Answer to Question #131073 in Thermal Power Engineering for Smith

Question

A train starts from rest at a station and accelerates at a rate of 2 m/s2
for 10 sec. It then runs at constant speed for 30 sec, decelerates at 4 m/s2
until it stops at the next station. Find the total distance covered.

Answer 1

The motion of train is divided into three parts.

In the first part, train accelerates with $a_1 = 2 \frac{m}{s^2}$ from rest for $t_1 = 10 s$ , hence the distance covered is $L_1 = \frac{a_1 t_1^2}{2} = 100 m$. At the end of this part, the train has speed $v_2 = a_1 t_1 = 20 \frac{m}{s}$.

In the second part, the train moves with constant speed $v_2$ for $t_2 = 30 s$, hence it covers $L_2 = v_2 t_2 = 600 m$.

In the third part, the train retards with $a_3 = -4 \frac{m}{s^2}$ from speed $v_2$ until full stop. For that, time $t_3 = \frac{v_2}{a_3}$ is required. The distance covered is $L_3 = v_2 t_3 - \frac{a_3 t_3^2}{2} = \frac{v_2^2}{2 a_3} = 50 m$.

Total distance covered is $L = L_1 + L_2 + L_3 = 750 m$.