<p>Hey guys, I have two problems that I need help on. I can’t solve the first but the second one I just need verification.</p>
<li><p>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.</p></li>
<li><p>A falling stone takes .30s to travel past a window 2.2 m tall. From what height above the window did the stone fall?</p></li>
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<p>I did all the problems assigned and thought I could do any kinematics problem, but then tried #1 (not assigned) and drew a blank. The answer to # 1 is 3.1s, I got 4.0 m for # 2 (answer not in book). Thanks</p>
<h1>1 The answer is obviously going to be less than 180 seconds, first of all.</h1>
<p>The runner has ran a total of 10000 - 1100 = 8900 m so far in 27 x 60 = 1620 seconds, so his velocity is 8900/1620 = 5.49 m/s. To achieve his goal, he has to run at a velocity of 1100m/(180s) = 6.11 m/s. So find the difference of the velocities, and divide by the acceleration to get (6.11 - 5.49)/0.2 = 3.1 s</p>
<p>I got a different answer for #2. How did you do this?</p>
<p>wow, I missed the crucial last step in #1. Found both velocities yet couldn't finish the problem, guess I didn't understand the problem too well :-. And #2 I'm already positive that i screwed up the signs, I'll have to redo it.</p>
<p>@ #2 hmm this time I got 1.8m. Taking downwards to be the positive direction, I first found the initial velocity of the 2.2 m drop by 2.2 m = Vo<em>(.30) + .5(9.8)(.30s)^2 => Vo = 5.86. Then I defined Vo to be Vfinal of the first part. Letting h be the height we want and starting from rest, h = Vavg</em>t, t = (5.86 - 0)/9.8 = .598 s. h = (5.86+0)/2 * .598 ~ 1.75m</p>
<p>I don't even know if my approach is correct/efficient but it took 3 minutes so if it works, awesome, but I'd like to see other solutions.</p>
<p>I got something along those lines</p>
<p>I could post another solution, but it uses the same general concept - find the initial velocity at the start of the 2.2 m drop then plug it back in to find the height.</p>