College Physics by Openstax Chapter 2 Problem 29

Freight trains can produce only relatively small accelerations and decelerations.

(a) What is the final velocity of a freight train that accelerates at a rate of 0.0500 m/s2 for 8.00 minutes, starting with an initial velocity of 4.00 m/s?

(b) If the train can slow down at a rate of 0.550 m/s2, how long will it take to come to a stop from this velocity?

(c) How far will it travel in each case?

Solution:

Part A

We are given the the following: a=0.0500 m/s2a=0.0500 \ \text{m/s}^2; t=8.00 minst=8.00 \ \text{mins}; and v0=4.00 m/sv_0=4.00 \ \text{m/s}.

The final velocity can be solved using the formula vf=v0+atv_f=v_0+at. We substitute the given values.

vf=v0+atvf=4.00 m/s+(0.0500 m/s2)(8.00 min×60 sec⁡1 min⁡)vf=28.0 m/s  (Answer)\begin{align*} v_f& = v_0+at \\ v_f & = 4.00\:\text{m/s}+\left(0.0500\:\text{m/s}^2\right)\left(8.00\:\text{min}\times \frac{60\:\sec }{1\:\min }\right) \\ v_f & = 28.0 \ \text{m/s} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \end{align*}

Part B

Rearrange the equation we used in part (a) by solving in terms of tt, we have

t=vf−v0at=0 m/s−28 m/s−0.550 m/s2t=50.91 sec⁡  (Answer)\begin{align*} t & =\frac{{v_f}-v_0}{a} \\ t & = \frac{0\:\text{m/s}-28\:\text{m/s}}{-0.550\:\text{m/s}^2} \\ t & = 50.91\:\sec\ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \end{align*}

Part C

The change in position for part (a), Δx \Delta x, or distance traveled is computed using the formula Δx=v0t+12at2 \Delta x=v_0 t+\frac{1}{2} at^2.

Δx=v0t+12at2Δx=(4.0 m/s)(480 s)+12(0.0500 m/s2)(480 s)2Δx=7680 m  (Answer)\begin{align*} \Delta x & =v_0 t+\frac{1}{2} at^2 \\ \Delta x & =\left(4.0\:\text{m/s}\right)\left(480\:\text{s}\right)+\frac{1}{2}\left(0.0500\:\text{m/s}^2\right)\left(480\:\text{s}\right)^2 \\ \Delta x & = 7680\:\text{m} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \end{align*}

For the situation in part (b), the distance traveled is computed using the formula Δx=vf2−v022a\Delta x=\frac{v_f^2-v_0^2}{2 a}.

Δx=(0 m/s)2−(28.0 m/s)22(−0.550 m/s2)Δx=712.73 m  (Answer)\begin{align*} \Delta x & =\frac{\left(0\:\text{m/s}\right)^2-\left(28.0\:\text{m/s}\right)^2}{2\left(-0.550\:\text{m/s}^2\right)} \\ \Delta x & =712.73\:\text{m} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \end{align*}
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