College Physics 2.29 – Acceleration of freight trains

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 $\Delta \text{x}=\frac{\text{v}^2-\text{v}_0^2}{2\text{a}}$

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

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