# Particle with mass m and force F(t). Show that x = x(t)

kafn8

## Homework Statement

A particle of mass m is initially at rest at x = 0.
It is acted upon by a force $F = A cosh (\beta t)$ (1)

A) Show that at very small values of t, the position is approximately given by $x(t) = \frac{1}{2}\frac{F_0}{m}t^2$ (2),
where $F_0$ is the force at $t =0$

## The Attempt at a Solution

If $F(t) = Acosh(\beta t) = ma(t)$
then $a(t) = \frac{A}{m}cosh(\beta t)$

Integrating twice yields the position such that
$x(t) = \frac{A}{\beta^2 m}\left[ cosh(\beta t) - 1 \right]$, (3)

Also, $F_0(t=0)=m\left[\frac{A}{m}cosh(\beta(0)) \right]=A$

With that out of the way, I've tried taking the limit of (3) as $t \rightarrow 0$ but end up with the following:
$$\lim_{t \rightarrow 0}x(t) = \lim_{t \rightarrow 0} \frac{A}{\beta^2 m}\left[ cosh(\beta (0)) - 1 \right]$$
$$= \frac{A}{\beta^2 m}\left[ (1) - 1 \right]$$
$$= 0$$

But all this says is that for very small values of time, the particle barely moves away from x=0. This does not directly confirm that (2) is a good approximation. Any guidance is much needed and greatly appreciated!