Separation at $t=\infty$

No maths needed for this! Just some reasoning. Jerry can't pull away from Tom as their speeds are the same. So that rules out infinity. Tom will initially get closer to Jerry and Jerry can't ultimately pull away from Tom so that rules out "d". Tom will always have to travel further than Jerry to catch up so that rules out 0. Leaves an answer <d and >0. Which is d/2.

Let us denote the coordinate of Tom at time $t$ by $\big(x(t), y(t)\big)$ , where its velocity vector makes an angle $\theta$ with the $+X$ axis. Then, from the given conditions of the problem, we have $\frac{dx}{dt}=u \cos(\theta), \frac{dy}{dt}= u \sin(\theta), \frac{y}{x-ut}=\tan(\theta).$ The last equation gives, $x-ut= y \cot(\theta)$ Differentiating both sides of the above equation w.r.t. the variable $y$ , we have $\frac{dx}{dy}-u \frac{dt}{dy}= \cot(\theta) - y \textrm{ cosec}^2(\theta) \frac{d\theta}{dy}.$ Using the previous set of equations, we have $\cot(\theta) - \textrm{ cosec}(\theta) = \cot(\theta) - y \textrm{ cosec}^2(\theta) \frac{d\theta}{dy}.$ i.e., $\frac{dy}{y}= \textrm{ cosec}(\theta) d \theta.$ Integrating and using the boundary condition ( $y=-d$ at $\theta =\pi/2$ ), we have $\ln(\frac{|y|}{d})=\ln(| \textrm{ cosec}(\theta)-\cot(\theta)|)= \ln(\tan(\theta/2)).$ i.e., $|y|=d (\tan(\theta/2)).$ Finally, $r(\theta) = |y|\textrm{ cosec}(\theta)=d (\tan(\theta/2))\textrm{ cosec}(\theta).$ It is clear that, $\theta \to 0$ as $t \to \infty$ . Hence, $\lim_{t \to \infty} r(t) = \lim_{\theta \to 0} r(\theta) = \frac{d}{2}. \blacksquare$

If Tom is daft he runs from A to B, then turns through a right angle and continues his pursuit in a straight line, then after his turn he will always trail Jerry by exactly d.

If Tom is smart he decides how long he wants to run, and then runs in a straight line towards the spot which Jerry will be at after this time. Pythagoras's theorem then gives the distance between them as

$\sqrt{d^2+(ut)^2}-ut$

When $ut>>d$ we can use the binomial theorem to approximate this by $ut(1+\frac{1}{2}(\frac{d}{ut})^2)-ut$ which tends to zero as $t \to \infty$ .

We conclude that since Tom is neither completely daft nor really smart, he follows a curved path that is substantially better than the right-angled dog-leg and poorer than the straight hypotenuse, and so

$0< \text{answer}<d$

The only available choice among the options is $\boxed{\frac{d}{2}}$

There is an accurate solution that does not actually involve calculus except nominally.

Consider J's horizontal x-displacement X relative to T. It is increasing at the rate of u(1-cosA), where A is the angle between T's velocity and horizontal x-axis. The limiting value of X is what we seek.

On the other hand, consider the distance r between J and T. It is being reduced at the rate of u(1-cosA), because while T is approaching J at rate of u, T is escaping from J at rate of cosA.

Because of the same rate, at t=infinity X= the remaining distance = d-X, namely X=d/2.