A body moving with uniform retardation covers $3 k m$ before its speed is reduced to half of its initial value. It comes to rest in another distance of :

1 k m

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2 k m

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3 k m

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\frac{1}{2} k m

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Solution

The correct option is A $1 k m$
Lets $u =$ initial velocity

$s = 3 k m = 3000 m$

$v = \frac{u}{2}$

3rd equation of motion,

$v^{2} - u^{2} = 2 a s$

$(\frac{u}{2})^{2} - u^{2} = 2 a \times 3000$

$- \frac{3}{4} u^{2} = 3000 \times 2 a$

$u^{2} = - 8000 a$ . . . . . .(1)

Now body comes to rest. 3rd equation of motion is

$0 - (\frac{u}{2})^{2} = 2 a s^{'}$

$\frac{- u^{2}}{4} = 2 a s^{'}$

$- \frac{- 8000 a}{4} = 2 a s^{'}$

$s^{'} = 1 k m$

The correct option is A.

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