Question

A bus is moving with a velocity of 72 km/hr. By applying brakes a deaccleration of 0.4 m/s$^{2}$ is produced. Find the distance covered by the bus befire it stops​

Answer 1

Answer:

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The distance covered by the bus before it stops is 500 m.

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Explanation:

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Given:

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Initial velocity (u) = 72 km/hr

$= \frac{72 \times 1000}{60 \times 60} = \: 20 \: \frac{m}{s}$

Final velocity (v) = 0

Accleration (a) = 0.4 m/s$^{2}$

Time (t) = x

To find:

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Distance covered by the bus.

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Solution:

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We know that,

2as = v$^{2}$ – u$^{2}$

$\implies$ 2 (0.4) s = 0 – (20)$^{2}$

$\implies$ – 0.85 = – 400

Now we know that,

Speed = Distance/Time

$\implies$ S = 400/0.8

= 500 m

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Thus, S = 500 m.