Lets say you have 2 points(adsbygoogle = window.adsbygoogle || []).push({}); AandB.Ais given as(x, y)andBis a point that revolves on a circle. You are given the radius (r), the initial angle (i), and the velocity that it travels at (v_b). You are also given the velocity that A travels at (v_a). The center of the circle is(0,0)and all points relate it that point.

http://img90.exs.cx/img90/8300/pic32.jpg

You want to find the shortest amount of time it takes forAto reachB.

To do it in the shortest amound of time,Amust travel in a straight line. In order to do that, you must know whereBwill be in the same amount of time it takesAto get there.

I've created 2 equations for distance that relate to time.

d(t) = sqrt( (x - rcos(at + i))^2 + (y - rsin(at + i))^2 )

d(t) = (v_a)t

These combine to make

(v_a)t = sqrt( (x - rcos(at + i))^2 + (y - rsin(at + i))^2 )

*note: sqrt = square root

d= distance (variable)

t= time (variable)

x= x-coordinate ofA(constant)

y= y-coordinate ofA(constant)

r= radius (distance from(0,0)toB) (constant)

i= initial angle (constant)

v_a= velocity ofA(constant)

v_b= velocity ofB(constant)

a= angular velocity (v_b/r) (constant)

The thing that is wrong with this equation is that you cannot isolatet, but you can use it to approximatet.

I've tried findingd'(t)and(d^2)''(t)to eliminatetfrom one of the functions (I'm not sure they arefunctions).

d'(t) = ( axrsin(at + i) - ayrcos(at + i) ) / sqrt( (x - rcos(at + i))^2 + (y - rsin(at + i))^2 )

d'(t) = v_a

(d^2)''(t) = 2(a^2)xrcos(at + i) + 2(a^2)yrsin(at + i)

(d^2)''(t) = 2(v_a)^2

But, these functions don't help in findingt.

The main purpose is to find the shorest amount of time it takes forAto get toB.

So, there's got to be another method, right?

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# Shortest time from a point to a moving point

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