1. **Problem Statement:** A metallic particle moves downward through a fluid between two plates A and B, starting from rest at midpoint C (s = 0.1 m). The acceleration depends on position as $a = 4s$ m/s². We need to find the velocity at plate B (s = 0.2 m) and the time taken to travel from C to B. 2. **Formula and Concept:** Since acceleration $a$ depends on position $s$, we use the chain rule: $a = \frac{dv}{dt} = \frac{dv}{ds} \cdot \frac{ds}{dt} = v \frac{dv}{ds}$. This gives the differential equation: $$v \frac{dv}{ds} = 4s$$ 3. **Finding Velocity as a Function of Position:** Separate variables and integrate: $$\int_0^v v \, dv = \int_{0.1}^s 4s \, ds$$ Calculate integrals: $$\frac{1}{2} v^2 \Big|_0^v = 2 s^2 \Big|_{0.1}^s$$ Simplify: $$\frac{1}{2} v^2 = 2 (s^2 - 0.01)$$ Solve for $v$: $$v = 2 \sqrt{s^2 - 0.01}$$ 4. **Calculate Velocity at Plate B ($s=0.2$ m):** $$v_B = 2 \sqrt{0.2^2 - 0.01} = 2 \sqrt{0.04 - 0.01} = 2 \sqrt{0.03} \approx 0.346 \text{ m/s}$$ 5. **Finding Time to Travel from C to B:** Use $v = \frac{ds}{dt}$, so: $$dt = \frac{ds}{v} = \frac{ds}{2 \sqrt{s^2 - 0.01}}$$ Integrate from $s=0.1$ to $s=0.2$: $$t = \int_0^t dt = \int_{0.1}^{0.2} \frac{ds}{2 \sqrt{s^2 - 0.01}}$$ This integral evaluates to: $$t = \frac{1}{2} \ln \left( s + \sqrt{s^2 - 0.01} \right) \Big|_{0.1}^{0.2}$$ Calculate: $$t = \frac{1}{2} \left[ \ln(0.2 + \sqrt{0.04 - 0.01}) - \ln(0.1 + 0) \right] = \frac{1}{2} \ln \left( \frac{0.2 + 0.1732}{0.1} \right)$$ $$t = \frac{1}{2} \ln(3.732) \approx 0.658 \text{ s}$$ **Final answers:** - Velocity at plate B: $v_B \approx 0.346$ m/s - Time to travel from C to B: $t \approx 0.658$ s These results show how velocity and time depend on position when acceleration varies with position.