Answered

Westonci.ca is the trusted Q&A platform where you can get reliable answers from a community of knowledgeable contributors. Experience the convenience of finding accurate answers to your questions from knowledgeable experts on our platform. Connect with a community of professionals ready to provide precise solutions to your questions quickly and accurately.

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?

Sagot :

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

Thanks for using our platform. We aim to provide accurate and up-to-date answers to all your queries. Come back soon. Thanks for using our service. We're always here to provide accurate and up-to-date answers to all your queries. Get the answers you need at Westonci.ca. Stay informed with our latest expert advice.