A drag racer, starting from rest, speeds up for 402 m with an acceleration of +18.0 m/s2. A parachute then opens, slowing the car down with an acceleration of -6.30 m/s^2. How fast is the racer moving 2.70 x 10^2 m after the parachute opens?

Answer 1

find the velocity after 402m

vf^2=vi^2+2ad solve for Vf
Then, using that velocity for the initial velocity vi for the deacceleration time,

Vf^2=Vi^2+2ad where a=-6.30m/s^2 and d=2700m