When brakes are applied to a bus, the retardation produced is $25 c m s^{- 2}$ and the bus takes $20 s$ to stop. Find the distance travelled by the bus during this time.

25 m

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30 m

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40 m

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50 m

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Solution

The correct option is D $50 m$
Final Velocity, $v = 0$

Retardation, $a = 0.25 m s^{- 2}$

Time, $t = 20 s$

Using equation, $v = u - a t$ ,
we get $u = 5 m s^{- 1}$

$∴$ distance, $s = u t - \frac{1}{2} a t^{2} = (5 \times 20) - (\frac{1}{2} \times 0.25 \times 20^{2}) = 50 m$

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When brakes are applied to a bus, the retardation produced is 25cms−2 and the bus takes 20s to stop. Find the distance travelled by the bus during this time.

The correct option is D 50mFinal Velocity, v=0Retardation, a=0.25ms−2Time, t=20sUsing equation, v=u−at, we get u=5ms−1∴ distance, s=ut−12at2=(5×20)−(12×0.25×202)=50m

When brakes are applied to a bus, the retardation produced is $25 c m s^{- 2}$ and the bus takes $20 s$ to stop. Find the distance travelled by the bus during this time.

The correct option is D $50 m$
Final Velocity, $v = 0$

Retardation, $a = 0.25 m s^{- 2}$

Time, $t = 20 s$

Using equation, $v = u - a t$ ,
we get $u = 5 m s^{- 1}$

$∴$ distance, $s = u t - \frac{1}{2} a t^{2} = (5 \times 20) - (\frac{1}{2} \times 0.25 \times 20^{2}) = 50 m$