Question

# A train is travelling at a speed of$90km{h}^{-1}$. Brakes are applied so as to produce a uniform acceleration of$-0.5m{s}^{-2}$. Find how far the train will go before it is brought to rest.

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Solution

## Step 1: GivenInitial speed of the train, $u=90km{h}^{-1}=90×\frac{5}{18}m{s}^{-1}=25m{s}^{-1}$Acceleration, $a=-0.5m{s}^{-1}$Final velocity, $v=0m{s}^{-1}$Let distance travelled by train before stop be$s$.Step 2: Formula used ${v}^{2}={u}^{2}+2as$Step 3: Calculation of distanceUsing third the equation of motion${v}^{2}={u}^{2}+2as\phantom{\rule{0ex}{0ex}}⇒0={\left(25m{s}^{-1}\right)}^{2}+2×\left(-0.5m{s}^{-2}\right)×s\phantom{\rule{0ex}{0ex}}⇒s=625m$Hence, the train will go $625m$ before it comes to rest.

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