A car traveling at 40 ft/sec decelerates at a constant 7 feet per second per second. How many feet does the car travel before coming to a complete stop?


2 Answers

Oddman Profile
Oddman answered
It takes the car (40 ft/s)/(7 ft/s^2) = 5 5/7 s to stop. In that time, it will have traveled
  (1/2)*(5 5/7 s)*(40 ft/s) = 114 2/7 ft
Consider a plot of velocity versus time. It starts at 40 ft/s and decreases linearly to 0 with a slope of -7 ft/s*2. The velocity will reach 0 at 5 5/7 seconds. The area of the triangle is the distance traveled. It is computed as 1/2*base*height = 1/2*(5 5/7 s)*(40 ft/s).
Richard Himmeler Profile

Vi = 40FT/SEC
a= -7FT/SEC2
vf = 0
we need  S
(vi)2 - (vf)2= 2as
(40)2-(0)2= -2*7*s
s= 1600/14= 114.28 feet, befor ethe vcar stops( ignore the minus sign of decelaration)
thanked the writer.
Oddman commented
If you write the formula correctly as vf^2 - vi^2 = 2as, then the sign of the acceleration is not ignored.

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