# College Physics 2.14 – A Football Quarterback’s Velocity

## Solution:

### Part A

The average velocity for each interval is computed using the formula

$\displaystyle \text{v}=\frac{\text{distance}}{\text{time}}$

For the first interval

$\displaystyle \text{v}_1=\frac{15.0\:\text{meters}}{2.50\:\sec }=6.00\:\text{m/s}$

For the second interval

$\displaystyle \text{v}_2=\frac{-3.00\:\text{meters}}{1.75\:\sec }=-1.71\:\text{m/s}$

For the third interval

$\displaystyle \text{v}_3=\frac{21.0\:\text{m}}{5.20\:\text{sec}}=4.04\:\text{m/s}$

### Part B

For the entire motion, we need displacement from the beginning to the end.

$\displaystyle \text{v}=\frac{\text{displacement}}{\text{time}}$

$\displaystyle \text{v}=\frac{15\:\text{m}-3\:\text{m}+21\:\text{m}}{2.50\:\text{s}+1.75\:\text{s}+5.20\:\text{s}}$

$\displaystyle \text{v}=\frac{33\:\text{m}}{9.45\:\text{s}}$

$\displaystyle \text{v}=3.49\:\text{m/s}$