Solved: PHYSICS Question for NEET

NEET PHYSICSEasy

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

$0 \text{ m/s}^2$

$5 \text{ m/s}^2$

$-4 \text{ m/s}^2$

$-8 \text{ m/s}^2$

Step-by-Step Solution

Concept: Acceleration is the second time derivative of the position vector. We can find the acceleration components ( $a_x$ and $a_y$ ) by differentiating the $x$ and $y$ coordinates twice with respect to time .
x-component:

Position: $x = 5t - 2t^2$
Velocity ( $v_x = dx/dt$ ): $v_x = \frac{d}{dt}(5t - 2t^2) = 5 - 4t$
Acceleration ( $a_x = dv_x/dt$ ): $a_x = \frac{d}{dt}(5 - 4t) = -4 \text{ m/s}^2$

y-component:

Position: $y = 10t$
Velocity ( $v_y = dy/dt$ ): $v_y = \frac{d}{dt}(10t) = 10$
Acceleration ( $a_y = dv_y/dt$ ): $a_y = \frac{d}{dt}(10) = 0 \text{ m/s}^2$

Total Acceleration: The acceleration vector is $\mathbf{a} = a_x \hat{i} + a_y \hat{j} = -4 \hat{i} + 0 \hat{j}$ . Since the question asks for the scalar value (magnitude usually, but options suggest a single component or value), the acceleration is $-4 \text{ m/s}^2$ (acting entirely in the x-direction).

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