A drag racer, starting from rest, speeds up for 402 m with an acceleration of +20.0 m/s2. A parachute then opens, slowing the car down with an acceleration of -5.20 m/s2. How fast is the racer moving 3.75 102 m after the parachute opens?

Answer 1

First compute the maximum speed using
V(max) = sqrt(2aX)
a is the acceleration and x = 402 m.

After the chute opens (which I will call t=0), the distance moved is
X = Vmax*t - 2.6 t^2

Solve for t when X = 375 m.
Then use that t to get the velocity at that time.

V(t) = Vmax - 5.20 t