Finding Acceleration Homework Solution

[gaobo9109]

Homework Statement

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.

Homework Equations

Sine rule
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.

 

[kuruman]

On line AB, draw point G marking the position of the fox at time dt. You get triangle DFG. Side DF is equal to L and side FG is v₁dt. Note that the hypotenuse DG of the triangle is the new direction of the hound's velocity vector.

Draw vectors showing the hound's initial and final velocities. Complete a new triangle by connecting the tips of the initial and final velocity vectors.

1. How should you label the vector connecting the tips of the velocity vectors? (Assume dt is very small)
2. How is the new triangle related to triangle DFG?

 

[gaobo9109]

Well, i think it's the math part that is giving me the problem

From Triangle FGD,i get

tanθ = v1dt/L

Then, from the vector diagram,i have

dv/sinθ = v2/ sin(90-θ/2)
dv/sinθ = v2/cos(θ/2)
dv = 2v2 sin(θ/2)

From here, how do i express sin(θ /2) in terms of v1 and L?

 

[kuruman]

Forget the trig functions. Note that the two triangles are similar. One right side of the velocity triangle is v₂. How should you label the other right side of that triangle? (Hint: It is the change in the hound's velocity).

Set the ratios of their right sides equal and see what you get.

 

[gaobo9109]

Okay, I got the answer alright. But there is one thing I don't understand. It is stated in the question that magnitude of the hound's velocity does not change. Then why would the vector diagram relating the v(intial), v(final) and dv be a right angle triangle. I thought it should be a isosceles triangle, since magnitude of intial and final veloctiy are the same?

 

[kuruman]

Okay, I got the answer alright. But there is one thing I don't understand. It is stated in the question that magnitude of the hound's velocity does not change. Then why would the vector diagram relating the v(intial), v(final) and dv be a right angle triangle. I thought it should be a isosceles triangle, since magnitude of intial and final veloctiy are the same?

Good question. You have to draw the triangles big to see what is going on, but don't forget that you are doing calculus here. Your drawing is actually in the limit that dt goes to zero. At the time given, the hound's velocity must change in a perpendicular direction and the size of that change is (a dt) in the limit that dt goes to zero. In that limit you get an isosceles triangle.