College Physics by Openstax, Problem 2.36

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?

SOLUTION:

Part a

We are given

v_0=22.0\:m/s

a=-0.150\:m/s^2

x=210\:m

We are required to solve for time, t. So we have

x=v_0t+\frac{1}{2}at^2

\left(210\:m\right)=\left(22.0\:m/s\right)t+\frac{1}{2}\left(-0.150\:m/s^2\right)t^2

If we rearrange the terms into a general quadratic equation, we have

0.075t^2-22t+210=0

Solve for t using the quadratic formula. We are given a=0.075;\:b=-22t;\:c=210

t=\frac{-b\pm \sqrt{b^2-4ac}}{2a}

t=\frac{-\left(-22\right)\pm \sqrt{\left(-22\right)^2-4\left(0.075\right)\left(210\right)}}{2\left(0.075\right)}

t=9.88\:s\:and\:t=283.46\:s

Discard t=283.46 s as it can be seen from the problem that this is not a feasible solution. So, we have

t=9.88\:s

Part b

We have the same given values from part a. So we have

v^2=v_0^2+2ax

v=\sqrt{v_0^2+2ax}

v=\sqrt{\left(22.0\:m/s\right)^2+2\left(-0.150\:m/s^2\right)\left(210\:m\right)}

v=20.6\:m/s

Part c

We are given the same values as in the previous parts, except that for the value of x since we should incorporate the length of the train. For the distance, x, we have

x=210\:m+130\:m

x=340\:m

Therefore, we have

x=v_ot+\frac{1}{2}at^2

You can solve this problem by the method presented in part a. But we can also rearrange the equation to solve for t, using also quadratic equation. We have

t=\frac{-v_0\pm \sqrt{v_0^2+2ax}}{a}

t=\frac{-\left(22.0\:m/s\right)\pm \sqrt{\left(22.0\:m/s\right)^2+2\left(-0.150\:m/s^2\right)\left(340\:m\right)}}{-0.150\:m/s^2}

t=16.4\:s

Part d

We have the same given values.

v^2=v_0^2+2ax

v=\sqrt{\left(22.0\:m/s\right)^2+2\left(-0.150\:m/s^2\right)\left(340\:m\right)}

v=19.5\:m/s

 

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