College Physics by Openstax Chapter 2 Problem 14


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

v=ΔxΔt\overline{v}=\frac{\Delta x}{\Delta t}

For the first interval

v1=15.0meters2.50sec=6.00m/s  (Answer) \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

v2=3.00meters1.75sec=1.71m/s  (Answer)\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

v3=21.0m5.20sec=4.04m/s  (Answer)\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.

v=ΔxΔt=15m3m+21m2.50s+1.75s+5.20s=33m9.45s=3.49m/s  (Answer)\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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