Answer to Question #131073 in Thermal Power Engineering for Smith

Question #131073
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.
1
Expert's answer
2020-09-04T05:55:41-0400

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".


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