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CGP EDU Academic Team
Published on: September 11, 2026
The displacement x of a particle moving in one dimension under the action of a constant force is related to the time t by the equation $t = \sqrt{x} + 3$ , where x is in meters and t is in seconds. The work done by the force in the first 6 seconds is
Text Solution
Verified by ExpertsThe correct answer is:
C
$x = (t - 3)^2$ ⇒ ⇒ $v = \frac{dx}{dt} = 2(t - 3)$
at $t=0$ ; $v_1 = -6\,m/s$ and at $t=6\,\mathrm{sec}$ , $v_{2} = 6 \, m/s$
so, change in kinetic energy $= W = \frac{1}{2} m v_2^2 - \frac{1}{2} m v_1^2 = 0$
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