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# Constant Acceleration Factors

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"A runner hopes to complete the 10,000-m run in less than 30.0 minutes. After exactly 27.0 minutes, 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?"

Here's what I've done:

So far, the runner has gone 8900 m in 1620 s, which should give her a velocity of 5.49 m/s going into the next leg of the race.

I've dealt with the next leg of the race separately, setting X(0) = 0, X = 1100m, V(0) = 5.49 m/s, and a = 0.20 m/s^2.

Since I can't find t directly from here (or at least, I don't know how to from the equation X - X(0) + V(0)t + 1/2at^2 because I don't know how to factor out the two different types of t's), I first find V.

V^2 = V(0)^2 + 2a(X - X(0)) = 5.49 m/s^2 + 2(.20)(1100) = 470; V = 21.68 (This number makes me suspicious of course, because 22 m is a lot of distance to cover in 1 second. But I trudge on...)

Then I use V = V(0) + at ; t = (V - V(0))/a = (21.68 - 5.49)/0.20 = 80.95 s, which is obviously not the answer. The answer is in fact 3.1 s, but I don't know how that is derived.

Can you help? (Can you also show me in detail how to use the X - X(0) + V(0)t + 1/2at^2 equation to find t. I don't know how to join together t^2 and t.)

##### Solution Summary

The constant acceleration is computed for a runner after 27 minutes is calculated. How many seconds the runner is accelerating in order to achieve the desired time is determined.

##### Solution Preview

Hi,
Upto 8900/1620, your approach is correct, but after that the problem is asking you a different thing:
It asks you : The runner has to accelerate only for a short time (t) so that his/her speed increases to some desired value (v) and then after he/she should run with that velocity (v) for remaining time (3*60 - t) so that total ...

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###### Education
• MSc , Pune University, India
• PhD (IP), Pune University, India
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##### Classical Mechanics

This quiz is designed to test and improve your knowledge on Classical Mechanics.