1. **State the problem:** We have a particle P moving along a line with displacement from a fixed point O given by $$s = 4t^2 + \frac{125}{t}$$ for time $t \geq 1$. We need to find the distance of P from O when the velocity $v = 0$. 2. **Recall the formula for velocity:** Velocity is the derivative of displacement with respect to time: $$v = \frac{ds}{dt}$$ 3. **Differentiate $s$ with respect to $t$:** $$s = 4t^2 + 125t^{-1}$$ $$v = \frac{d}{dt}(4t^2) + \frac{d}{dt}(125t^{-1}) = 8t - 125t^{-2}$$ 4. **Set velocity to zero and solve for $t$:** $$0 = 8t - \frac{125}{t^2}$$ Multiply both sides by $t^2$ to clear the denominator: $$0 = 8t^3 - 125$$ $$8t^3 = 125$$ $$t^3 = \frac{125}{8}$$ $$t = \sqrt[3]{\frac{125}{8}} = \frac{5}{2} = 2.5$$ 5. **Find displacement $s$ at $t=2.5$:** $$s = 4(2.5)^2 + \frac{125}{2.5} = 4(6.25) + 50 = 25 + 50 = 75$$ **Final answer:** The distance of P from O when $v=0$ is **75 metres**.