PYQ NEET- Motion In A Plane L-3

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) 0

B) $5 \mathrm{~m} / \mathrm{s}^2$

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

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

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

Sol:#

Given, $x=5 t-2 t^2$

Velocity of the particle,

$$ v_x=\frac{d x}{d t}=\frac{d}{d t}\left(5 t-2 t^2\right)=5-4 t $$

Acceleration, $a_x=\frac{d}{d t} v_x=-4 \mathrm{~ms}^{-2}$ Also, $$ y=10 t $$

Velocity, $$ v_y=\frac{d y}{d t}=10 $$

$\therefore$ Acceleration $a_y=\frac{d v_y}{d t}=0$

$\therefore$ Net acceleration of the particle,

$$ \mathbf{a}{\text {net }}=a_x \hat{\mathbf{i}}+a_y \hat{\mathbf{j}}=\left(-4 \mathrm{~ms}^{-2}\right) \hat{\mathbf{i}} $$ or $\quad \mathbf{a}{\text {net }}=-4 \hat{\mathbf{i}} \mathrm{ms}^{-2}$