# College Physics 2.23 – Acceleration of a train

## Solution:

### Part A

We use the formula $\displaystyle \text{t}=\frac{\text{v}-\text{v}_0}{\text{a}}$

We need to convert 80.0 km/h to m/s first so that we have a unit uniformity for all the given values. $\displaystyle 80\:\text{km/hr}=\left(80\:\text{km/hr}\right)\left(\frac{1000\:\text{m}}{1\:\text{km}}\right)\left(\frac{1\:\text{hr}}{3600\:\text{s}}\right)=22.2222\:\text{m/s}$

Substituting the given values into the formula, we have $\displaystyle \text{t}=\frac{22.2222\:\text{m/s}-0\:\text{m/s}}{1.35\:\text{m/s}^2}=16.5\:\text{s}$

### Part B

For this problem, we still use the formula used in Part (a). This time, the values of the final and initial velocities interchange and the value of the given deceleration is negative of acceleration. That is, $\displaystyle \text{t}=\frac{\text{v}-\text{v}_0}{\text{a}}=\frac{0\:\text{m/s}-22.2222\:\text{m/s}}{-1.65\:\text{m/s}^2}=13.5\:\text{s}$

### Part C

For this problem, we will use the formula $\displaystyle \text{a}=\frac{\text{v}-\text{v}_0}{\text{t}}$

Substituting all the given values into the formula, we have $\displaystyle \text{a}=\frac{0\:\text{m/s}-22.2222\:\text{m/s}}{8.30\:\text{s}}=2.68\:\text{m/s}^2$