College Physics Problem 2.29

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/s^2 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/s, 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 v=v_0+at. We substitute the given values.

     v=v_0+at

     v=4.00\:m/s+\left(0.0500\:m/s^2\right)\left(8.00\:min\:\times \frac{60\:sec}{1\:minute}\right)

     v=28.0\:m/s

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

     t=\frac{v-v_0}{a}

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

     t=50.91\:sec

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

     \Delta x=\left(4.0\:m/s\right)\left(480\:s\right)+\frac{1}{2}\left(0.0500\:m/s^2\right)\left(480\:s\right)^2

     \Delta x=7680\:m

For the situation in part (b), the distance traveled is computed using the formula \Delta x=\frac{v^2-v_0^2}{2a}

     \Delta x=\frac{\left(0\:m/s\right)^2-\left(28.0\:m/s\right)^2}{2\left(-0.550\:m/s^2\right)}

     \Delta x=712.73\:m

    

    

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