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.

Answer 1

Step 1: Given

Initial speed of the train, $u=90km{h}^{-1}=90\times \frac{5}{18}m{s}^{-1}=25m{s}^{-1}$

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 distance

Using third the equation of motion

$$\begin{gathered} v^2=u^2+2as \\ \Rightarrow 0={\left(25m{s}^{-1}\right)}^2+2\times (-0.5m{s}^{-2})\times s \\ \Rightarrow s=625m \end{gathered}$$

Hence, the train will go $625m$ before it comes to rest.