A particle moves along a straight line such that its displacement x varies with time t as x = sqrt{t} + 1. The acceleration of the particle is propor…

Given $x = t^{1/2} + 1$. Velocity $v = \frac{dx}{dt} = \frac{1}{2}t^{-1/2}$. Acceleration $a = \frac{dv}{dt} = \frac{1}{2}\left(-\frac{1}{2}\right)t^{-3/2} = -\frac{1}{4}t^{-3/2}$. Since $v = \frac{1}{2}t^{-1/2}$, we have $t^{-1/2} = 2v$, so $t^{-3/2} = (2v)^3 = 8v^3$. Thus, $a = -\frac{1}{4}(8v^3) = -2v^3$, meaning $a \propto v^3$.