A motorcyclist who is moving along an x- axis has an acceleration given by a = (6.60 - 2.29t) for 0 =< t =< 6.0s

time-interval
velocity-position

#1

A motorcyclist who is moving along an x- axis has an acceleration given by a = (6.60 - 2.29t) for 0 =< t =< 6.0s, where a is in m/s^2 and t is in seconds. At t = 0, the velocity and position of the cyclist are 3.20 m/s and 7.30 m. For the time interval from 0 =< t =< 6.0 s, what is the difference between the maximum speed and the average speed of the motorcyclist? The answer is 3.45 m/s but can someone post the work for me? thanks!

Answer:

I hope you know integration and differentiation.

a=dv/dt (acceleration is rate of change of velocity)
Integrating both sides

image

Now, acceleration is actually decreasing with time and soon becomes negative.
So, max velocity is attained at the point where a = 0.
so equating a = 0 = 6.6 - 2.29 t
t=2.88s
Just put this value of time in the velocity equation, you will get a max velocity of 12.71092 m/s

Average speed = Total distance / Total time
total time = 6 s

Use the position equation to find its position at the end of t=6 s
it will be 62.86m.
Total distance = 62.86m - 7.3m = 55.56m (bcoz he started from initial position of 7.3m not 0m)
average speed = 55.56m / 6s = 9.26 m/s

Thus, difference in the max velocity and average speed = 3.45m/s