Question

A train travels between two stations ½ mile apart in a minimum time of 41 sec. If the train accelerates and decelerates at 8 ft/sec^2, starting from rest at the first station and coming to a stop at the end of the station, what is its maximum speed in mph? how long does it travel at this top speed?

Answer 1

Answer on Question #61192, Physics / Mechanics | Relativity

A train travels between two stations $\frac{1}{2}$ mile apart in a minimum time of 41 sec. If the train accelerates and decelerates at 8 ft/sec^2, starting from rest at the first station and coming to a stop at the end of the station, what is its maximum speed in mph? how long does it travel at this top speed?

Solution:

The mile is an English unit of length of linear measure equal to 5,280 feet.

So, the halfway between two stations is

$$d_1 = \frac{1}{4} \text{ mile} = \frac{5280 \text{ ft}}{4} = 1320 \text{ ft}$$

Let's say that the train takes $t_1$ time to reach the max. speed $v$ and then it travels at this top speed distance $d_2$ at time $t_2$.

Use the kinematic equation

$$d_1 = \frac{a t_1^2}{2} + \frac{v t_2}{2}$$

The time is

$$t_1 + \frac{t_2}{2} = \frac{t}{2} = \frac{41 \text{ s}}{2} = 20.5 \text{ s}$$

The equation for speed is

$$v = a t_1$$

Thus, substituting in first equation

$$d_1 = \frac{a t_1^2}{2} + \frac{a t_1 t_2}{2}$$ $$1320 = \frac{8 t_1^2}{2} + \frac{8 t_1 (41 - 2 t_1)}{2}$$ $$330 = t_1^2 + 41 t_1 - 2 t_1^2$$ $$t_1^2 - 41 t_1 + 330 = 0$$ $$(t_1 - 30) (t_1 - 11) = 0$$

The physical solution is

$$t_1 = 11 \text{ s}$$

Hence,

$$v = a t_1 = 8 \cdot 11 = 88 \text{ ft/s}$$

1 Foot per Second = 0.681818 Miles per Hour

Thus,

$$v = 88 \cdot 0.681818 = 60 \text{ mph}$$

The distance that it travels at this top speed is

$$d_2 = v t_2 = v (41 - 2 t_1) = 88 (41 - 22) = 1672 \text{ ft}$$

Answer: 60 mph; 1672 ft.

http://www.AssignmentExpert.com