Question
A particle executes the motion described by $\text{x}(\text{t})=\text{x}_0(1-\text{e}^{-\gamma\text{t}});\text{t}\ge0,\text{x}_0>0.$ Where does the particle start and with what velocity?
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Answer
$\text{x}(\text{t})=\text{x}_0[1-\text{e}^{-\gamma\text{t}}]\ \ \ ...(\text{i})$ $\text{v}(\text{t})=\frac{\text{dx(t)}}{\text{dt}}=\frac{\text{d}}{\text{dt}}[\text{x}_0(1-\text{e}^{\gamma\text{t}})]=+\text{x}_0\gamma\text{e}^{-\gamma\text{t}}\ \ \ ...(\text{ii})$ $\text{a}(\text{t})=\frac{\text{dv}}{\text{dt}}=\frac{\text{d}[+\text{x}_0\gamma^2\text{e}^{-\gamma^4}]}{2}=-\text{x}_0\gamma^2\text{e}^{-\gamma\text{t}}\ \ \ ..(\text{iii})$ $(\text{i})\text{At, t}=0\ \ \text{x}(0)=\text{x}_0[1-\text{e}^0]=\text{x}_0(1-1)=0$ $\text{v}(0)=\text{x}_0\gamma\text{e}^0=\text{x}_0\gamma$ Hence, the particle start from x = 0 with velocity $\text{v}_0=\text{x}_0\gamma.$
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