Question

Blythe and Geoff compete in a 1.00-km race. Blythe's strategy is to run the first 600 m of the race at a constant speed of 4.10 m/s, and then accelerate to her maximum speed of 7.40 m/s, which takes her 1.00 min, and then finish the race at that speed. Geoff decides to accelerate to his maximum speed of 8.40 m/s at the start of the race and to maintain that speed throughout the rest of the race. It takes Geoff 3.00 min to reach his maximum speed. Assume all accelerations are constant.

a. Calculate the time of Blythe's run. (Express your answer to three significant figures.)
b. Calculate the time of Geoff s run. (Express your answer to three significant figures.)

Answer 1

Blythe's time in the race is approximately 260.39 seconds, while Geoff's time is approximately 119.05 seconds.

Step-by-step explanation:

To calculate Blythe's time in the race, we need to find the time taken for each leg of her strategy.

For the first 600 m at a constant speed of 4.10 m/s, the time taken is given by distance divided by speed:

Time = 600 m / 4.10 m/s = 146.34 s

Next, she accelerates to her maximum speed of 7.40 m/s in 1.00 min (60 s). The distance covered during this time is given by initial velocity multiplied by time, plus half the acceleration multiplied by time squared:

Distance = (4.10 m/s * 60 s) + (0.5 * 0.5 m/s^2 * (60 s)^2) = 315 m

Finally, she runs the remaining 400 m at her maximum speed of 7.40 m/s:

Time = 400 m / 7.40 m/s = 54.05 s

Adding up the times for each leg, we get:

Time = 146.34 s + 60 s + 54.05 s = 260.39 s

Therefore, Blythe's time in the race is approximately 260.39 seconds.

To calculate Geoff's time in the race, since he accelerates to his maximum speed at the start and maintains it for the entire race, we only need to consider the distance covered at constant speed. He runs the entire 1.00 km at a constant speed of 8.40 m/s:

Time = 1000 m / 8.40 m/s = 119.05 s

Therefore, Geoff's time in the race is approximately 119.05 seconds.

Answer 2

a.
`t_b=213.774\ s`

b.
`t_g=209.0476\ s`

Step-by-step explanation:

Given:

• total distance to be covered,
  `d=1000\ m`

• distance covered by Blythe in the first span of the race,
  `s_(b1)=600\ m`

• speed of Blythe in the first stage,
  `v_(b1)=4.1\ m.s^(-1)`

• time of acceleration of Blythe,
  `t_(ab)=60\ s`

• maximum velocity after acceleration of Blythe,
  `v_(b2)=7.4\ m.s^(-1)`

• initial speed of Geoff,
  `v_(g1)=0\ m.s^(-1)`

• maximum speed of Geoff,
  `v_(g2)=8.4\ m.s^(-1)`

• time taken to reach the maximum speed of Geoff,
  `t_(ag)=180\ s`

Acceleration of Blythe:

Since acceleration is the rate of change in velocity.

`a_b=(v_(b2)-v_(b1))/(t_(ab))`

`a_b=(7.4-4.1)/(60)`

`a_b=0.055\ m.s^(-2)`

Distance covered during the acceleration:

`s_(ab)=v_(b1).t_(ab)+(1)/(2) a.(t_(ab))^2`

`s_(ab)=4.1* 60+0.5* 0.055* 60^2`

`s_(ab)=345\ m`

Now the remaining distance:

`\Delta d=d-(s_(b1)+s_(ab))`

`\Delta d=100-(600+345)`

`\Delta d=55\ m`

Time of Blythe's run:

`\rm time=(distance)/(speed)`

`t_b=(s_(b1))/(v_(b1)) +s_(ab)+(\Delta d)/(v_(b2))`

`t_b=(600)/(4.1) +60+(55)/(7.4)`

`t_b=213.774\ s`

b.

Acceleration of Geoff:

`a_g=(v_(g2)-v_(g1))/(t_(ag))`

`a_g=(8.4-0)/(180)`

`a_g=0.047\ m.s^(-2)`

distance covered by Geoff while acceleration:

`s_(ag)=v_(g1).t_(ag)+(1)/(2).a_g.(t_(ag) )^2`

`s_(ag)=0+0.5* 0.047* 180^2`

`s_(ag)=756\ m`

Now the remaining distance:

`\Delta d_g=d-s_(ag)`

`\Delta d_g=1000-756`

`\Delta d_g=244\ m.s^(-1)`

Total time taken to complete the race:

`t_g=t_(ag)+(\Delta d_g)/(v_(g2))`

`t_g=180+29.0476`

`t_g=209.0476\ s`