Question #169846

A train accelerates uniformly from rest to reach 54km/h in 200 seconds after which the speed remains constant for 300 seconds. At the end of this time the train decelerates to rest in 150 seconds.find the total distance travelled


1
Expert's answer
2021-03-08T08:11:02-0500

Let's first find the distance traveled by the train when it accelerates:


d1=12(vi+vf)t=12(0+15 ms)200 s=1500 m.d_1=\dfrac{1}{2}(v_i+v_f)t=\dfrac{1}{2}\cdot(0+15\ \dfrac{m}{s})\cdot200\ s=1500\ m.

Then, let's find the distance traveled by the train when it moves with constant speed:


d2=vt=15 ms300 s=4500 m.d_2=vt=15\ \dfrac{m}{s}\cdot300\ s=4500\ m.

Let's find the distance traveled by the train when it decelerates:


d3=12(vi+vf)t=12(15 ms+0)150 s=1125 m.d_3=\dfrac{1}{2}(v_i+v_f)t=\dfrac{1}{2}\cdot(15\ \dfrac{m}{s}+0)\cdot150\ s=1125\ m.

Finally, we can find the total distance traveled by the train:


dtot=d1+d2+d3=1500 m+4500 m+1125 m=7125 m.d_{tot}=d_1+d_2+d_3=1500\ m+4500\ m+1125\ m=7125\ m.

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