Solved: PHYSICS Question for NEET

NEET PHYSICSEasy

The displacement of a particle, moving in a straight line, is given by $s = 2t^2 + 2t + 4$ where $s$ is in metres and $t$ in seconds. The acceleration of the particle is:

2 m/s²

4 m/s²

6 m/s²

8 m/s²

Step-by-Step Solution

Velocity ( $v$ ): Instantaneous velocity is defined as the rate of change of displacement with respect to time ( $v = \frac{ds}{dt}$ ) . Given $s = 2t^2 + 2t + 4$ . Differentiating with respect to $t$ : $v = \frac{d}{dt}(2t^2 + 2t + 4) = 4t + 2$ .
Acceleration ( $a$ ): Instantaneous acceleration is defined as the rate of change of velocity with respect to time ( $a = \frac{dv}{dt}$ ) . Differentiating $v$ with respect to $t$ : $a = \frac{d}{dt}(4t + 2) = 4$ .
Result: The acceleration is constant at $4 \text{ m/s}^2$ .

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