Constant velocity object catching up to an accelerating object

[maccam912]

Homework Statement

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?

Homework Equations

x = 1/2*a*t²
x = v*t

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.

 

[ehild]

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


ehild

 

[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?

 

[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

 

[rwisz]

I would approach this problem by first identifying the relevant variables and equations that describe the motion of the train and the girl. In this case, we have the train's constant acceleration (0.41 m/s2) and the time it takes for the girl to reach the point next to the track (6.4 seconds).

Using the equations x = 1/2*a*t2 and x = v*t, we can determine the position of the train at 6.4 seconds and the position of the girl at any given time t.

To find the slowest constant speed at which the girl can run and still catch the train, we need to find the point where the two positions are equal. This can be done by setting the two equations equal to each other and solving for v.

1/2*a*t2 = v*t

Plugging in the values for a and t, we get:

1/2*0.41*(6.4)2 = v*(6.4)

Simplifying, we get:

v = 2.05 m/s

This is the slowest constant speed at which the girl can run and still catch the train. Any speed slower than this would result in the girl being unable to catch up to the train.