Calculating Time Difference in Pursuit Situations

[Saitama]

I'm curious (like my name suggests), but if you didn't solve that d.e., how exactly did you arrive at the time for Sven? I solved that d.e., and that's how I found that time. That curve I plotted is exact, and uses that equation.

I did solve the D.E but when I plot the result, it doesn't look like your sketch. Solving the D.E and using x=100, y=0
$$y=\frac{1}{30}(x-300)\sqrt{x}+100$$
but plotting this on Wolfram Alpha doesn't give the sketch as shown in your attachment for Sven's path.

EDIT: Got it, I have some mistakes in the above equation. It should be 200/3 instead of 100. Thank you!

 

[haruspex]

Fwiw, I felt this question might be easier using local coordinates.
u = pursuer's speed, v = target's speed, s = distance between them, theta = angle between their current directions. Initially (s, theta) = (L, pi/2).
$s \dot \theta = - v \sin \theta$, $\dot s = v \cos \theta - u$
Whence $\frac{\dot s}s = \frac{\dot \theta}{\sin \theta} ( \frac uv - \cos \theta)$
$\ln s = \frac u{2v} \ln \left(\frac{1-\cos\theta}{1+\cos\theta}\right) + \ln L$
$s = \frac L{\sin\theta}\left(\frac{1-\cos\theta}{1+\cos\theta}\right)^{\frac u{2v}}$
Using (1) again: $\dot \theta \frac L{\sin^2\theta}\left(\frac{1-\cos\theta}{1+\cos\theta}\right)^{\frac u{2v}} = - v$
Since u = 2v,
$vT = L\int_0^{\pi/2} \frac{(1-\cos \theta).d\theta}{(1+\cos\theta)\sin^2\theta}$
$= L\int^{\pi/2}_0 \frac{d\theta}{(1+\cos\theta)^2}$
Writing 2x = theta
$= L\int^{\pi/4}_0 \frac{dx}{2 \cos^4x}$
$= L\frac16 \left[\tan^3x+3\tan x\right]^{\pi/4}_0 = 2L/3$

 

[Saitama]

$s \dot \theta = - v \sin \theta$, $\dot s = v \cos \theta - u$

Sorry but I really can't follow this, can you please explain how you got those two equations?

$\dot{s}$ represents the velocity but how do you get the RHS of the second equation? And I don't have any idea about the first equation. $s\dot{\theta}$ has dimensions of velocity but I am confused.

 

[haruspex]

Sorry but I really can't follow this, can you please explain how you got those two equations?

$\dot{s}$ represents the velocity but how do you get the RHS of the second equation? And I don't have any idea about the first equation. $s\dot{\theta}$ has dimensions of angular velocity but I am confused.

In the line joining the two, target is moving away from pursuer with speed v cos θ, while pursuer advances with speed u: $\dot s=vcosθ−u$. In the perpendicular direction, the target moves with speed $v \sin \theta$. From the pursuer's perspective that's a tangential speed, tending to reduce theta. The 'radius' is s: $v \sin \theta = -s \dot \theta$ (as in the familiar v = rω).

 

[Saitama]

In the line joining the two, target is moving away from pursuer with speed v cos θ, while pursuer advances with speed u: $\dot s=vcosθ−u$. In the perpendicular direction, the target moves with speed $v \sin \theta$. From the pursuer's perspective that's a tangential speed, tending to reduce theta. The 'radius' is s: $v \sin \theta = -s \dot \theta$ (as in the familiar v = rω).

Thanks a lot haruspex!