Solved: PHYSICS Question for NEET

NEET PHYSICSEasy

The distance travelled by a particle is proportional to the squares of time, then the particle travels with:

Uniform acceleration

Uniform velocity

Increasing acceleration

Decreasing velocity

Mathematical Relationship: Given that the distance ( $s$ ) is proportional to the square of time ( $t^2$ ), we can express this as $s = kt^2$ , where $k$ is a constant.
Velocity: The instantaneous velocity ( $v$ ) is the first derivative of distance with respect to time ( $v = \frac{ds}{dt}$ ). Differentiating $s = kt^2$ , we get $v = 2kt$ .
Acceleration: The acceleration ( $a$ ) is the rate of change of velocity ( $a = \frac{dv}{dt}$ ). Differentiating $v = 2kt$ , we get $a = 2k$ .
Conclusion: Since $k$ is a constant, $2k$ is also a constant. This implies the acceleration is uniform (constant) throughout the motion. This aligns with the kinematic equation for a body starting from rest ( $u=0$ ): $s = \frac{1}{2}at^2$ , where $s \propto t^2$ .

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