Question

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 meter and $t$ is in units of sec.)

• A. $5$ meter
• B. $5e$ meter
• C. $\frac{5}{e}$ meter
• D. $10$ meter

Answer 1

Answer: (C)

The position of the particle is given as $y(t)=10 t e^{-2 t}$
Velocity of the particle is given as
$v =\frac{d}{d t} y(t)=\frac{d}{d t}\left(10 t e^{-2 t}\right) $
$=10\left[t e^{-2 t}(-2)+e^{-2 t} \cdot 1\right]$
$=10\left[-2 t e^{-2 t}+e^{-2 t}\right]=10 e^{-2 t}(1-2 f)$
When body stops, then
${\left[v=0 \Rightarrow e^{-2 t}(1-2 f)=0 \Rightarrow 1-2 t=0\right.}$
$t=\frac{1}{2} s$
From Eq. (i), displacement of the particle from origin is given as
$y\left(\frac{1}{2}\right)=10 \times \frac{1}{2} \times e^{-2 \times \frac{1}{2}}=5 e^{-1}=\frac{5}{e} m$
$\Rightarrow \frac{\sec \left(\frac{y}{x}\right)}{x y}=C_{1} $
$\Rightarrow \sec \left(\frac{y}{x}\right)=C_{1} x y $
$\Rightarrow \frac{1}{\cos \left(\frac{y}{x}\right)}=\frac{C}{x y}$
$\left(\because \text { where } \frac{1}{C_{1}}=C\right)$