I've been stuck on this problem of Modified pursuit curve, in which the dog chases the cat with a constant acceleration $a$, starting from rest. The cat moves horizontally with a uniform speed of $v_0$, and the dog wants to catch the cat at all points of time. The dog is at a distance of $h$ from the cat initially.
What I'm confused about:
What would be optimal for the dog, acceleration vector always towards the cat, or something else?
What I did: I assumed acceleration vector always towards the cat, and let the dog catch the cat in time $T$. And this is what I'm getting for the equations: $$v_0 T = \int_{0}^{T}{(v \cos{\alpha}) dt}$$ $$h = \int_{0}^{T}{(v \sin{\alpha}) dt} = \int_{0}^{T}{(v_0 \cos{\alpha} - v)}dt$$ I am a beginner in calculus. So can someone correct me/help me with the equations?
The dog should want to travel the least distance, so should compute the point it will catch the cat if it travels on a straight line towards that point. I assume the cat starts at the origin and travels along the $+x$ axis, while the dog starts at $(0,h)$ and can travel anywhere that is within the allowed acceleration.
The cat's position at time $t$ is $(v_0t,0)$. In time $t$ the dog can travel $\frac 12at^2$. For the intercept, we must have $\frac 12at^2=\sqrt{(v_0t)^2+h^2}$ because the dog can cover the distance on the left and needs to cover the distance on the right. Solve for $t$, then have the dog follow a straight line to the intercept point. No integral is required.
If the dog always heads at the cat, let $(x(t),y(t))$ be the position of the dog at time $t$. The cat is at $(v_0t,0)$. The dog's speed is $v=at$. The slope of the dog's trajectory is $\frac {dy}{dx}=\frac y{v_0t-x}$. We have $$\frac {dy}{dt}=\frac {dy}{dx}\frac {dx}{dt}=\frac y{v_0t-x}\frac {dx}{dt}\\ at=\sqrt{\left(\frac{dy}{dt}\right)^2+\left(\frac {dx}{dt}\right)^2}\\ at=\frac {dx}{dt}\sqrt {1+\left(\frac y{v_0t-x}\right)^2}$$ but I don't know how to go further