Physics: A 77.0 kg rider sitting on a 7.3 kg bike is riding along at 9.3 m/s in the positive direction. The rider drags a foot on the ground and slows down to 6.0 m/s still in the positive direction. A) What is the change in momentum of the rider and bike? B) What is the impulse delivered by the ground to the rider s foot? C) What force is acting on the bike and rider if slowing down took 13.2 seconds? D) And, how far did the bike and rider travel during these 13.2 seconds?

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A 77.0 kg rider sitting on a 7.3 kg bike is riding - id770735537738, floresvanexel4pejp1p

noahgampp

21.01.2022

Answer with Explanation:

We are given that

Mass of rider=m=77 kg

Mass of bike =m'=7.3 kg

Initial velocity,u=9.3 m/s

Final velocity,v=6 m/s

A.Change in velocity=v-u=6-9.3=-3.3 m/s

Total mass,M=m+m'=77+7.3=84.3 kg

Change in momentum= M(v-u)=84.3(-3.3)=-278.19 kgm/s

B.Impulse=Ft=Change in momentum=-278.19kg m/s

C.Time,t=13.2 s

v=u+at

Using the formula

6=9.3+13.2a

13.2a=6-9.3=-3.3

$a=-\frac{3.3}{13.2}=-0.25 ms^{-2}$

F=84.3(-0.25)=-21.075 N

D. $S=ut+\frac{1}{2}at^2$

$S=9.3(13.2)+\frac{1}{2}(-0.25)(13.2)^2$

S=100.98 m

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