Problem. There is a uniform string with one end fixed to a peg, and the other end getting pulled away at a constant speed, say $u$. At time $0$, the string has length $a$ and is taut. At time $t$, a point of the string has a distance $x$ away from the fixed end. What is the speed of this point at time $t$?
My Attempt. One thing I am certain about is that all points of the string cannot have the same speed, because if so, the whole string would be moving towards some direction, which is not possible when one end of it is fixed. Another thing I am certain about is that for a point, the further away it is from the fixed point, the faster it moves, because if it were the other way around, the string would be at least partially slack, but at least intuitively it cannot happen given the stated scenario. Unfortunately, here is where I feel difficult to tackle further.
Comments. I know the answer is $\frac {xu}{a+ut}$, but why exactly? I don't know if I should ask this in the physics forum, but personally I feel like the potential analysis is more mathematical. Thanks in advance!
So this is not a string, really it is a rubber band. There are some hidden assumptions in this problem. The main assumption is that the band is "evenly deformed", i.e a group of points that start evenly spaced, will remain evenly spaced as we stretch the band.
Let's get some intuition for the problem. The end of the band that we are pulling will, clearly, move at the same rate at which we are pulling it, and, the fixed end of the band will not move at all.
Let $L(t)$ be the length of the band at time $t$. In your question, we are stretching the band at a constant rate, but I will actually keep $L$ general for now, other than the initial condition $L(0)=a$. We introduce the dimensionless quantity $s(t)=x(t)/L(t)$ where $x(t)$ is the path traced by some point on the band. This describes how far along the band a point is, i.e $s=0.25$ signifies the point is a quarter of the way down the band.
Now, the key insight - because of our "evenly spaced" assumption, a particle that starts in the middle of the band will remain at the middle of the band, in other words, $\dot{s}=0$ ! Expanding, using the product rule, $$\dot{s}(t)=\frac{\dot{x}(t)L(t)-\dot{L}(t)x(t)}{L(t)^2}=0$$ So we arrive at the first order linear IVP $$\dot{x}(t)-\frac{\dot{L}(t)}{L(t)}x(t)=0 \\ x(0)=x_0$$ One can see either by inspection, or by using the method of integrating factors, that $x(t)=\frac{x_0}{a}L(t)$ solves the equation. In the case that $L(t)=a+vt$, then $x(t)=x_0+\frac{vx_0}{a}t=x_0+vs_0t$ and its speed is $\dot{x}(t)=vs_0$. We can also remark that $$\frac{x(t)v}{a+vt}=\frac{vs_0L(t)}{L(t)}=vs_0=\dot{x}(t)$$ Which is the answer you state in your question. This is just a fancy way of restating the differential equation we found, $$\dot{x}=\frac{\dot{L}}{L}x$$ Just with the special case $L(t)=a+vt$.