Respuesta :
Answer:
a. [tex]t_b=213.774\ s[/tex]
b. [tex]t_g=209.0476\ s[/tex]
Explanation:
Given:
- total distance to be covered, [tex]d=1000\ m[/tex]
- distance covered by Blythe in the first span of the race, [tex]s_{b1}=600\ m[/tex]
- speed of Blythe in the first stage, [tex]v_{b1}=4.1\ m.s^{-1}[/tex]
- time of acceleration of Blythe, [tex]t_{ab}=60\ s[/tex]
- maximum velocity after acceleration of Blythe, [tex]v_{b2}=7.4\ m.s^{-1}[/tex]
- initial speed of Geoff, [tex]v_{g1}=0\ m.s^{-1}[/tex]
- maximum speed of Geoff, [tex]v_{g2}=8.4\ m.s^{-1}[/tex]
- time taken to reach the maximum speed of Geoff, [tex]t_{ag}=180\ s[/tex]
Acceleration of Blythe:
Since acceleration is the rate of change in velocity.
[tex]a_b=\frac{v_{b2}-v_{b1}}{t_{ab}}[/tex]
[tex]a_b=\frac{7.4-4.1}{60}[/tex]
[tex]a_b=0.055\ m.s^{-2}[/tex]
Distance covered during the acceleration:
[tex]s_{ab}=v_{b1}.t_{ab}+\frac{1}{2} a.(t_{ab})^2[/tex]
[tex]s_{ab}=4.1\times 60+0.5\times 0.055\times 60^2[/tex]
[tex]s_{ab}=345\ m[/tex]
Now the remaining distance:
[tex]\Delta d=d-(s_{b1}+s_{ab})[/tex]
[tex]\Delta d=100-(600+345)[/tex]
[tex]\Delta d=55\ m[/tex]
Time of Blythe's run:
[tex]\rm time=\frac{distance}{speed}[/tex]
[tex]t_b=\frac{s_{b1}}{v_{b1}} +s_{ab}+\frac{\Delta d}{v_{b2}}[/tex]
[tex]t_b=\frac{600}{4.1} +60+\frac{55}{7.4}[/tex]
[tex]t_b=213.774\ s[/tex]
b.
Acceleration of Geoff:
[tex]a_g=\frac{v_{g2}-v_{g1}}{t_{ag}}[/tex]
[tex]a_g=\frac{8.4-0}{180}[/tex]
[tex]a_g=0.047\ m.s^{-2}[/tex]
distance covered by Geoff while acceleration:
[tex]s_{ag}=v_{g1}.t_{ag}+\frac{1}{2}.a_g.(t_{ag} )^2[/tex]
[tex]s_{ag}=0+0.5\times 0.047\times 180^2[/tex]
[tex]s_{ag}=756\ m[/tex]
Now the remaining distance:
[tex]\Delta d_g=d-s_{ag}[/tex]
[tex]\Delta d_g=1000-756[/tex]
[tex]\Delta d_g=244\ m.s^{-1}[/tex]
Total time taken to complete the race:
[tex]t_g=t_{ag}+\frac{\Delta d_g}{v_{g2}}[/tex]
[tex]t_g=180+29.0476[/tex]
[tex]t_g=209.0476\ s[/tex]
Acceleration is the ratio of the change of the velocity to the time.
- The time of Blythe's run is 211 seconds.
- The time of Geoff's run is 209 seconds.
What is acceleration?
Acceleration is the ratio of the change of the velocity to the time.
Given information-
Blythe and Geoff compete in a 1.00-km race.
Blythe runs first 600 m with constant speed 4.10 m/s.
Blythe accelerated to her maximum speed of 7.40 m/s, to complete the race in 1 min.
- a. Calculate the time of Blythe's run.
Acceleration is the ratio of change of velocity to the time. Thus the acceleration of Blythe's is,
[tex]a=\dfrac{7.40-4.10}{60}\\a=0.055 \rm m/s^2[/tex]
The distance traveled in the 1 minute when he accelerated her speed is,
[tex]s=v\times t+\dfrac{1}{2} at^2\\s=4.1\times60+\dfrac{1}{2} 0.055\times60^2\\\\s=345\rm m[/tex]
As the total distance was 1000 m thus the distance remained is,
[tex]d=1000-600-345\\d=55[/tex]
As the time is the ratio of distance to the speed . thus,
[tex]t=\dfrac{600}{4.1} +\dfrac{345}{7.4} +60\\t=146.34+4.6621+60\\t=211[/tex]
Thus the time of Blythe's run is 211 seconds.
- b. Calculate the time of Geoff's run.
Acceleration is the ratio of change of velocity to the time. Thus the acceleration of Geoff's is,
[tex]a=\dfrac{8.4-0}{180}\\a=0.047 \rm m/s^2[/tex]
The distance traveled Geoff's while accelerated his speed is,
[tex]s=v\times t+\dfrac{1}{2} at^2\\s=0+\dfrac{1}{2} 0.047\times180^2\\\\s=756\rm m[/tex]
As the total distance was 1000 m thus the distance remained is,
[tex]d=1000-600-756\\d=244[/tex]
As the time is the ratio of distance to the speed. thus,
[tex]t=180 +\dfrac{244}{8.4} +60\\t=209[/tex]
Thus the time of Geoff's is 209 seconds.
Hence,
- The time of Blythe's run is 211 seconds.
- The time of Geoff's run is 209 seconds.
Learn more about the acceleration here;
https://brainly.com/question/11021097
