A bicycle racer sprints at the end of a race to clinch a victory. The racer has an initial velocity of 12.5 m/s and accelerates at the rate of 0.450 m/s2 for 7.00 s.

Question

(a) What is his final velocity?

(b) The racer continues at this velocity to the finish line. If he was 300 m from the finish line when he started to accelerate, how much time did he save?

(c) One other racer was 5.00 m ahead when the winner started to accelerate, but he was unable to accelerate and traveled at 12.8 m/s until the finish line. How far ahead of him (in meters and in seconds) did the winner finish?
distance

Answer 1

a. V1 = Vo + a*t=12.5 + 0.450*7=15.7m/s.
d1 = 12.5*7 + 0.225*7^2 = 98.5 m.

b. d2 = V1*T2 = 300-98.5 = 201.5
15.7*T2 = 201.5
T2 = 12.8 s.
T1+T2 = 7 + 12.8 = 19.8 s. To travel 300 m with acceleration.

d = Vo*T3 = 300 m.
12.5*T3 = 300.
T3 = 24 s. To travel 300 m without accelerating.

T = T3-(T1+T2) = 24 - 19.8 = 4.2s Saved.

c. d = V*T = 300-5 = 295 m.
12.8 * T = 295.
T = 23 s. To finish.

23 - 19.8 = 3.2 s. Behind the winner.

300m/19.8s * 3.2s. = 48.5 m. Behind the
winner.

Answer 2

THIS IS A CORRECTION FOR PART C, WHEN YOU ARE SUPPOSED TO CONVERT SECONDS TO METERS!!! THE REST IS CORRECT THOUGH.

c.

12.8 m/s * 3.2s = 40.96 (NOT 48.5)

Answer 3

woops nvm, I was looking at another problem.

Answer 4

Actually Jasmine was correct on her Correction. the answer for c. os 40.96m