1. **State the problem:** We are given velocity values $V$ of a body at different times $t$ and need to find the acceleration at $t=1.0$ seconds. 2. **Recall the formula:** Acceleration $a$ is the rate of change of velocity with respect to time, mathematically given by $$a = \frac{dV}{dt}$$ Since we have discrete data points, we approximate acceleration using the average rate of change (difference quotient): $$a(t) \approx \frac{V(t+\Delta t) - V(t)}{\Delta t}$$ 3. **Identify values:** At $t=1.0$, velocity $V=43.1$. At $t=1.1$, velocity $V=47.7$. The time difference $\Delta t = 1.1 - 1.0 = 0.1$ seconds. 4. **Calculate acceleration:** $$a(1.0) = \frac{47.7 - 43.1}{0.1} = \frac{4.6}{0.1}$$ 5. **Simplify:** $$a(1.0) = 46$$ 6. **Interpretation:** The acceleration at $t=1.0$ seconds is $46$ units of velocity per second (e.g., meters per second squared if velocity is in m/s). **Final answer:** $$\boxed{a(1.0) = 46}$$