A particle is moving along the Y-axis. The po | vedclass

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(C) The position of the particle is given by the function $y(t) = 10 t e^{-2 t}$.To find when the particle stops momentarily,we calculate its velocity

Vedclass · Accepted Answer

$\frac{5}{e} \ m$

Vedclass · Answer

(C) The position of the particle is given by the function $y(t) = 10 t e^{-2 t}$.To find when the particle stops momentarily,we calculate its velocity $v$ by differentiating $y(t)$ with respect to time $t$:$v = \frac{dy}{dt} = \frac{d}{dt} (10 t e^{-2 t})$.Using the product rule,$v = 10 [t \cdot (-2 e^{-2 t}) + e^{-2 t} \cdot 1] = 10 e^{-2 t} (1 - 2 t)$.The particle stops momentarily when $v = 0$,which implies $1 - 2 t = 0$,so $t = \frac{1}{2} \ s$.Now,substitute $t = \frac{1}{2} \ s$ into the position equation to find the distance from the origin:$y(\frac{1}{2}) = 10 	imes (\frac{1}{2}) 	imes e^{-2 	imes (\frac{1}{2})} = 5 	imes e^{-1} = \frac{5}{e} \ m$.

$A$ particle is moving along the $Y$-axis. The position of the particle from the origin as a function of time $(t)$ is given as $y(t) = 10 t e^{-2 t}$. How far is the particle from the origin when it stops momentarily? ($y$ is given in units of metre and $t$ is in units of second)

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