I've never been so confused (Application of Integral Calculus) Here's a problem on Application of Integral calculus to find the work done in moving a particle. I was able to 'reach' the 'right answer'. But I'm totally confused and utterly dissatisfied with the way I did it. It was like I found the way to end up at given answer and not the other way round.
The Question:
A particle of mass $m$ starts from rest at time $t=0$ and is moved along the $x$-axis with constant acceleration $a$ from $x=0$ to $x=h$ against a variable force of magnitude $F(t)=t^2$. Find the work done.
The answer to this question is given as
$$
\frac{4h\sqrt{3mh}}{3}
$$
My work:
The way to reach the given answer is simple. Let
$$
\begin{aligned}
F(t)&=t^2 \\
&=ma \\
\implies a=\frac{t^2}{m}
\end{aligned}
$$
Integrating this from $0$ to $t$, we get the velocity as a function of $t$ and doing it again, we get displacement as a function of t.
$$
v(t)=\frac{t^3}{3m} \\
x(t)=\frac{t^4}{12m}
$$
Time at which the the particle is at $x=h$ would be
$$
t_h=\sqrt[4]{12mh}
$$
Work done on the particle is the change in its kinetic energy between $t=0$ and $t=t_h$
$$
\begin{aligned}
W &= \frac{1}{2}m\big[v(t_h)\big]^2-\underbrace{\frac{1}{2}m\big[v(0)\big]^2}_{=0} \\
&= \frac{1}{2}m \left\{\frac{t_h^3}{3m}\right\}^2 \\
&= \frac{1}{2}m \left\{\frac{(\sqrt[4]{12mh})^3}{3m}\right\}^2 \\
&= \frac{4h\sqrt{3mh}}{3}
\end{aligned}
$$
My Confusions:
The first equation that I've used:
$$
a=\frac{t^2}{m}
$$
goes against the problem statement. In the problem, $a$ is constant. But here I've taken it as a function of time and consequently, variable (since mass is constant).
My line of thought goes like this: If the particle is moving at a constant acceleration $a$ against the force $F(t)=t^2$, then there must be another force, say $F_1$, which should also be variable with time, that is overwhelming $F(t)$ to make the particle move with that constant acceleration. Right?
$$
ma=F_1-F(t)=ma_1-t^2
$$
where $a_1$ should be variable with time.
If we continue with this, then we won't end up with the answer that is given. I'm really, really confused. Can someone help?
 A: You're right to be suspicious; the question is just all sorts of wrong. The question says

A particle of mass $m$ starts from rest at time $t=0$ and is moved along the $x$-axis with constant acceleration $a$ from $x=0$... blablablablablablabla

Once I read this first part of the sentence, without even thinking, I immediately translate this into an initial value problem for a second order ODE:
\begin{align}
\begin{cases}
x''(\cdot)&=a\\
x'(0)&=0\\
x(0)&=0
\end{cases}
\end{align}
The solution is immediately given as $x(t)=\frac{1}{2}at^2$, and so it takes a total of $t_h=\sqrt{\frac{2h}{a}}$ units of time to travel a displacement of $h$. It doesn't matter how many forces or what kinds of forces act on the particle. If the particle has a constant acceleration of $a$, then this is the trajectory of the particle. That's it.
Next, the net work done along the path from $x=0$ to $x=h$ is simply the change in kinetic energy (as you've correctly identified), and in this case, it is
\begin{align}
\frac{m}{2}[x'(t_h)]^2-\frac{m}{2}[x'(0)]^2=\frac{m}{2}[at_h]^2-\frac{m}{2}(0)^2=mah.
\end{align}
If you wish to calculate only the work done by the force $F$, then it is
\begin{align}
\text{work done by $F$ on the particle}&=\int_0^{t_h}F(t)\cdot x'(t)\,dt=\int_0^{t_h}t^2\cdot at\,dt=\frac{at_h^4}{4},
\end{align}
which again is not the "intended" answer. In the current form of how the question is phrased, you're right to introduce a new force $F_1$, but even the work done by $F_1$ is not equal to the "intended" answer. i.e none of the forces $F, F_1, F_1-F$ have work equal to the "intended" answer. This just goes to show how poorly the question is worded.

If one wants to obtain the "intended" answer, here's how the question should have been phrased:

Consider a particle of mass $m$ which undergoes motion along the $x$-axis; let $x(t)$ denote the position of the particle at time $t$. Suppose that the particle is under the influence of a force $F$, such that $F(x(t))=kt^2$ for some constant $k>0$ (to ensure that $F$ has the right units) and that the particle starts off at rest at the origin. Find the work done by the force $F$ on the particle as it is displaced from $0$ to some final displacement $h$.

I'm pretty sure here the answer works out to be $\frac{4}{\sqrt{3}}\sqrt{mh^3k}$ (the same as what you have, up to the factor of $k$).
