# Homework Help: Constant velocity object catching up to an accelerating object

1. Sep 15, 2010

### maccam912

1. The problem statement, all variables and given/known data

A train pulls away from a station with a constant acceleration of 0.41 m/s2. A passenger arrives at a point next to the track 6.4 s after the end of the train has passed the very same point. What is the slowest constant speed at which she can run and still catch the train?

2. Relevant equations

x = 1/2*a*t2
x = v*t

3. The attempt at a solution

I think the 0.41 m/s2 belongs in the first equation, and gives a graph of position of the end of the train vs time. The 6.4 seconds can be used in the second equation, turing it into x = v*(t-6.4) to graph the motion of the girl running to catch up to the train. I can find the derivative of the first equation to find the slope that the line of the second equation must match, but I always get strange results when I try to tell that line to also go through the point (6.4,0). All I need is the value of v in the second equation, but I always end up eliminating it or getting strange results when simplifying.

2. Sep 15, 2010

### ehild

Eliminate x from your two equations and solve for v(t). Find the minimum of this v(t) function.

ehild

3. Sep 15, 2010

### maccam912

I'm still having trouble. I eliminate x by setting the equations equal to each other: v*t = (1/2)*0.41*t2. Then I solve for V by dividing both sides by t, ending up with

$$v = \frac{(1/2)*0.41*t^{2}}{t}$$

The t on the bottom cancels out 1 of the t's on the top. When I try to graph this equation, I end up with a straight line with a positive slope, and no minimum. Where did I go wrong here?

4. Sep 15, 2010

### ehild

The time of the runner is 6.4 less than the time of the train. So you have v(t-6.4) = 1/2 *0.41*t^2.

ehild