# 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?
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.

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