A fox is running at constant velocity v1 along a straight line AB. A hound is chasing the fox at constant velocity v2, always facing the fox. At a certain instant, the fox is at point F, the hound is at poind D, where FD is perpendicular to AB and FD = L. What is the acceleration of the hound at this instant? The answer is (v1)(v2)/L, but i don't know how to arrive at that answer.
S = vt
The Attempt at a Solution
I tried to use calculus to solve this question. Suppose after time Δt, the distance moved by the fox would be s= v1Δt. The direction of the hound's velocity would have changed by angle θ, where tanθ=v1Δt/L.
Then i tried to relate the intial and final velocity of the hound, as well as its acceleration using a vector diagram.
The equation i get is a/sinθ = v1/sin(θ/2). But i don't know how to continue from here.