Question

Two subway stops are separated by 1500 m. If a subway train accelerates at 1.2 m/s^2 from rest through the first half of the distance and decelerates at -1.2 m/s^2 through the second half, what are (a) its travel time and (b) its maximum speed?

Answer 1

Answer on Question #51380, Physics, Mechanics, Kinematics, Dynamics

Question:

Two subway stops are separated by 1500 m. If a subway train accelerates at 1.2 m/s^2 from rest through the first half of the distance and decelerates at -1.2 m/s^2 through the second half, what are (a) its travel time and (b) its maximum speed?

Answer:

a) Time of acceleration equals:

$$\frac{d}{2} = \frac{a t_a^2}{2}$$ $$t_a = \sqrt{\frac{d}{a}}$$

For deceleration $t_d = t_a$. Therefore, total time equals:

$$t = 2 t_a = 2 \sqrt{\frac{d}{a}} = 2 \sqrt{\frac{1500 \, m}{1.2 \, \frac{m}{s^2}}} = 70.7 \, s$$

b) Maximum speed equals:

$$v = a t_a = a \sqrt{\frac{d}{a}} = \sqrt{d a} = \sqrt{1500 \, m \, 1.2 \, \frac{m}{s^2}} = 42.4 \, \frac{m}{s}$$

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