Race: Blythe and Geoff compete in a 1-km race. Blythe’s strategy is to run the first 600 m of the race at a constant speed of 4 mm/ss, and then accelerate with constant acceleration to her maximum speed of 7.5 mm/ss, which takes her 1 min, and finish the race at that speed. Geoff decides to accelerate with constant acceleration to his maximum speed of 8 m/s at the start of the race and to maintain that speed throughout the rest of the race. It takes Geoff 3 min to reach his maximum speed. Who wins the race?

Respuesta :

Answer:

From the value of each person's time to finish the race, it can be concluded that Geoff will win the race.

Explanation:

Given;

total distance to be covered, d = 1000 m

first distance covered by Blythe, [tex]d_b_1[/tex] = 600 m

constant speed of Blythe at the first stage, [tex]v_b_1[/tex] = 4 m/s

final speed of Blythe, [tex]v_b_f[/tex] = 7.5 m/s

time of Blythe's constant acceleration, [tex]t[/tex] = 1 min = 60 s

initial speed of Geoff, [tex]v_g_1[/tex] = 0

final speed of Geoff, [tex]v_g_f[/tex] = 8 m/s

time of Blythe's constant acceleration, [tex]t[/tex] = 3 min = 180 s

The acceleration of Blythe is given as;

[tex]a_b = \frac{dv}{dt} = \frac{7.5 - 4}{60} = 0.0583 \ m/s^2[/tex]

distance covered by Blythe during this acceleration is given as;

[tex]d_{a_b} = v_b_1t + \frac{1}{2} a_b t^2\\\\d_{a_b} = 4*60 + \frac{1}{2}(0.0583) (60)^2\\\\d_{a_b} = 344.94 \ m[/tex]

The remaining distance covered by 7.5 m/s is given as;

[tex]d_f = d - (600 + 344.94)\\\\d_f = 1000- (944.94)\\\\d_f = 55.06 \ m[/tex]

The total time for Blythe to covere the entire distance = time taken to cover 600 m + time taken to cover 344.94 m + time taken to cover 55.06 m

[tex]t_b = \frac{600 \ m}{4 \ m/s}\ +\ 60 s \ + \ \frac{55.06 \ m}{7.5 \ m/s} \\\\t_b = 150 s \ + \ 60s \ + \ 7.34 s\\\\t_b = 217.34 s[/tex]

The acceleration of Geoff  is given as;

[tex]a_g = \frac{dv}{dt} = \frac{8 -0}{180} = 0.044 \ m/s^2[/tex]

distance covered by Geoff during this acceleration is given as;

[tex]d_{a_g} = v_g_1 + \frac{1}{2}a_gt^2\\\\ d_{a_g} =0 + \frac{1}{2}(0.044)(180)^2\\\\ d_{a_g} = 712.8 \ m[/tex]

The remaining distance covered by 8 m/s is given a;

[tex]d_f = 1000 \ m - 712.8 \ m\\\\d_f = 287.2 \ m[/tex]

The total time for Geoff to covere the entire distance = time taken to cover 712.8 m + time taken to cover 287.2 m

[tex]t_g = 180s + \frac{287.2 \ m}{8 \ m/s} \\\\t_g = 180s + 35.9 s\\\\t_g = 215.9 s[/tex]

Therefore, from the value of each person's time to finish the race, it can be concluded that Geoff will win the race

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