**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

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