Answer:
t_man = 10.16 s, t_horse = 10.73 s, the winner is the man
Explanation:
To solve this problem we are going to find the time of each one separately.
Man we look for distance and time during acceleration
x₁ = v₀ t₁ + ½ a₁ t₁²
as it comes out of rest its initial velocity is zero
x₁ = ½ a₁ t₁²
x₁ = ½ 6.0 1.8²
x₁ = 9.72 m
at this point its speed is
v₁ = v₀ + a t
v₁ = 0 + 6 1.8
v₁ = 10.8 m / s
From here on it continues at constant speed, the distance that the man needs to travel from the point where the man leaves at 100m is
x₂ = 100 - x₁
x₂ = 100- 9.72
x₂ = 90.28 m
the time for this part is
v₁ = x₂ / t₂
t₂ = x₂ / v₂
t₂ = 90.28 / 10.8
t₂ = 8.36 s
the total time for the man is
t_man = t₁ + t₂
t _man = 1.8 + 8.36
t_man = 10.16 s
We repeat the calculation for the horse
distance traveled during the acceleration period
x₃ = v₀ t + ½ a₂ t₃²
as part of rest its initial velocity is zero
x₃ = ½ a₂ t₃²
x₃ = ½ 5.0 4.8²
x₃ = 57.6 m
the velocity at this point is
v₃ = v₀ + a₂ t₃
v₃ = 0 + 5 4.8
v₃ = 24 m / s
the rest of the route is at constant speed, the remaining distance
x₄ = 200 - x₃
x₄ = 200 - 57.6
x₄ = 142.4 m
the time to go through it is
t₄ = x₄ / v₃
t₄ = 142.4 / 24
t₄ = 5.93 s
the total time for the horse is
t_horse = t₃ + t₄
t_horse = 4.8 + 5.93
t_horse = 10.73 s
when we compare the times we see that the man arrives a little before the horse, the winner is the man
