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 $y(t)=10t{e}^{-2t}$. 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 $y(t)=10t{e}^{-2t}$
Velocity of the particle is given as
$v=\frac{d}{dt}y(t)=\frac{d}{dt}\left(10t{e}^{-2t}\right)$
$=10\left[t{e}^{-2t}(-2)+e^{-2t}\cdot 1\right]$
$=10\left[-2t{e}^{-2t}+e^{-2t}\right]=10e^{-2t}(1-2f)$
When body stops, then
$[v=0\Rightarrow e^{-2t}(1-2f)=0\Rightarrow 1-2t=0$
$t=\frac{1}{2}s$
From Eq. (i), displacement of the particle from origin is given as
$y\left(\frac{1}{2}\right)=10\times \frac{1}{2}\times e^{-2\times \frac{1}{2}}=5e^{-1}=\frac{5}{e}m$
$\Rightarrow \frac{\sec \left(\frac{y}{x}\right)}{xy}=C_1$
$\Rightarrow \sec \left(\frac{y}{x}\right)=C_{1}xy$
$\Rightarrow \frac{1}{\cos \left(\frac{y}{x}\right)}=\frac{C}{xy}$
$\left(\because \text{ where }\frac{1}{C_1}=C\right)$