Time for Distance with variable velocity

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!

[Teja Muppirala]

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]

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]

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!

Answer 1

[the cyclist]

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).

[Steve]

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]

Pretty sure that your answer and mine are equivalent. (I just used different variable names.)

[Steve]

I just checked it.. you're right. The equation for t is

t = ((1/x)-(1/L))/k.

Thanks very much!