Inspired by Mads Carlsen

The rate at which the height increases is inversely proportional to the surface area of the water: $\frac{dh}{dt} \propto \frac 1A.$ (This follows from the constant rate of increase in volume: $\text{const} = dV/dt = A\:dh/dt$ implies $dh/dt = \text{const}/A$ .)

From the drawing we surmise that the container is a truncated cone, with its vertex somewhere below the cup. There is nothing to stop us from measuring the height from this vertex; then the surface area is a circle with radius $r \propto h$ , so that $A \propto h^2$ and $\frac{dh}{dt} \propto \frac 1{h^2}.$ The solution is of the form $h(t) \propto \sqrt[3]t,$ for properly scaled time coordinate $t$ . Taking the derivative two and three times, we find the acceleration and the jerk $\frac{d^2h}{dt^2} t^{1/3} \propto -t^{-5/3};\ \ \ \ \ \ \ \ \ \frac{d^3h}{dt^3} t^{1/3} \propto t^{-8/3}.$ Since the acceleration is negative but the jerk is positive, we conclude that the magnitude of acceleration $\boxed{\text{decreases}}$ .

Observe that speed of the waterline is always positive ( $v>0$ ). If the negative acceleration was constant or increasing in magnitude, at one point in time, and no later than $T \leq v_0/|a_0|$ , the speed of the waterline would become negative. This, would mean that, while pouring water in the cup, the cup somehow is emptying itself.

EDIT: I agree with the moderator. It is well possible that acceleration could initially increase in magnitude, and afterwards decrease (or even oscillate), depending on the shape of container.

Practical Observation

Overall height of water level is increasing however, due to increase in the cross sectional area of cup the height gained by water level at any instant(phase) is less than that of previous phase which results decrease in displacement of height of water level with respect to time where curve will be concave downward resulting decrease in the non-uniform speed and slowing down of water level which further tells that acceleration is negative and magnitude is decreasing.

Note: Negative acceleration doesn't infer that acceleration is decreasing rather it's tells speed is decreasing.

---

To prove our claim let us take a cone (since it shows similar phenomenon by extending lines backward in the given cup) where $R$ and $H$ be the radius and height whereas, $r$ and $h$ be the radius and height of water level at any time $t$ . To calculate the volume of cone we have, $V = \dfrac{1}{3}\pi r^2h$ . If we differentiate with respect to time $t$ , the right hand side will involve the quantity $\dfrac{dr}{dt}$ . Since we have no direct information about $\dfrac{dr}{dt}$ , it is wise idea to eliminate $r$ from formula we have . Using the similarity of triangle, we see $\dfrac{r}{h} = \dfrac{R}{H}\implies r=\dfrac{Rh}{H}$ . Substituting in the formula we have $V= \left(\dfrac{\pi R^2}{3H^2}\right)h^3$ Differentiating with respect to time $t$ we obtain $\dfrac{dV}{dt} = \dfrac{R^2 \pi}{3H^2}\left(3h^2\dfrac{dh}{dt}\right)\Rightarrow \dfrac{dh}{dt} = \dfrac{X}{h^2}$ Now, with further differentiation with respect to time $a(t) = \dfrac{d^2h}{dt^2} = - \dfrac{X'}{h^3}\implies$ that acceleration $a$ is negative(implying speed is decreasing which is true) and magnitude of acceleration is falling with increase in height of water level meaning $a$ is not constant. Hence, this conclude the magnitude of acceleration is decreasing since $\frac{\,a(t)}{dt}>0$ .

$\pm$ Acceleration Versus Time graph :

Magnitude of Accelaration of the water's height is starting to tend towards 0 as we consider huge cases, when the drip of water wouldn't change the water level at all.

Okay, let's do some calculus, just out of curiosity.

The volume of the cup is given by: $V = \int_{h_{o}}^{h_{f}}A(h)\, dh,$ where $A(h) = \pi(ah)^{2}$ is the cross-section area as a function of $h$ . (There is a linear dependence between height and radius of the cross-section)

By chain rule, we have: $\begin{aligned} \frac{\partial V}{\partial t} &= \frac{\partial V}{\partial h}\frac{\partial h}{\partial t} = Q \\ \implies \frac{\partial h}{\partial t}&= \dfrac{Q}{\pi a^{2}h^{2}},\end{aligned}$ where $Q$ is a constant.

Applying chain rule further, we get: $\begin{aligned} \frac{\partial^2 h}{\partial t^2} &= \frac{\partial^{2} h}{\partial t\, \partial h}\frac{\partial h}{\partial t} = -\dfrac{2Q}{\pi a^{2}h^{3}}\dfrac{Q}{\pi a^{2}h^{2}} = -\dfrac{2Q^{2}}{\pi^{2} a^{4}h^{5}} \\ \frac{\partial^3 h}{\partial t^3} &= \frac{\partial^{3} h}{\partial t^{2}\, \partial h}\frac{\partial h}{\partial t} = \dfrac{10Q^{2}}{\pi^{2} a^{4}h^{6}}\dfrac{Q}{\pi a^{2}h^{2}} = \boxed{\dfrac{10Q^{3}}{\pi^{3} a^{6}h^{8}}>0} \\ &~~\vdots \\ \frac{\partial^n h}{\partial t^n} &= \frac{\partial^{n} h}{\partial t^{n-1}\, \partial h}\frac{\partial h}{\partial t} = (-1)^{n-1}\dfrac{\left( \prod_{k=2}^{n}3k-1\right )Q^{n}}{\pi^{n}a^{2n}h^{3n-1}}\end{aligned}$

We see that third-order derivative of height with respect to time, which captures changing of height acceleration, is positive, implying that acceleration is increasing. But since the acceleration is negative, the magnitude of acceleration is decreasing .

I'd try a numerical approach to show how this can be done without calculus (and because I felt like it). Simplify things by letting the cup be a cone with $r=h$ , then $V=\frac{1}{3}\pi r^{3}$ . Solve for $r$ and pull out the constant terms which can be ignored and $r=C \sqrt[3]{V}$ . Since the volume is increasing at a constant rate, let's call it time.

I rounded to two decimals, but it's clear enough that number in the acceleration row are getting closer to zero: The magnitude of acceleration is decreasing.

As the water level rises, the trickle rate $\dot{V} = k$ becomes insignificant next to the volume of the water. If we take the limit as the drip rate becomes very small relative to volume ( $k/V \rightarrow 0$ ) we can see that the rate of change of the water level's velocity should approach zero.

Why? Since we're adding water, the water level's velocity certainly won't ever become negative, but the water level's acceleration is strictly negative. For both to be true, it must be that the water level's downward acceleration is approaching zero as well.

Edit : as it is totally unnecessary to do a calculation here, I've only left the intuitive part of my solution.

V ~ $h^{3}$ - k ~ t (where k is the volume of the imagined cone that would have been below the bottom)

V'(t) ~ 3 $h^{2}$ h'(t) and V'(t) = constant

h'(t) ~ $h^{-2}$ > 0

h''(t) ~ -2 $h^{-3}$ h'(t) ~ - $h^{-5}$ < 0

Since the acceleeration is always negative, the magnitude of acceleration is -h''(t).

h'''(t) ~ 5 $h^{-6}$ h'(t) ~ $h^{-8}$ ~ $(t+k)^{-8/3}$ > 0

Since h''(t) is always increasing, -h''(t) is always decreasing.

It helps to think in terms of extremes. I did it that way without calculations

Let $h$ be the height. $h'$ is inversely proportional to $r^2$ , the area of the surface, so $h'\sim r^{-2}$ , so $h''\sim-2r^{-3}r'$ . $r'>0$ is constant because the cup's a cone, so $h''\sim-2r^{-3}$ . $h''$ changes at rate $6r^{-4}r'$ , which is positive, so $h''$ is decreasing in magnitude.