A drag racer, starting from rest, speeds up for 402 m with an acceleration of +17 m/s^2. A parachute then opens, slowing the car down with an acceleration of -6.10 m/s^2. How fast us the racer moving 3.50*10^2 m after the parachute opens?

Answer 1

Use 1 of your kinematic equations twice. vfinal^2 = vinitial^2 + 2ad.

First you want to find the speed of the drag racer right when the parachute opens.

vfinal^2 = vinitial^2 + 2ad.
Substitute 0 for vinitial^2 and take the square root of 2(17)(402).

That becomes your initial velocity for when the parachute opens. Now you have to find the final velocity so sub what you get in for vfinal as vinitial^2 in

vfinal^2 = vinitial^2 + 2ad
vfinal^2 = vinitial^2 + 2(-6.10)(3.50x10^2).

Answer 2

64.4

Answer 3

96.94

Answer 4

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