Question

A train accelerated from 54 km/h to 90km/h in 20 seconds. How much
distance does it cover in this period? Assume that the tracks are straight.

Answer 1

• Initial velocity=u=54km/h
• Final velocity=v=90km/h
• Time=t=20s
• Acceleration=a
• Distance=s=?

First convert velocities to m/s

$\\ \LARGE\sf\longmapsto u=54\times \dfrac{5}{18}=15m/s$

$\\ \LARGE\sf\longmapsto v=90\times \dfrac{5}{18}=25m/s$

We know

$\boxed{\LARGE{\sf Acceleration=\dfrac{v-u}{t}}}$

$\\ \LARGE\sf\longmapsto Acceleration=\dfrac{25-15}{20}$

$\\ \LARGE\sf\longmapsto Acceleration=\dfrac{10}{20}$

$\\ \LARGE\sf\longmapsto Acceleration=0.5m/s^2$

Using 2nd equation of kinematics

$\boxed{\LARGE{\sf s=ut+\dfrac{1}{2}at^2}}$

$\\ \LARGE\sf\longmapsto s=15(20)+\dfrac{1}{2}(0.5)(20)^2$

$\\ \LARGE\sf\longmapsto s=300+\dfrac{1}{2}(0.5)(400)$

$\\ \LARGE\sf\longmapsto s=300+(0.5)(200)$

$\\ \LARGE\sf\longmapsto s=300+100$

$\\ \LARGE\sf\longmapsto s= 400m$

Hence

Distance travelled by the train is 400m.