Question

A point moving on a straight horizontal
line with an initial velocity of 60 fps to the right
is given an acceleration of 12 fps2 to the left for
8 sec.
Determine (a) the total distance traveled during
the 8-sec interval; (b) the displacement during
the 8-sec interval.

Answer 1

At the beginning of the 8 s interval, the velocity is 60 ft/s.

At the end of the 8s interval, the velocity is 60 - 12*8 = -36 ft/s

The average velocity during the interval is (60-36)/2 = 12 ft/s
You can multiply that by 8 s for the total displacement.

(a) X = 60 t - 6 t^2
Maximum X value occurs @ dX/dt = 0, where t = 5 s. The value of X at that time is 300 - 150 = 150 ft. Then it goes backwards until t = 8, when
X(final) = 480 - 384 = 96 ft.

Total distance traveled, regardless of direction, is 150 + 96 = 246 ft

(b) Displacement from starting position = 96 feet. We already found that out part (a)