A runner hopes to complete the 10,000m run in less than 30.0 minutes. After running at constant speed for exactly 27.0 minutes there are still 1100 m to go. The runner must then accelerate at 0.20 m/s² for how many seconds to achieve the desired time? I chose to approach this one graphically, rather than numerically (initially, anyway). I began by drawing a velocity vs. time graph. We could choose to draw a graph of the entire thirty minutes, or merely focus on the last three, because that is really what we’re interested in; for clarity’s sake, let’s go with a graph of the whole thirty minutes. So, for the initial 27 minutes (1620 seconds), we have that the runner’s velocity is 8900 m 27 min 5.4938 m

s= . Then, from 1620 st = until 1620 At t= + (which we shall define as the length of time for which the runner is accelerating), the velocity increases at a rate of 20.20 m

s;

that is, the slope of the velocity-time curve is 0.20. From 1620 At t= + until (30 minutes), the slope of the velocity-time curve is once again zero.

1800 st =

You may not immediately see where I’m going with this, so bear with me; it should become abundantly clear soon. So let’s draw this graph now (as it would turn out, MS Paint is not a good tool for drawing graphs). Remember that this graph is not to scale, because that would obscure the important parts.

So we have this graph. We now must make the key insight that position is the integral of velocity. In layman’s terms, the total distance traveled by the runner is equal to the area beneath the velocity vs. time curve. That is, the sum of the regions A + B + C must be 10,000 meters. We can now develop an expression for the area of each region. Note that A and C are rectangles, while B is a trapezoid. Make note that 5.49 0.20f Av t= + (the final velocity) and make sure you see why this is the case. [ ] (5.49)(1620) 8900A = =

1 1 11 22 2 2[ ] ( )( ) (5.49 (5.49 0.20 ))( ) (10.98 0.20 )( )A A A AB b b h t t t t= + = + + = +

[ ] (1800 (1620 ))(5.49 0.20 ) (180 )(5.49 0.20 )A A AC t t t= − + + = − + AtWe thus see that , so that: 8900 [ ] [ ] 10,000B C+ + =[ ] [ ] 1100B C+ = 12 (10.98 0.20 )( ) (180 )(5.49 0.20 ) 1100A A A At t t t+ + − + = We now have an equation of one variable, tA, which we can solve to find the total length of time for which the runner accelerated. Liberally apply a graphing calculator, and find that 3.1133 sAt = . Round as much as you’d like, and there you have it.