College Physics 2.29 – Acceleration of freight trains


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

The final velocity can be solved using the formula \text{v}=\text{v}_0+\text{at}. We substitute the given values.

\text{v}=\text{v}_0+\text{at}

\displaystyle \text{v}=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)

\text{v}=28.0\:\text{m/s}

Part B

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

      \displaystyle \text{t}=\frac{\text{v}-\text{v}_0}{\text{a}}

      \displaystyle \text{t}=\frac{0\:\text{m/s}-28\:\text{m/s}}{-0.550\:\text{m/s}^2}

      \text{t}=50.91\:\sec

Part C

The change in position for part (a), \Delta \text{x}, or distance traveled is computed using the formula \Delta \text{x}=\text{v}_0\text{t}+\frac{1}{2}\text{at}^2

     \Delta \text{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 \text{x}=7680\:\text{m}

For the situation in part (b), the distance traveled is computed using the formula \Delta \text{x}=\frac{\text{v}^2-\text{v}_0^2}{2\text{a}}

     \displaystyle \Delta \text{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 \text{x}=712.73\:\text{m}