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,

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

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