1. **Problem Statement:** We are given the position $x$ of a slider at various times $t$ and need to compute the velocity and acceleration at $t=0.4$ seconds using the Centered Difference Approximation with step sizes $h=0.1$ and $h=0.2$. 2. **Formulas:** - Velocity (first derivative) using centered difference: $$v(t) \approx \frac{x(t+h) - x(t-h)}{2h}$$ - Acceleration (second derivative) using centered difference: $$a(t) \approx \frac{x(t+h) - 2x(t) + x(t-h)}{h^2}$$ 3. **Given data:** \begin{align*} t & : 0, 0.1, 0.2, 0.3, 0.4, 0.5, 0.6 \\ x(t) & : 30.13, 31.62, 32.87, 33.64, 33.95, 33.81, 33.24 \end{align*} 4. **Calculate for $h=0.1$:** - Velocity at $t=0.4$: $$v(0.4) \approx \frac{x(0.5) - x(0.3)}{2 \times 0.1} = \frac{33.81 - 33.64}{0.2} = \frac{0.17}{0.2} = 0.85\ \text{cm/s}$$ - Acceleration at $t=0.4$: $$a(0.4) \approx \frac{x(0.5) - 2x(0.4) + x(0.3)}{(0.1)^2} = \frac{33.81 - 2(33.95) + 33.64}{0.01} = \frac{33.81 - 67.9 + 33.64}{0.01} = \frac{-0.45}{0.01} = -45\ \text{cm/s}^2$$ 5. **Calculate for $h=0.2$:** - Velocity at $t=0.4$: $$v(0.4) \approx \frac{x(0.6) - x(0.2)}{2 \times 0.2} = \frac{33.24 - 32.87}{0.4} = \frac{0.37}{0.4} = 0.925\ \text{cm/s}$$ - Acceleration at $t=0.4$: $$a(0.4) \approx \frac{x(0.6) - 2x(0.4) + x(0.2)}{(0.2)^2} = \frac{33.24 - 2(33.95) + 32.87}{0.04} = \frac{33.24 - 67.9 + 32.87}{0.04} = \frac{-1.79}{0.04} = -44.75\ \text{cm/s}^2$$ **Final answers:** - For $h=0.1$: velocity $=0.85$ cm/s, acceleration $=-45$ cm/s$^2$ - For $h=0.2$: velocity $=0.925$ cm/s, acceleration $=-44.75$ cm/s$^2$