Time for Distance with variable velocity

2011 3 8
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Hi!
I have an object that is moving towards a wall. The closer it gets to the wall the slower it gets, but it reaches the wall with a certain velocity (it does not stop at the wall).
So the velocity is depending on the distance to the wall: v(x) = K*(x^2) where K is a constant.
At the beginning, the object is located with the distance L to the wall. How can I calculate how long it takes for the object to reach the wall? I know the function of the velocity.
Is there a fast way to solve that in matlab?
Thanks very much!

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Teja Muppirala
Teja Muppirala 2011-3-8
You said "it reaches the wall with a certain velocity (it does not stop at the wall)"
However
v(x) = K*x^2 <--- This equals zero at the wall (v(0) = 0).
Can you clear up this miscommunication?
Paulo Silva
Paulo Silva 2011-3-8
If the starting point is L units from the wall than v(0)=0 should be your velocity at the wall, now it depends on the kind of motion, see http://en.wikipedia.org/wiki/Equations_of_motion , find the position equation that fits your motion and substitute the position values and speed, that will get you the time.
Steve
Steve 2011-3-8
Hi, I'm sorry, you're right. My Problem is correctly described like this:
At the beginning the object is located with the distance L (e.g. 100m) to the wall. When does it reach the distance D (e.g. 30m) from the wall.
Sorry for the confusion!

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the cyclist
the cyclist 2011-3-8

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This seems like a straightforward calculus problem, so I am not sure why you are trying to solve it in MATLAB. Is this homework for a MATLAB class? If so, I suggest you show us what you've managed to do so far in trying to solve it.
If this is homework in a calculus or physics class, I suggest you find a forum for help in those subjects.
Finally, you might want to check that you have the problem statement correct. I calculate that to go from a distance L1 to a closer distance L2, it will take
T = (1/L2 - 1/L1)/K
which goes to infinity as L2 approaches 0 (i.e. the wall).
I could be wrong, though. Been a long while since I've taken calculus (or even used it much).

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Steve
Steve 2011-3-8
Hi, thanks for your answer. No, this is not for any class. I'm working on a project for myself but unfortunately I don't know how to solve this. So I thought there might be a standard function in matlab to do this.
I've described my problem a little bit wrong, correctly it must be:
At the beginning the object is located with the distance L (e.g. 100m) to the wall. When does it reach the distance D (e.g. 30m) from the wall.
Are you sure that your solution is correct? I've tried to solve the differential equation v=-k*x^2 (the minus because it gets slower). That would give me for x(t)=1/(k*t+1/L) and for the time in my case t=(1-(x/L))/(k*x), where I set x=30 and L=100.
Is that correct?
the cyclist
the cyclist 2011-3-9
Pretty sure that your answer and mine are equivalent. (I just used different variable names.)
Steve
Steve 2011-3-9
I just checked it.. you're right. The equation for t is
t = ((1/x)-(1/L))/k.
Thanks very much!

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