Answer to Question #180056 in Physics for jomari galdiano

Question

 subway train starts from rest at a station and accelerates at a rate of 1.60m/s2 for 14s. It 

runs at a constant speed for 70s and slows down at a rate of 3.50m/s2 until it stops at the

next station. Find the total distance covered by the train

Answer 1

Let's first find the velocity of the subway train after 14 s:

$$v=v_0+at=0+1.6\ \dfrac{m}{s^2}\cdot14\ s=22.4\ \dfrac{m}{s}.$$

Then, we can find the distance covered by the subway train during these 14 s:

$$d_1=v_0t+\dfrac{1}{2}at^2,$$ $$d_1=0+\dfrac{1}{2}\cdot1.6\ \dfrac{m}{s^2}\cdot(14\ s)^2=156.8\ m.$$

Let's find the distance covered by the subway train during the next 70 s:

$$d_2=vt=22.4\ \dfrac{m}{s}\cdot70\ s=1568\ m.$$

Let's find the time that the subway train takes to stop at the next station:

$$v=v_0+at,$$ $$t=\dfrac{v-v_0}{a}=\dfrac{0-22.4\ \dfrac{m}{s}}{-3.50\ \dfrac{m}{s^2}}=6.4\ s.$$

Then, we can find the distance covered by the subway train during the last 6.4 s:

$$d_3=\dfrac{1}{2}(v+v_0)t,$$ $$d_3=\dfrac{1}{2}\cdot(0+22.4\ \dfrac{m}{s})\cdot6.4\ s=71.68\ m.$$

Finally, we can find the total distance covered by the subway train:

$$d=d_1+d_2+d_3,$$ $$d=156.8\ m+1568\ m+71.68\ m=1796.48\ m.$$