How to correctly integrate when acceleration is time dependent?

Question

The problem is as follows: A ball moving in a straight line is experiencing acceleration $a(t)=kt$ until it arrives at a certain length $l$ when some time $t_f$ has passed. The initial speed and position are both $0$

In order to solve this equation I tried using the alternative form $\int v dv = \int a dx$

In which I found the final speed via $\int_{0}^{v_f}v dv=\int_{0}^{l} k t dx$ result being $\frac{1}{2} {v_f}^2 = l k t_f $

This is wrong.

My questions are:

1.How does one go about integrating acceletarion with respect to position, when acceletarion is time dependant.

2.Why does the aformentioned process work when acceletarion is constant.

Answer 1

Just use the definition of acceleration as first derivative of speed:

\begin{equation} a=\frac{dv}{dt}. \end{equation}

Integrating both sides we have

\begin{equation} \int_0^v dv' = \int_0^t dt'\,a(t')=\int_0^t dt'\,kt'=\frac{kt^2}{2}. \end{equation}

Hence, the speed is $v(t)=\frac{kt^2}{2}$. Use the definition of speed as first derivative of position

\begin{equation} v=\frac{dx}{dt}, \end{equation}

and integrate both sides:

\begin{equation} l=\int_0^l dx = \int_0^{t_f} dt\,v(t)=\int_0^{t_f} dt\,\frac{kt^2}{2}=\frac{kt^3_f}{6} \end{equation}

from which $t_f^3=\frac{6l}{k}$. Plug now $t_f$ into $v(t_f)$ and you get the final speed as a function of $l$:

\begin{equation} v_f=\frac{kt^2_f}{2}=\frac{k}{2}\left(\frac{6l}{k}\right)^{\frac{2}{3}}=\frac{1}{2}\left(6l\sqrt{k}\right)^{\frac{2}{3}} \end{equation}