# College Physics 2.20 – Acceleration of an Olympic-class sprinter

## Solution:

We are given $\displaystyle \text{a}=4.50\:\text{m/s}^2,\:\text{t}=2.40\:\sec ,\:\text{and}\:\text{v}_0=0\:\text{m/s}$

### Part A

The unknown is vf. The formula in solving for vf is

$\displaystyle \text{v}_{\text{f}}=\text{v}_0+\text{at}$

Substituting the given values,

$\displaystyle \text{v}_{\text{f}}=0\:\text{m/s}+\left(4.50\:\text{m/s}^2\right)\left(2.40\:\text{s}\right)=108\:\text{m/s}$

### Part B

The relationship between position and time can be calculated using the formula

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

Then, with the given, we can express position in terms of time

$\displaystyle \text{x}=0+\frac{1}{2}\left(4.50\:\text{m/s}^2\right)\left(\text{t}^2\right)=2.52\text{t}^2$

The values of the position given the time is tabulated below

The values are plotted in the coordinate axes