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helpmeplease12 · Sep 29, 2011, 08:03 PM

A runner leaves the start line travelling at a constant velocity of 3.25 m/s. When he is 50m away from the start line a second runner leaves the start line accelerating from rest at 0.30 m/s^2. How long does it take the second runner to catch up to the first runner?

jcaron2 · Sep 29, 2011, 08:43 PM

You need to use two different formulas for this problem:

1) $x = x_0 +vt$

2) $x = \frac 12 at^2 +v_0t + x_0$

If you take t=0 to be the time when the second runner takes off from the starting blocks, you can use formula (1) to find the position of the first runner as a function of time:

$x_1 = 3.25t + 50$

Meanwhile, the position of the second runner can be found with formula (2):

$x_2 = \frac 12 \cdot 0.3t^2$

Since you're looking for the time when the second runner has caught up to the first, that means that at that point in time they'll be in the same position. Thus, if you set x1 and x2 equal to each other (in other words, set the right-hand sides of the above equations equal to each other), you can then solve for the time t at which this occurs. That's your answer!

By the way, note that since the equation you'll be solving is a quadratic, you'll actually get two answers for t. In such a case, you need to look at the two answers and decide which one makes sense and which one doesn't. In this case, one of the answers will be positive and one will be negative. Obviously the second runner will catch the first after some positive amount of time has elapsed, so the positive answer is the right one.