# College Physics by Openstax, Problem 2.36

#### (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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