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?


Part (a). The final velocity can be solved using the formula v=v_0+at. We substitute the given values.


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


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




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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