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December 20, 2014

December 20, 2014

Posted by **John** on Friday, January 21, 2011 at 3:40pm.

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.

- physics -
**drwls**, Friday, January 21, 2011 at 4:47pmAt 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)

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