(a) $\sqrt{x} = t + 1$ Squaring both sides, we get $x = (t + 1)^{2} = t^{2} + 2t + 1$ Differentiating it w.r.t time $t$, we get $\frac{dx}{dt} = 2t + 2$ Velocity, $v = \frac{dx}{dt} = 2t + 2$
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