# Velocity of approach which changes continuously [closed]

There are two particles A and B which are moving with constant speed $$v$$ and $$u$$ such that $$v$$ is always directed towards B. At $$t = 0$$, the separation between A and B is $$l$$ and $$u$$ is perpendicular to the line joining A and B. The Velocity of B is constant i.e., it does not change its direction. Also $$v>u$$. Find the time when they will collide.

Here's my approach

The angle which A makes with the horizontal changes continuously from $$0$$ to $$\theta$$(let). At any time the angle which $$v$$ makes with horizontal will be a function of time. So $$\alpha = f(t)$$ where $$\alpha$$ lies from $$0$$ to $$\theta$$. Let time when they will collide be $$t$$ sec. So, $$\int_0^t v\cos(f(t)) dt = l$$ And $$\int_0^t v\sin(f(t)) dt = ut$$ Now I am stuck what to do next

## closed as off-topic by Kyle Kanos, ACuriousMind♦May 1 at 14:01

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## 1 Answer

I think you have the right idea about setup but as long as you are fixated on angles and endpoints per se you will have trouble. What you actually want to write would seem to be the coupled equations $$\dot x(t) = v\cos\alpha=v\frac{x-ut}{\sqrt{y^2 + (x-u t)^2}}\\ \dot y(t) = v\sin\alpha=-v\frac{y}{\sqrt{y^2 + (x-u t)^2}}$$ And then maybe you can solve for this. So all I have done here is substituted the definition of sine and cosine for this triangle in.

That looks complicated, though, and I wonder if it would be much easier to consider a particle with variable velocity $$v(t)$$ traveling toward the origin, then transform this into the desired reference frame to just get an equation for $$v(t)$$ which makes the transformed velocity constant.