College Physics by Openstax Chapter 2 Problem 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/s2 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 the following: v0=22.0 m/sv_0=22.0\:\text{m/s}; a=−0.150 m/s2a=-0.150\:\text{m/s}^2; and Δx=210 m\Delta x=210\:\text{m}

We are required to solve for time, tt. We are going to use the formula

Δx=v0t+12at2\Delta x=v_0t+\frac{1}{2}at^2

Substituting the given values, we have

Δx=v0t+12at2210 m=(22.0 m/s)t+12(−0.150 m/s2)t2\begin{align*} \Delta x & =v_0t+\frac{1}{2}at^2 \\ 210\:\text{m} & =\left(22.0\:\text{m/s}\right)t+\frac{1}{2}\left(-0.150\:\text{m/s}^2\right)t^2 \end{align*}

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

0.075t2−0.22t+210=00.075t^2-0.22t+210=0

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

t=−b±b2−4ac2at=−(−22)±(−22)2−4(0.075)(210)2(0.075)\begin{align*} 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)}\\ \end{align*}

There are two values of tt that can satisfy the quadratic equation.

t=9.88 sandt=283.46 st=9.88\ \text{s} \qquad \text{and} \qquad t=283.46 \ \text{s}

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

t=9.88 s  (Answer)t=9.88\:\text{s} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)

Part B

We have the same given values from part a. We are going to solve vfv_f in the formula

(vf)2=(v0)2+2aΔx\left(v_f\right)^2=\left(v_0\right)^2+2a\Delta x

So we have

(vf)2=(v0)2+2aΔxvf=(v0)2+2aΔxvf=(22.0 m/s)2+2(−0.150 m/s2)(210 m)vf=20.6 m/s  (Answer)\begin{align*} \left(v_f\right)^2 & =\left(v_0\right)^2+2a\Delta x \\ v_f & =\sqrt{\left(v_0\right)^2+2a\Delta x} \\ v_f & =\sqrt{\left(22.0\:\text{m/s}\right)^2+2\left(-0.150\:\text{m/s}^2\right)\left(210\:\text{m}\right)} \\ v_f & =20.6\:\text{m/s} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \end{align*}

Part C

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

Δx=210 m+130 mΔx=340 m\begin{align*} \Delta x & =210\:\text{m}+130\:\text{m} \\ \Delta x & = 340 \ \text{m} \end{align*}

To solve for time tt, we are going to use the formula

Δx=vot+12at2\Delta x=v_ot+\frac{1}{2}at^2

If we rearrange the formula into a general quadratic equation and solve for tt using the quadratic formula, we come up with

t=−v0±v02+2axat=\frac{-v_0\pm \sqrt{v_0^2+2ax}}{a}

Substituting the given values:

t=−v0±v02+2axat=−(22.0 m/s)±(22.0 m/s)2+2(−0.150 m/s2)(340 m)−0.150 m/s2t=16.4 s  (Answer)\begin{align*} t & =\frac{-v_0\pm \sqrt{v_0^2+2ax}}{a} \\ t & =\frac{-\left(22.0\:\text{m/s}\right)\pm \sqrt{\left(22.0\:\text{m/s}\right)^2+2\left(-0.150\:\text{m/s}^2\right)\left(340\:\text{m}\right)}}{-0.150\:\text{m/s}^2} \\ t &=16.4\:\text{s} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \end{align*}

Part D

We have the same given values. We are going to solve for vfv_f in the equation

(vf)2=(v0)2+2aΔx\left(v_f\right)^2=\left(v_0\right)^2+2a\Delta x

Substituting the given values:

(vf)2=(v0)2+2aΔxvf=(v0)2+2aΔxvf=(22.0 m/s)2+2(−0.150 m/s2)(340 m)vf=19.5 m/s  (Answer)\begin{align*} \left(v_f\right)^2 & =\left(v_0\right)^2+2a\Delta x \\ v_f & =\sqrt{\left(v_0\right)^2+2a\Delta x} \\ v_f & =\sqrt{\left(22.0\:\text{m/s}\right)^2+2\left(-0.150\:\text{m/s}^2\right)\left(340\:\text{m}\right)} \\ v_f & =19.5\:\text{m/s} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \end{align*}
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