Question

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?

Answer 1

Answer:

given,

initial velocity = 11.5 m/s

acceleration = 0.5 m/s²

time = 7 s

a) using equation of motion

v = u + a t

v = 11.5 + 0.5 × 7

v = 15 m/s

b) $s = u t + \dfrac{1}{2}at^2$

$s = 11.5\times 7 + \dfrac{1}{2}\times 0.5\times 7^2$

s = 92.75 m

distance traveled by racer with final velocity

d = 300 - 92.75

   = 207.25 m

time spend to travel 207.25

$t = \dfrac{207.25}{15}$

t = 13.82 s

total time spent to travel 300m

t = 7 + 13.82

t = 20.82 s

time required to travel 300 m with initial velocity

$t_0 =\dfrac{d}{v_0}$

$t_0 =\dfrac{300}{11.5}$

      = 26.09 s

time saved due to acceleration

t = 26.09 - 20.82 = 5.27 s

c) second racer is 5 m apart from the first one

distance traveled by the second racer till first one finishes the race

$d_2 = v_2t$

$d_2 = 11.8\times 20.82 = 245.676 m$

distance behind the leader

d = 300 -5 -245.676 = 49.34 m

time behind the winner

$t_f = \dfrac{d}{v_2} =\dfrac{49.34}{11.8} = 4.18 s$