a(t) = 6.60-2.29t
v(t) = 6.60t - 1.145t^2 + c
v(0) = 3.20, so c = 3.20, and
v(t) = 3.20 + 6.60t - 1.145t^2
s(t) = 3.20t + 3.30t^2 - 0.3817t^3 + c
s(0) = 7.30, so c = 7.30
s(t) = 7.30 + 3.20t + 3.30t^2 - 0.3817t^3
max speed occurs when a=0, or at t=2.882
v(2.882) = 12.7109
s(6.0) = 62.8528
avg speed is s(6)/6 = 10.4755
vmax - vavg = 2.235 m/s
If your answer agrees with this (should have shown your work), then we can be pretty sure the supposed answer is bogus or there's a typo in the problem.
If we disagree, double-check my math and yours.
A motorcyclist who is moving along an x-axis has an acceleration given by a=(6.60-2.29t) 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?
I tried finding the position and velocity functions and plugging in 6secs for both functions to find total distance and max speed, and then finding the average velocity, but when I subtracted the maximum speed and the average velocity I keep getting the wrong answers. The answer was supposed to be 3.45m/s, and I'm always off by a few numbers.
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