A car is traveling at 50 mi/h when the brakes are fully applied, producing a constant deceleration of 22 ft/s^2.
what is the distant traveled before the car comes to a stop?
v' = -22 ft/s^2,
so:
v = -22t +v0
at t=0 v=50*5280/60^2 ~= 73.33 ft/s
Hence:
v = -22 t + 73.33
and the vehicle comes to a rest when v=0, that is when:
-22 t + 73.33 = 0,
or t = 3.33 s
now if x is the distanced travelled since breaking started then:
x' = v = -22 t + 73.33,
so:
x = -11 t^2 +73.33 t + x0,
but x0=0, and at t= 3.33 s:
x = 122.21 ft
RonL