1. **Problem statement:** A particle moves along a straight line from a fixed point O with acceleration $a$ proportional to the square of its displacement $r$ from O, given by $$a = \frac{3}{8} r^2.$$ We need to find the relation between the velocity $v$ of the particle and its displacement $r$. 2. **Formula and approach:** Acceleration $a$ is the derivative of velocity $v$ with respect to time $t$, i.e., $$a = \frac{dv}{dt}.$$ Velocity $v$ is the derivative of displacement $r$ with respect to time, $$v = \frac{dr}{dt}.$$ Using the chain rule, acceleration can be expressed as $$a = \frac{dv}{dt} = \frac{dv}{dr} \cdot \frac{dr}{dt} = v \frac{dv}{dr}.$$ This allows us to write the differential equation: $$v \frac{dv}{dr} = \frac{3}{8} r^2.$$ 3. **Solving the differential equation:** Multiply both sides by $dr$: $$v dv = \frac{3}{8} r^2 dr.$$ 4. **Integrate both sides:** $$\int v dv = \int \frac{3}{8} r^2 dr.$$ 5. **Perform the integrations:** $$\frac{v^2}{2} = \frac{3}{8} \cdot \frac{r^3}{3} + C,$$ where $C$ is the constant of integration. 6. **Simplify the right side:** $$\frac{v^2}{2} = \frac{3}{8} \cdot \frac{r^3}{3} + C = \frac{r^3}{8} + C.$$ 7. **Multiply both sides by 2:** $$v^2 = \cancel{2} \cdot \frac{r^3}{8} \cdot \cancel{1} + 2C = \frac{r^3}{4} + 2C.$$ 8. **Final relation:** $$v^2 = \frac{r^3}{4} + K,$$ where $K = 2C$ is a constant determined by initial conditions. **Interpretation:** The square of the velocity is proportional to the cube of the displacement plus a constant. This relation connects velocity and displacement for the particle under the given acceleration.