# Answer to Question #16121 in Mechanics | Relativity for seniha

Question #16121

A runner hopes to complete the 10,000- m} run in less than 30.0 min}. After running at constant speed for exactly 27.0 min}, there are still 1100 m}} to go.The runner must then accelerate at 0.20 m{s}}^2 for how many seconds in order to achieve the desired time?

Expert's answer

Let V0 be the initial speed. Then V0 = (10000 - 1100) / (27 * 60) = 5

m/s.

Let t be the period of time (in seconds) when the runner ran with

acceleration.

After t seconds he reached the speed V = V0 + a*t, where a =

0.2 m/(s^2).

The equation for the distance covered in t seconds is:

S = V0

* t + a * t^2 / 2.

The remaining part (1100 - S) m the runner ran with a

constant speed V:

1100 - S = V * (3 * 60 - t).

or

1100 - S = (V0 + a*t)

* (3*60 - t).

After substitution of S we get:

1100 - V0 * t - a * t^2

/ 2 = (V0 + a*t) * (3*60 - t) - quadratic equation for t.

1100 - 5t - 0.1*t^2

= (5 + 0.2*t)(180 - t)

0.1 * t^2 - 36*t + 200 = 0

D = 36^2 - 80 =

1216

t = (36-sqrt(1216)) / 0.2 = 5.64 s.

So, the runner should run

5.64 s with acceleration 0.2 m/(s^2) and then with a constant speed to achieve

the desired result.

m/s.

Let t be the period of time (in seconds) when the runner ran with

acceleration.

After t seconds he reached the speed V = V0 + a*t, where a =

0.2 m/(s^2).

The equation for the distance covered in t seconds is:

S = V0

* t + a * t^2 / 2.

The remaining part (1100 - S) m the runner ran with a

constant speed V:

1100 - S = V * (3 * 60 - t).

or

1100 - S = (V0 + a*t)

* (3*60 - t).

After substitution of S we get:

1100 - V0 * t - a * t^2

/ 2 = (V0 + a*t) * (3*60 - t) - quadratic equation for t.

1100 - 5t - 0.1*t^2

= (5 + 0.2*t)(180 - t)

0.1 * t^2 - 36*t + 200 = 0

D = 36^2 - 80 =

1216

t = (36-sqrt(1216)) / 0.2 = 5.64 s.

So, the runner should run

5.64 s with acceleration 0.2 m/(s^2) and then with a constant speed to achieve

the desired result.

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