Previous Year NEET Question- Optics L-4

Question: The $x$ and $y$ coordinates of the particle at any time are $x=5 t-2 t^2$ and $y=10 t$ respectively, where $x$ and $y$ are in metres and $t$ in seconds. The acceleration of the particle at $t=2 \mathrm{~s}$ is#

A) $5 \mathrm{~m} \mathrm{~s}^{-2}$

B) $-4 \mathrm{~m} \mathrm{~s}^{-2}$

C) $-8 \mathrm{~m} \mathrm{~s}^{-2}$

D) 0

Answer: $-4 \mathrm{~m} \mathrm{~s}^{-2}$#

Sol:#

$$ \begin{aligned} & \mathrm{x}=5 \mathrm{t}-2 \mathrm{t}^2 \text { and } \mathrm{y}=10 \mathrm{t} \ & \frac{d x}{d t}=5-4 \mathrm{t}, \frac{d y}{d t}=10 \ & \therefore \mathrm{v}{\mathrm{x}}=5-4 \mathrm{t}, \mathrm{v}{\mathrm{y}}=10 \ & \frac{d v_x}{d t}=-4, \frac{d v_y}{d t}=0 \ & \therefore a_x=-4, a_y=0 \ & \vec{a}=a_x \hat{i}+a_y \hat{j}=-4 \hat{j} \mathrm{~m} / \mathrm{s}^2 \end{aligned} $$ $\therefore$ Acceleration of the particle at $\mathrm{t}=2 \mathrm{~s}$ is $-4 \mathrm{~m} / \mathrm{s}^2$