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 a1=2ms2a_1 = 2 \frac{m}{s^2} from rest for t1=10st_1 = 10 s , hence the distance covered is L1=a1t122=100mL_1 = \frac{a_1 t_1^2}{2} = 100 m. At the end of this part, the train has speed v2=a1t1=20msv_2 = a_1 t_1 = 20 \frac{m}{s}.

In the second part, the train moves with constant speed v2v_2 for t2=30st_2 = 30 s, hence it covers L2=v2t2=600mL_2 = v_2 t_2 = 600 m.

In the third part, the train retards with a3=4ms2a_3 = -4 \frac{m}{s^2} from speed v2v_2 until full stop. For that, time t3=v2a3t_3 = \frac{v_2}{a_3} is required. The distance covered is L3=v2t3a3t322=v222a3=50mL_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=L1+L2+L3=750mL = L_1 + L_2 + L_3 = 750 m.


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