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(D) Given the displacement equation: $\sqrt{x} = t + 7$.Squaring both sides,we get: $x = (t + 7)^2 = t^2 + 14t + 49$.The velocity $v$ is the rate of c

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The particle moves with constant acceleration

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(D) Given the displacement equation: $\sqrt{x} = t + 7$.Squaring both sides,we get: $x = (t + 7)^2 = t^2 + 14t + 49$.The velocity $v$ is the rate of change of displacement with respect to time: $v = \frac{dx}{dt} = \frac{d}{dt}(t^2 + 14t + 49) = 2t + 14$.The acceleration $a$ is the rate of change of velocity with respect to time: $a = \frac{dv}{dt} = \frac{d}{dt}(2t + 14) = 2 \, m/s^2$.Since the acceleration $a = 2 \, m/s^2$ is a constant value,the particle moves with constant acceleration.

If the displacement of a particle varies with time as $\sqrt{x} = t + 7$,then

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