(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

A runner hopes to complete the 10,000-m run in less than 30.0 min. After 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?

2. Relevant equations

v_{f}= v_{o}+ at

avg velocity = (v_{f}+ v_{o}) /2

d = v_{o})t + (1/2) at^{2}

v_{f}^{2}= v_{o}^{2}+ 2ad

3. The attempt at a solution

Most of my solutions just wind up at 0 = 0 or a = a.... In other words, I'm not doing anything wrong, but it's that I can't think of any other way to solve it.

d_{1}= 8900 m

t = 1620 s

d2 = 1100 m

avg velocity = 8900 / 1620 = A

(I took this route so I could find the instantaneous velocity at t = 1620s, which I though would lead up to another part that could solve the problem)

A = (v_{f}+ v_{o}) / 2

2A - v_{f}= v_{o}

a = [v_{f}- (2A - v_{f})] / t

a = (2vf - 2A) /t

v_{f}^{2}= (2A - v_{f})^{2}+ 2[ (2v_{f}-2A) /t] d_{1}

(which results in)

0 = 0

I've been thinking about this problem for a loooong time, but I can't figure out another way to solve it.

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# Homework Help: Runner's motion in one dimension

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