7.1. The graph displays velocity over time, so the distance covered by "him" is equal to the unsigned (positive) area under the curve. (In contrast, the signed area represents displacement.) Finding this area is just an exercise in basic geometry.
• From time 0 to 3 s, the distance is equal to the area of a triangle with height 15 m/s and length 3 s:
1/2 (15 m/s) (3 s) = 22.5 m
• From 3 to 5.5 s, the distance is the area of a rectangle with height 15 m/s and length 5.5 s - 3 s = 2.5 s:
(15 m/s) (2.5 s) = 37.5 m
• From 5.5 to 6.5 s, you have a trapezoid with "bases" 15 m/s and 5 m/s, and "height" 6.5 s - 5.5 s = 1 s:
1/2 (15 m/s + 5 m/s) (1 s) = 10 m
• From 6.5 to 8 s, you have a triangle with height 5 m/s and length 8 s - 6.5 s = 1.5 s:
1/2 (5 m/s) (1.5 s) = 3.75 m
• From 8 to 9 s, another triangle with height 13 m/s and length 9 s - 8 s = 1 s:
1/2 (13 m/s) (1 s) = 6.5 m
• From 9 to 13 s, a rectangle with height 13 m/s and length 13 s - 9 s = 4 s:
(13 m/s) (4 s) = 52 m
• From 13 to 16.5 s, a triangle with height 13 m/s and length 16.5 s - 13 s = 3.5 s:
1/2 (13 m/s) (3.5 s) = 22.75 m
Add up the distances to get the total:
22.5 m + 37.5 m + 10 m + 3.75 m + 6.5 m + 52 m + 22.75 m = 155 m
7.2. The velocity is non-zero for any given time interval, so "he" is never at rest. (True, his velocity is 0 at 8 s, but only instantaneously.)
7.3. Given the plot of velocity, the acceleration is negative wherever the slope of the tangent line to the curve is negative. This happens in the interval from 5.5 to 9 s.
7.4. Similarly, positive acceleration corresponds to a positively-sloped tangent line. This happens from 0 to 3 s, and again from 13 to 16.5 s.
7.5. Where the velocity curve is horizontal, the accleration is zero, so you can ignore those intervals.
• From 0 to 3 s, the acceleration is
(15 m/s - 0 m/s)/(3 s - 0 s) = 5 m/s²
• From 5.5 to 6.5 s, it is
(5 m/s - 15 m/s)/(6.5 s - 5.5 s) = -10 m/s²
• From 6.5 to 8 s, it is
(0 m/s - 5 m/s)/(8 s - 6.5 s) ≈ -3.3 m/s²
• From 8 to 9 s, it is
(-13 m/s - 0 m/s)/(9 s - 8 s) = -13 m/s²
• From 13 to 16.5 s, it is
(0 m/s - (-13 m/s))/(16.5 s - 13 s) ≈ 3.7 m/s²
The clear winner is the interval from 8 to 9 s, where the acceleration has a magnitude of 13 m/s².
8. The magnitude of the velocity of the ball decreases until it reaches zero at its maximum height, then increases as it falls back down. Acceleration is constant and pointing downward the entire time.