A motorcyclist who is moving along an x-axis has an acceleration given by a=(6.47-2.15t) for 0 >= t <= 6.0 s, where a is in m/s2 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?
Taking the integral of acceleration I got
v= -1.145t^2 + 6.60t +3.20
x= -.5725t^3 +3.3t^2 +3.20t + 7.30 (position function)
Not sure what to do from here.
To find the maximum speed of the motorcyclist, we need to find the maximum value of the velocity function v(t) over the time interval from t = 0 to t = 6.0 s.
First, let's find the derivative of the velocity function to find the critical points:
v'(t) = -2.29t + 6.60
Setting v'(t) = 0 to find the critical points:
-2.29t + 6.60 = 0
Solving for t, we get:
t = 6.60 / 2.29
t ≈ 2.88 s
Next, let's substitute this critical point into the velocity function to find the maximum speed:
v(2.88) = -1.145(2.88)^2 + 6.60(2.88) + 3.20
v(2.88) ≈ 11.28 m/s
So, the maximum speed of the motorcyclist is approximately 11.28 m/s.
Now, to find the average speed, we need to calculate the total displacement and divide it by the total time interval.
The total displacement is the difference between the final position and the initial position:
Δx = x(6.0) - x(0)
= [-(.5725)(6.0)^3 + 3.3(6.0)^2 + 3.20(6.0) + 7.30] - [-(.5725)(0)^3 + 3.3(0)^2 + 3.20(0) + 7.30]
= -19.58 m
The average speed is total displacement divided by the total time interval:
average speed = Δx / Δt
= -19.58 m / 6.0 s
≈ -3.26 m/s
The difference between the maximum speed and the average speed is:
11.28 m/s - (-3.26 m/s)
= 14.54 m/s
Therefore, the difference between the maximum speed and the average speed of the motorcyclist is approximately 14.54 m/s.