posted by John .
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
Determine (a) the total distance traveled during
the 8-sec interval; (b) the displacement during
the 8-sec interval.
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)