Question

An express train passes through a station. It enters with an initial velocity of 22.0 m/s and decelerates at a rate of 0.150 m/s^{2} as it goes through. The station is 210 m long. (a) How long is the nose of the train in the station? (b) How fast is it going when the nose leaves the station? (c) If the train is 130 m long, when does the end of the train leave the station? (d) What is the velocity of the end of the train as it leaves?

Answer 1

Answer:

a) $t \approx 9.879\,s$, b) $v = 20.518\,\frac{m}{s}$, c) $t = 16.368\,s$, d) $v = 19.545\,\frac{m}{s}$

Explanation:

a) Since train is only translating in a straight line and experimenting a constant deceleration throughout the station, whose length is 210 meters. The time required for the nose of the train to reach the end of the station can be found with the help of the following motion formula:

$210\,m = \left(22\,\frac{m}{s}\right) \cdot t + \frac{1}{2}\cdot \left(-0.150\,\frac{m}{s^{2}} \right) \cdot t^{2}$

The following second-order polynomial needs to be solved:

$-0.075\cdot t^{2} + 22\cdot t - 210 = 0$

Whose roots are presented herein:

$t_{1}\approx 283.455\,s$ and $t_{2} \approx 9.879\,s$

Both solutions are physically reasonable, although second roots describes better the braking process of the train.

b) The speed of the nose leaving the station is given by this expression:

$v = 22\,\frac{m}{s} + \left(-0.150\,\frac{m}{s^{2}}\right)\cdot (9.879\,s)$

$v = 20.518\,\frac{m}{s}$

c) First, it is required to calculate the time when nose of the train reaches a distance of 130 meters.

$130\,m = \left(22\,\frac{m}{s}\right) \cdot t + \frac{1}{2}\cdot \left(-0.150\,\frac{m}{s^{2}} \right) \cdot t^{2}$

$-0.075\cdot t^{2} + 22\cdot t - 130 = 0$

Roots of the second-order polynomial are:

$t_{1} \approx 287.300\,s$ and $t_{2} \approx 6.033\,s$

Both solutions are physically reasonable, although second roots describes better the braking process of the train. Now, the speed experimented by the train at this instant is:

$v = 22\,\frac{m}{s} + \left(-0.150\,\frac{m}{s^{2}}\right)\cdot (6.033\,s)$

$v = 21.095\,\frac{m}{s}$

The distance traveled by the end of the train throughout station is modelled after the following equation:

$210\,m = \left(21.095\,\frac{m}{s}\right) \cdot t + \frac{1}{2}\cdot \left(-0.150\,\frac{m}{s^{2}} \right) \cdot t^{2}$

$-0.075\cdot t^{2} + 21.095\cdot t - 210 = 0$

Roots of the second-order polynomial are:

$t_{1} \approx 270.932\,s$ and $t_{2} \approx 10.335\,s$

Both solutions are physically reasonable, although second roots describes better the braking process of the train. The instant when the end of the train leaves the station is:

$t = 6.033\,s + 10.335\,s$

$t = 16.368\,s$

d) The velocity experimented by the end of the train is:

$v = 21.095\,\frac{m}{s} + \left(-0.150\,\frac{m}{s^{2}} \right)\cdot (10.335\,s)$

$v = 19.545\,\frac{m}{s}$