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

a)
`v_(1)=14.29m/s\\v_(2)=9.25m/s\\v_(3)=6.36m/s`

b)
`v=+9.97m/s`

Step-by-step 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=(x_(2)-x_(1)  )/(t_(2)-x_(1)  )`

a) For the three intervals:

`v_(1)=(x_(2)-x_(1)  )/(t_(2)-t_(1)  )=((-3-15)m)/((1.74-3)s)=14.29m/s`

`v_(2)=(x_(3)-x_(2)  )/(t_(3)-t_(2)  )=((29-(-3))m)/((5.20-1.74)s)=9.25m/s`

`v_(3)=(x_(3)-x_(1)  )/(t_(3)-t_(1)  )=((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=(v_(1)+v_(2)+v_(3)   )/(n) =((14.29+9.25+6.36)m/s)/(3)=+9.97m/s`