# College Physics 2.25 – Deceleration of a race runner

## Solution:

### Part A

For this part, we use the formula $\text{x}=\text{x}_0\text{t}+\text{v}_0\text{t}+\frac{1}{2}\text{a}\text{t}^2$.

$\text{x}=\text{x}_0\text{t}+\text{v}_0\text{t}+\frac{1}{2}\text{a}\text{t}^2$

$\text{x}=0\:\text{m}+\left(9.00\:\text{m/s}\right)\left(5.00\:\text{s}\right)+\frac{1}{2}\left(-2.00\:\text{m/s}^2\right)\left(5.00\:\text{s}\right)^2$

$\text{x}=20\:\text{meters}$

### Part B

The final velocity can be determined using the formula $\text{v}=\text{v}_0+\text{at}$

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

$\text{v}=9.00\:\text{m/s}+\left(-2.00\:\text{m/s}^2\right)\left(5.00\:\text{s}\right)$

$\text{v}=-1\:\text{m/s}$

### Part C

The result says that the runner starts at the rate of 9 m/s and decelerates at 2 m/s2. After some time, the velocity is already negative. This does not make sense because if the velocity is negative, that means that the runner is already running backwards.