Question

A football wide receiver rushes 16 m straight down the playing field in 2.9 s (in the positive direction). He is then hit and pushed 2.5 m straight backwards in 1.65 s. He breaks the tackle and runs straight forward another 24 m in 5.2 s.Part (a) Calculate the wide receiver's average velocity in the horizontal direction during the first interval, in meters per second.Part (b) Calculate the wide receiver's average velocity in the horizontal direction during the second interval, in meters per second.Part (c) Calculate the wide receiver's average velocity in the horizontal direction during the third interval, in meters per second.Part (d) Calculate the wide receiver's average velocity in the horizontal direction for the entire motion, in meters per second.

Answer 1

Answer:

a) $v_1=5.5172\ m.s^{-1}$

b) $v_2=-1.5152\ m.s^{-1}$

c) $v_3=4.6154\ m.s^{-1}$

d) $v_{avg}=3.8462\ m.s^{-1}$

Explanation:

Given:

• distance down the field in the first interval, $d_1=16\ m$

• time duration of the first interval, $t_1=2.9\ s$

• distance down the field in the second interval, $d_2=-2.5\ m$

• time duration of the second interval, $t_2=1.65\ s$

• distance down the field in the third interval, $d_3=24\ m$

• time duration of the third interval, $t_3=5.2\ s$

a)

velocity in the first interval:

$v_1=\frac{d_1}{t_1}$

$v_1=\frac{16}{2.9}$

$v_1=5.5172\ m.s^{-1}$

b)

velocity in the second interval:

$v_2=\frac{d_2}{t_2}$

$v_2=\frac{-2.5}{1.65}$

$v_2=-1.5152\ m.s^{-1}$

c)

velocity in the third interval:

$v_3=\frac{d_3}{t_3}$

$v_3=\frac{24}{5.2}$

$v_3=4.6154\ m.s^{-1}$

d)

We know that the average velocity is given as the total displacement per unit time.

$v_{avg}=\frac{d_1+d_2+d_3}{t_1+t_2+t_3}$

$v_{avg}=\frac{16-2.5+24}{2.9+1.65+5.2}$

$v_{avg}=3.8462\ m.s^{-1}$