A bicycle racer sprints at the end of a race to clinch a victory. The racer has an initial velocity of 11.5 m/s and accelerates at the rate of 0.500 m/s2 for 7.00 s. (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 11.8 m/s until the finish line. How far ahead of him (in meters and in seconds) did the winner finish?

Step-by-step solution 38PE Step 1 of 7 The formula for final velocity from equation of motion is, Here, u is the initial ve locity, is the final ve locity, is the acceleration and t is time the racer takes to reach final velocity. Step 2 of 7 (a) From the first equation of motion the final velocity of the sprinter is, Substitute 11.5 m/s for u , 0.500 m/s2 for a and 7.00 s for t . Hence, the velocity of the racer after 7.00 s is . Step 3 of 7 (b) The formula for time taken when velocity is constant is, Substitute 300 m for d and 11.5 m/s for v. Step 4 of 7 The distance s traveled by racer during the last 7.00 s of race is, Substitute 11.5 m/s for u, 0.500 m/s2 for a and 7.00 s for t . The racer traveled 92.75 m distance in 7.00 s with accelerated motion. He traveled the rest of the distance with increased but constant velocity of 15.0 m/s. The time taken by the racer at the constant final velocity to complete the race is,