Question

A cyclist sprints at the end of a race to clinch a victory. She has an initial velocity of 11.3 m/s and accelerates at a rate of 0.610 m/s for 7.85 s. (a) What is her final velocity? (Enter the magnitude in m/s.) 16.089 m/s (b) The cyclist continues at this velocity to the finish line. If she is 300 m from the finish line when she starts to accelerate, how much time did she save (in s)? (c) The second-place winner was 5.00 m ahead when the winner started to accelerate, but he was unable to accelerate, and traveled at 11.6 m/s until the finish line. What was the difference in finish time in seconds between the winner and runner-up? S How far back (in m) was the runner-up when the winner crossed the finish line? m

Answer 1

a) deitial relority of the cyclist, U= 13 m/s acceleration - 0't mis time t-7.853 Final velocity, Vi ut at 1:3 +0%648785) m/s à 16:09 ms 6 Distance Traversed in above 7.855. szett tat = 113 x 7-85 + 220/61x7.85 m F107.50 m. After allaining v distance remain is, 8' 3 (300- 10705) z 1925 m.

Time saved, t- time takente cover 300m with rel.n. - 1785+ time taken to cover is with relor). 1925 3 300 7-85+ +16009) s 13 à 6.735. when the winner o At the time started to place winner was accelerate, the second 295 m away from the finishing mark. : Finishing time difference, a 562 s. Distance carered by the second place rimer in the time the winner starts to accelerate & finishes line is, s'116 (7*85+69289) ñ 229.84 m. The second place womer was. (225-229-844) on=6516 om aray the winner race. when she finished the om