Question

A football quarterback runs 15.0 m straight down the playing field in 3.00 s. He is then hit and pushed 3.00 m straight backward in 1.74 s. He breaks the tackle and runs straight forward another 29.0 m in 5.20 s. (a) Calculate his average velocity (in m/s) for each of the three intervals. (Assume the quarterback's initial direction is positive. Indicate the direction with the sign of your answer.)
v1= m/s, v2= m/s, v3= m/s
(b) Calculate his average velocity (in m/s) for the entire motion. (Assume the quarterback's initial direction is positive. Indicate the direction with the sign of your answer.)
m/s

Answer 1

Answer:

a) $v_{1}=14.29m/s\\v_{2}=9.25m/s\\v_{3}=6.36m/s$

b) $v=+9.97m/s$

Explanation:

From the exercise we know that

$x_{1} =15m, t_{1}=3s$

$x_{2} =-3m, t_{1}=1.74s$

$x_{3} =29m, t_{3}=5.20s$

From dynamics we know that the formula for average velocity is:

$v=\frac{x_{2}-x_{1} }{t_{2}-x_{1} }$

a) For the three intervals:

$v_{1}=\frac{x_{2}-x_{1} }{t_{2}-t_{1} }=\frac{(-3-15)m}{(1.74-3)s}=14.29m/s$

$v_{2}=\frac{x_{3}-x_{2} }{t_{3}-t_{2} }=\frac{(29-(-3))m}{(5.20-1.74)s}=9.25m/s$

$v_{3}=\frac{x_{3}-x_{1} }{t_{3}-t_{1} }=\frac{(29-15)m}{(5.20-3)s}=6.36m/s$

b) The average velocity for the entire motion can be calculate by the following formula:

$v=\frac{v_{1}+v_{2}+v_{3} }{n} =\frac{(14.29+9.25+6.36)m/s}{3}=+9.97m/s$