A motorcycle is moving at 34 m/s when the rider applies the brakes, giving the motorcycle a constant deceleration. During the 4.0 s interval immediately after braking begins, the speed decreases to 12 m/s. What distance does the motorcycle travel from the instant braking begins until it comes to rest?
Ok first, we must find the acceleration of the motorcycle. Vf Vi + at 12 34 + a(4) -22 4a -5.5 m/s? a Then, we can find the distance of the motorcycle for we now know the acceleration of the motorcycle. (Vf)? (Vi)? + 2ad Vi 12 m/s Vf 0 [since the motorcycle comes to rest] (0)? (12)? + 2(-5.5)d Solving for d. d 13 1/11 meters Then the distance covered from 34 m/s to 12 m/s is d (Vi + Vf)/2 x t d (34 + 12)/2 x 4 d 92 m 13 1/11 + 92 105 1/11 meters Total distance covered!
34-12 is the change in acceleration you multiply that by the seconds and that tells you the distance in meters Think about canceling out the signs, (22 m/s)*4s this cancels the seconds and ends up with just meters
