Answer:
\[\text{x}=\text{1}0\text{t}{{\text{e}}^{-\text{1}}}\]
Instantaneous velocity, \[\text{v}=\frac{dx}{dt}=\text{
1}0\text{t}(\text{
1}){{\text{e}}^{-\text{t}}}+\text{1}0{{\text{e}}^{-\text{t}}}=\text{
}{{\text{e}}^{-\text{t}}}\left[ \text{1}0\text{ }\text{ 1}0\text{t} \right]\]
For the body to be at rest, v = 0 ; so \[{{\text{e}}^{-\text{t}}}\left(
\text{1}0\text{1}0\text{t} \right)=0\]
t = ¥ or 1 sec.
But t = ¥ is not allowed; so t = 1 second.
Thus \[\text{x}=\text{1}0\times \text{1}\times
{{\text{e}}^{-\text{1}}}=\text{1}0/\text{e metres}\].
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