Answer to Question #169846 in Physics for Boluwatife

Question

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

Answer 1

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

$$d_1=\dfrac{1}{2}(v_i+v_f)t=\dfrac{1}{2}\cdot(0+15\ \dfrac{m}{s})\cdot200\ s=1500\ m.$$

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

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

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

$$d_3=\dfrac{1}{2}(v_i+v_f)t=\dfrac{1}{2}\cdot(15\ \dfrac{m}{s}+0)\cdot150\ s=1125\ m.$$

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

$$d_{tot}=d_1+d_2+d_3=1500\ m+4500\ m+1125\ m=7125\ m.$$