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

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`

Step-by-step 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 =(d)/(v)`, and if we replace we got:

`d_a = (x_t)/(v_i) =(300 m)/(11.5 m/s) = 26.09 s`

Now we can find the distance covered whn she accelerates

`x = v_i t + (1)/(2) a t^2 = 11.5 m/s * 7s +(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 = (x_f)/(v_f) = (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 = (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`