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?
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Expert's answer
2012-10-08T11:38:16-0400
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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