u²/8 × s

- u² = -2u²s/8

s = -8u² /- 2u²

s = 4

Therefore the car takes 4 km from start to rest.

If the car had taken 3km, then the rest of the distance was 4km - 3km = 1km