5. The jogger's velocity is a constant 3.55 m/s between t = 4 s and t = 8 s.
6. Given a linear plot of velocity, the acceleration is determined by the slope of the line. Take any two points on the part of the plot after t = 8 s - for instance, we see it passes through (8 s, 3.5 m/s) and (10 s, 4 m/s) - and compute the slope:
(4 m/s - 3.5 m/s)/(10 s - 8 s) = (0.5 m/s)/(2 s) = 0.25 m/s^2
7. This amounts to finding the area between the velocity function and the time axis and between t = 4 s and t = 8 s. During this time, the velocity is 3.5 m/s. The time interval lasts 4 s. So the distance covered is
(3.5 m/s)*(4 s) = 14 m
8. After 4 seconds, Jimmy's speed decreases from 30.0 m/s to 27.2 m/s, so his acceleration (assuming it was constant) was
a = (27.2 m/s - 30.0 m/s)/(4 s) = -0.200 m/s^2
It's unclear what is meant by "rate of acceleration", since the acceleration is itself a rate. But maybe they just mean to ask for the acceleration, or possibly the magnitude?
