Drag racing Physics

Question

A drag racer, starting from rest, speeds up for 401.0 m with an acceleration of +15.2 m/s². A parachute then opens, slowing the car down with an acceleration of -6.26 m/s². How fast is the racer moving 334.0 m after the parachute opens?

Answer 1

At the end of the 401.0 meters, the velocity of the car can be calculated from:

v_f² = v_i² + 2ax where v_f is that final velocity, v_i is the initial (starting) velocity which in this case is zero, "a" is the acceleration (15.2 m/s2), and "x" is the distance traveled (401.1 m):

v_f² = 0² + 2(15.2)(401.0)

v_f² = 12431

v_f = 111.494 m/s

During the second portion of travel, we can apply the same equation except v_i = 111.494 m/s, a = -6.26 m/s², and x = 334.0:

v_f² = 111.494² + 2(-6.26)(334.0)

v_f² = 8249.32

v_f = 90.8 m/s