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A train is travelling at a speed of $90 km / hr$ . Brakes are applied so as to produce a uniform acceleration of $- 0.5 m / s^{2}$ . The train will go ____ $m$ before it is brought to rest.

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Detailed Solution

The train will go

625 m

before it is brought to rest.
Given:
Initial velocity,

u = 90 km / hr

We know,

1 km / hr = \frac{5}{18} m / s

\Rightarrow

= 90 \times \frac{5}{18} m / s

\Rightarrow

= 25 m / s

u = 25 m / s

Acceleration,

a = - 0.5 m / s^{2}

Final velocity,

v = 0

From the third equation of motion,

{2 as = v}^{2} - u^{2}

s = \frac{v^{2} - u^{2}}{2 a}

where

s

is the distance.

\Rightarrow

= \frac{0 - (25 m / s)^{2}}{2 (- 0.5 m / s^{2})}

s = 625 m

Thus, the train will go

625 m

further, after applying brakes.

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