Question

A cyclist sprints at the end of a race to clinch a victory. She has an initial velocity of 11.5 m/s and accelerates at a rate of 0.500 m/s² for 7.00 s. (a) What is her final velocity? (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? (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 11.8 m/s until the finish line. What was the difference in finish time in seconds between the winner and runner-up? How far back was the runner-up when the winner crossed the finish line?

Answer 1

Answer:

a) $v_f = 11.5 m/s + 0.5 m/s^2 (7s) = 15 m/s$

b) $t_s = 26.09 s -20.817 s =5.273 s$

c) $\delta t = 25s - 20.817 =4.183 s$

$x_d= 4.183 m/s *11.8 m/s = 49.359 m$

Explanation:

For this case we have the following data given:

$v_i = 11.5 m/s$ represent the initial velocity

$t = 7$ the time when we have acceleration

$a = 0.5 m/s^2$ the acceleration

$x = 5m$ represent the distance

$v_r = 11.8 m/s$ the velocity for the second place winner

$x_t = 300 m$ represent the total distance

Part a

For this case we can use the following equation:

$v_f = v_i +at$

And if we replace we got:

$v_f = 11.5 m/s + 0.5 m/s^2 (7s) = 15 m/s$

Part b

For this case first let's calculate the time taken to end the race with the initial velocity provided $t =\frac{d}{v}$, and if we replace we got:

$d_a = \frac{x_t}{v_i} =\frac{300 m}{11.5 m/s} = 26.09 s$

Now we can find the distance covered whn she accelerates

$x = v_i t + \frac{1}{2} a t^2 = 11.5 m/s * 7s +\frac{1}{2} (0.5 m/s^2) (7s)^2 = 92.75m$

Now we can calculate the distance for the final velocity:

$x_f = 300-92.75 m =207.25m$

And the time to complet this distance is:

$t_f = \frac{x_f}{v_f} = \frac{207.25 m}{15 m/s}=13.817 s$

And the total time would be then:

$t_{total}= t +t_f = 7 + 13.817 s= 20.817s$

And we can calculate the tame saved taking the difference from the total tim and $t_{total}$

$t_s = 26.09 s -20.817 s =5.273 s$

Part c

For this case we have the velocity for the second player;

$v_r = 11.8 m/s$

The total distanc for this case woudl be 300--5 = 295 m we can calculate the time to travel this distance at 11.8 m/s like this:

$t_r = \frac{295 m}{11.8m/s}=25 s$

And if we find the difference between this time and the $t_{total}$ we got:

$\delta t = 25s - 20.817 =4.183 s$

And the distance would be:

$x_d= 4.183 m/s *11.8 m/s = 49.359 m$