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 \ \text{m/s}^2; t=8.00 \ \text{mins}; and v_0=4.00 \ \text{m/s}.
The final velocity can be solved using the formula v_f=v_0+at. We substitute the given values.
\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 t, we have
\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), \Delta x, or distance traveled is computed using the formula \Delta x=v_0 t+\frac{1}{2} at^2.
\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 \Delta x=\frac{v_f^2-v_0^2}{2 a}.
\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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