A car traveling at 43 ft/sec decelerates at a constant 8 feet per second per second. How many feet does the car travel before coming to a complete stop? .

Answer 1

vo=43 ft/sec

a= -8 ft/s^2

v(t) = 43-8t = 0 when stopped

t = 43/8= 5 3/8 seconds to stop

s(t) = -4t^2 +43t= distance travelled in t seconds

s(43/8) = -4(43/8)^2 + 43(43/8) = 115.5625 feet