# Two subway stops are separated by 1200 m. If a subway train accelerates at 1.4 m/s^2 from rest through the first half of the distance and decelerates at -1.4 m/s^2 through the second half, what are (a) its travel time and (b) its maximum speed ?

Sep 8, 2015

(a). $58.6 \text{s}$

(b). $41 \text{m/s}$

#### Explanation:

For the 1st part of the journey:

${v}^{2} = {u}^{2} + 2 a s$

${v}^{2} = 0 + \left(2 \times 1.4 \times 600\right)$

${v}^{2} = 1680$

$v = 41 \text{m/s}$

This gives the answer to (b) as it is the maximum speed.

To get the time for the 1st part:

$v = u + a t$

$41 = 0 + 1.4 t$

$t = \frac{41}{1.4} = 29.3 \text{s}$

The 2nd part of the journey is a mirror image of the 1st part so total time $= 29.3 \times 2 = 58.6 \text{s}$ which is the answer to (a).

Sep 8, 2015

Travel time: $\text{58.5 s}$
Maximum speed: $\text{41 m/s}$

#### Explanation:

The idea here is that you need to determine the time and velocity of the train for the first half of the distance.

Since the train is starting its motion from rest, you can say that

$\frac{d}{2} = {\underbrace{{v}_{0}}}_{\textcolor{b l u e}{= 0}} \cdot t + \frac{1}{2} \cdot a \cdot {t}^{2} \text{ }$, where

$d$ - the distance between the two stops;

This means that you have

t^2 = d/a implies t = sqrt(d/a) = sqrt((1200color(red)(cancel(color(black)("m"))))/(1.4color(red)(cancel(color(black)("m")))/"s"^2)) = "29.3 s"

The speed of the train right before it starts to decelerate, which of course will be its maximum speed, is

$v = {\underbrace{{v}_{0}}}_{\textcolor{b l u e}{= 0}} + a \cdot t$

v = 1.4"m"/"s"^color(red)(cancel(color(black)(2))) * 29.3color(red)(cancel(color(black)("s"))) = color(green)("41 m/s")

The train decelerates at a rate equal to that with which it accelerated, and the distance is the same in both cases, which means that the time it takes for the train to cover the second half of the distance is

${t}_{2} = t = \text{29.3 s}$

The total time of travel will be

t_"total" = 2 * t = 2 * "29.3 s" = color(green)("58.6 s")