1. **State the problem:** We need to estimate Jeremy's acceleration at $t=10$ seconds and his average speed from $t=0$ to $t=40$ seconds based on the velocity-time graph. 2. **Recall formulas:** - Acceleration $a$ is the rate of change of velocity with respect to time: $$a = \frac{\Delta v}{\Delta t}$$ - Average speed over a time interval is total distance divided by total time. Since velocity is given, average speed is the average value of velocity over the interval. 3. **Estimate acceleration at $t=10$ seconds:** From the graph description, velocity rises steeply from 0 at $t=0$ to about 50 m/s near $t=40$ seconds, leveling off near 50 m/s. To estimate acceleration at $t=10$ seconds, approximate the slope of the velocity curve near $t=10$. Assuming velocity at $t=0$ is 0 m/s and at $t=10$ is about 40 m/s (since velocity rises steeply), $$a \approx \frac{40 - 0}{10 - 0} = \frac{40}{10} = 4\ \text{m/s}^2$$ 4. **Estimate average speed from $t=0$ to $t=40$ seconds:** Velocity increases from 0 to about 50 m/s over 40 seconds, then remains constant. Average speed is approximately the average of initial and final velocities over this interval: $$\text{Average speed} = \frac{0 + 50}{2} = 25\ \text{m/s}$$ Rounded to two significant figures: 25 m/s. **Final answers:** - a) Acceleration at $t=10$ seconds is approximately $4.0\ \text{m/s}^2$. - b) Average speed from $t=0$ to $t=40$ seconds is approximately $25\ \text{m/s}$.