(021) The acceleration at 1 second is the slope of the line plotted over the first 2 seconds. This line passes through the points (0 s, 0 m/s) and (2 s, 5 m/s), so its slope and thus the (average) acceleration is
(5 m/s - 0 m/s) / (2 s - 0 s) = 5/2 m/s² = 2.5 m/s²
(022) At time 2 seconds, the velocity curve passes through the point (2 s, 5 m/s), so the velocity is 5 m/s
(023) The distance traveled by this object after 2 seconds is equal to the area under the velocity function over the first 2 seconds. This region is a triangle with base 2 s and height 5 m/s, so the area is
1/2 (2 s) (5 m/s) = 5 m
The object's initial position is 10 m, so its final position after 2 seconds is
10 m + 5 m = 15 m
(024) Similar to (021): compute the slope of the line connecting the point (2 s, 5 m/s) and (6 s, 7 m/s):
(7 m/s - 5 m/s) / (6 s - 2 s) = (2 m/s) / (4 s) = 1/2 m/s² = 0.5 m/s²
(025) We know the distance traveled in the first 2 seconds. Over the next 4 seconds, the object travels an additional distance equal to the area of a trapezoid with "height" 6 s - 2 s = 4 s, and "bases" 5 m/s and 7 m/s. The area of this trapezoid is
1/2 (4 s) (5 m/s + 7 m/s) = 24 m
The net distance traveled after 6 s is then
5 m + 24 m = 29 m
so the object's position after 6 s is
10 m + 29 m = 39 m
(026) Again, compute the slope of a line, this time through the point (6 s, 0 m/s) and (9 s, -1 m/s) :
(-1 m/s - 0 m/s) / (9 s - 6 s) = (-1 m/s) / (3 s) ≈ -0.333 m/s²
(027) If the velocity at 6 s is 0 m/s, and after each second the velocity decreases by 0.333 m/s, then after 2 more seconds the velocity would be
0 m/s + (2 s) (-0.333 m/s²) = -0.666 m/s
(028) The distance traveled between the 6th second and 8th second corresponds to the area of another trapezoid, with "height" 8 s - 6 s = 2s and "bases" 0 m/s and 0.666 m/s - 0 m/s = 0.666 m/s. So the distance traveled in this interval is
1/2 (2 s) (0 m/s + 0.666 m/s) = 0.666 m
However, the velocity is negative for this duration, so the object has turned around and is moving in the opposite direction. This means it covers a net distance after a total of 8 s of
5 m + 24 m - 0.666 m = 28.333 m
and so its position after 8 s is
10 m + 28.333 m = 38.333 m
