# Position of a Particle

A particle of mass m is subject to a force $$F(t)=me^{-bt}$$. The initial position and speed are zero. Find $x(t)$.

Solution: In the given problem $$\ddot{x}=e^{-bt}$$
Integrating this with respect to time gives $$v(t)=-\frac{e^{-bt}}{b}+A$$ ( A is the constant of integration)
We integrate again with respect to x.
$$x(t)=\frac{e^{-bt}}{b^2}+At+B$$ ( B is the constant of integration)
The initial condition $v(0)=0$,gives $\frac{-1}{b}+A=0$ $$\Rightarrow A= \frac{1}{b}$$
The intial condition $$x(0)=0$$, gives $$\frac{1}{b^2}+B=0$$ $$\Rightarrow B=-\frac{1}{b^2}$$

Hence,

$$x(t)=\frac{e^{-bt}}{b^2}+\frac{1}{b}t-\frac{1}{b^2}$$