question_answer

A particle moving along the x-axis has a position given by \[\text{x}=\text{1}0\text{t}{{\text{e}}^{-\text{1}}}\]metres where t is in seconds. How far is the particle from the origin when it momentarily stops? (Do not consider its stop at infinity).

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 \[\text{x}=\text{1}0\times \text{1}\times {{\text{e}}^{-\text{1}}}=\text{1}0/\text{e metres}\].