Question

A train is traveling at a speed of 90 km h^-1. Brakes are applied so as to produce a uniform acceleration of –0.5 ms^-2. Find how far the train will go before it is brought to rest?

Answer 1

Given data:

1. The initial velocity(u) of the train is $90\mathrm{km}/\mathrm{hr}=25\mathrm{m}/\mathrm{s}$.
2. The final velocity(v) of the train is zero as it comes to rest finally.
3. Acceleration of the train is $-0.5\mathrm{m}/{\mathrm{s}}^2$.

Finding the distance traveled by the train:

By using the 3rd equation of motion:

${\mathrm{v}}^2={\mathrm{u}}^2+2\mathrm{aS}$

Where u and v are the final and initial velocities, a is a constant acceleration, and S is the distance traveled.

$$\begin{gathered} \Rightarrow 0^2={25}^2+2(-0.5)\mathrm{S} \\ \Rightarrow 625=1\times \mathrm{S} \\ \Rightarrow \mathrm{S}=625\mathrm{m} \end{gathered}$$

Hence, the train will travel 625 m before it comes to a rest position.