1. **State the problem:** We have a particle P moving along a line with displacement from point O given by $$s = 4t^2 + \frac{125}{t}$$ for $$t \geq 1$$ seconds. 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 the displacement function:** $$s = 4t^2 + 125t^{-1}$$ Using the power rule: $$\frac{ds}{dt} = 8t - 125t^{-2} = 8t - \frac{125}{t^2}$$ 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: $$0 = 8t^3 - 125$$ Rewrite: $$8t^3 = 125$$ Divide both sides by 8: $$t^3 = \frac{125}{8}$$ Take the cube root: $$t = \sqrt[3]{\frac{125}{8}} = \frac{\sqrt[3]{125}}{\sqrt[3]{8}} = \frac{5}{2} = 2.5$$ 5. **Find the displacement at $$t=2.5$$:** $$s = 4(2.5)^2 + \frac{125}{2.5} = 4(6.25) + 50 = 25 + 50 = 75$$ 6. **Conclusion:** The distance of P from O when velocity is zero is $$\boxed{75}$$ metres.