A football quarterback runs 15.0 m straight down the playing field in 2.50 s. He is then hit and pushed 3.00 m straight backward in 1.75 s. He breaks the tackle and runs straight forward another 21.0 m in 5.20 s. Calculate his average velocity
(a) for each of the three intervals and
(b) for the entire motion.
Solution:
Part A
The average velocity for each interval is computed using the formula
\overline{v}=\frac{\Delta x}{\Delta t}For the first interval
\overline{v_1}=\frac{15.0\:\text{meters}}{2.50\:\sec }=6.00\:\text{m/s} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)For the second interval
\overline{v_2}=\frac{-3.00\:\text{meters}}{1.75\:\sec }=-1.71\:\text{m/s} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)For the third interval
\overline{v_3}=\frac{21.0\:\text{m}}{5.20\:\text{sec}}=4.04\:\text{m/s} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)Part B
For the entire motion, we need displacement from the beginning to the end.
\begin{align*}
\overline{v}& =\frac{\Delta x}{\Delta t} \\
& =\frac{15\:\text{m}-3\:\text{m}+21\:\text{m}}{2.50\:\text{s}+1.75\:\text{s}+5.20\:\text{s}} \\
& =\frac{33\:\text{m}}{9.45\:\text{s}} \\
& =3.49\:\text{m/s} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
\end{align*}
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