A man’s higher initial acceleration means that a man can outrun a horse over a very short race. A simple – but plausible – model for a sprin

Question

A man’s higher initial acceleration means that a man can outrun a horse over a very short race. A simple – but plausible – model for a sprint by a man and a horse uses the following assumptions: The man accelerates at 6.0 m/s2for 1.8 s and then runs at a constant speed. A horse accelerates at a more modest 5.0 m/s2 but continues accelerating for 4.8 s and then continues at a constant speed. A man and a horse are competing in a 200 m race. The man is given a 100 m head start, so he begins 100 mfrom the finish line.

How much time does it take the man to finish the race? The horse? Who wins?

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Ladonna 4 years 2021-09-01T01:03:51+00:00 1 Answers 48 views 0

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    2021-09-01T01:04:52+00:00

    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

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