Constant velocity object catching up to an accelerating object

  • Thread starter maccam912
  • Start date
  • #1
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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.
 

Answers and Replies

  • #2
ehild
Homework Helper
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Eliminate x from your two equations and solve for v(t). Find the minimum of this v(t) function.


ehild
 
  • #3
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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

[tex]v = \frac{(1/2)*0.41*t^{2}}{t}[/tex]

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
ehild
Homework Helper
15,521
1,898
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
 

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