1. **Problem statement:** Find the acceleration of a particle moving along the x-axis at the time when it reaches its maximum positive displacement, given velocity $v_x = 32.0 t - 2.00 t^3$ for $t > 0$. 2. **Key concepts:** - Velocity $v_x$ is the derivative of position $x$ with respect to time $t$: $v_x = \frac{dx}{dt}$. - Acceleration $a_x$ is the derivative of velocity with respect to time: $a_x = \frac{dv_x}{dt}$. - Maximum displacement occurs when velocity changes from positive to negative, i.e., when $v_x = 0$ and velocity is about to decrease. 3. **Find the time $t$ when displacement is maximum:** Set velocity to zero: $$ 32.0 t - 2.00 t^3 = 0 $$ Factor out $t$: $$ t(32.0 - 2.00 t^2) = 0 $$ Solutions: - $t = 0$ (initial time, ignore since $t > 0$) - $32.0 - 2.00 t^2 = 0 \Rightarrow 2.00 t^2 = 32.0 \Rightarrow t^2 = 16 \Rightarrow t = 4$ s (positive root) 4. **Calculate acceleration at $t=4$ s:** Differentiate velocity: $$ a_x = \frac{dv_x}{dt} = \frac{d}{dt}(32.0 t - 2.00 t^3) = 32.0 - 6.00 t^2 $$ Substitute $t=4$: $$ a_x = 32.0 - 6.00 \times 4^2 = 32.0 - 6.00 \times 16 = 32.0 - 96.0 = -64.0 \text{ m/s}^2 $$ 5. **Interpretation:** The acceleration at the time of maximum positive displacement is $-64.0$ m/s$^2$, indicating the particle is decelerating as it reaches the peak position. **Final answer:** $$a_x = -64.0 \text{ m/s}^2 \text{ at } t=4 \text{ s}$$