Question

A train starting from rest attains a velocity of $72\mathrm{km}{\mathrm{h}}^{-1}$ in $5$ minutes. Assuming that the acceleration is uniform. Find the distance traveled by train for attaining this velocity.

Answer 1

Step 1: Given information

1. A train starts moving from rest. So initial velocity$u=0$.
2. The train moves for the time of $5min$=$5\times 60=300s$.
3. The train attains a velocity of $72km{h}^{-1}$.
4. So final velocity$v=72km/h=72\times \frac{1000}{3600}m/s$ =$72\times \frac{5}{18}=20m/s$.

Step 2: Calculating distance traveled by train

Here,

$u=0$

$v=20m/s$

$t=300s$

So, acceleration $a=\frac{v-u}{t}$$\Rightarrow a=\frac{20}{300}=\frac{1}{15}m/s^2$

Now,

$v^2-u^2=2as$

$\Rightarrow {20}^2-0=2\times \frac{1}{15}\times s$

$\Rightarrow 400=2\times \frac{1}{15}\times s$

$\Rightarrow s=\frac{400\times 15}{2}=200\times 15$ m

$\Rightarrow s=3000m=3km$

Hence, the distance covered by the train is $3km$.