Answer to Question #144674 in Mechanics | Relativity for Raphael Moy

Question

3. A subway train is at rest at station A
and must depart for station B. They are
1000 m apart from each other, and the
train can accelerate at no greater than
2 m/s2
(or less than −2 m/s2
). The

train must begin at rest (0 m/s) at sta-
tion A and end at rest at station B.

a. Design a subway trip consisting of three phases: 1) the train speeds up from rest, 2) the
train travels at a constant velocity, and 3) the train slows down to a stop. Your trip must
be exactly 1000 m in total displacement, and the magnitude of the train’s acceleration
cannot exceed 2 m/s2

(but you may use a smaller acceleration if you wish). You may
draw a graph, a diagram of the train itself, or simply provide calculations. It may help
to imagine your own experiences on a subway trip.

b. For each of the three phases of the train’s motion, how much time does each phase take
under your design? What is the maximum velocity of your train?

Answer 1

(a) The distances that the train accelerates and decelerates are the same, assume we consider motion from 0 to 40 m/s at 2 m/s²:

$$d=\frac{v^2}{2a}=\frac{40^2}{2\cdot2}=400\text{ m}.$$

This is the distance taken to accelerate and decelerate.

(b) Now, 200 m are left for constant speed motion, which will take time

$$t_c=\frac{200}{40}=5\text{ s}.$$

The time of accelerated/decelerated motion is

$$t_a=t_d=\frac{2\cdot400}{40}=10\text{ s}.$$