A car braked with a constant deceleration of 16ft/s^2, producing a marks measuring 200ft before coming to a stop.
How fast was the car traveling when the brakes were first applied?
Let the initial speed be v0, then under braking we have:
v' = -16 ft/s^2
so:
v = -16 t + v0
and so the vehicle comes to rest at t=v0/16 s.
Now let x be the distance travelled since braking started then:
x' = v = -16 t + v0
so:
x = -8 t^2 + v0 t + x0,
but x0=0 here so:
x = -8 t^2 + v0 t
As the vehicle stops after 200ft we know that:
200 = -8 (v0/16)^2 + v0 (v0/16)
simplifying gives v0^2 = 6400, and so v=80 ft/s
RonL