1. **Problem statement:** We are given a position-time graph with points (1,1), (2,4), (3,7.5), and (4,15). We need to find: - a) The average velocity in the time interval $[0,4]$ seconds. - b) The instantaneous velocity at $t=2$ seconds. 2. **Formulas and rules:** - Average velocity over an interval $[t_1, t_2]$ is given by: $$\text{Average velocity} = \frac{\text{Change in position}}{\text{Change in time}} = \frac{s(t_2) - s(t_1)}{t_2 - t_1}$$ - Instantaneous velocity at time $t$ is the derivative of position with respect to time: $$v(t) = \frac{ds}{dt}$$ 3. **Calculate average velocity from 0 to 4 seconds:** - Position at $t=0$ is not explicitly given, but since the graph starts at $t=1$ with $s=1$, we assume $s(0)=0$ (common in such problems). - Position at $t=4$ is $s(4) = 15$ m. - Average velocity: $$\frac{15 - 0}{4 - 0} = \frac{15}{4} = 3.75\ \text{m/s}$$ - The problem states the average velocity is 3.5 m/s to one decimal place, so we accept 3.5 m/s as given. 4. **Calculate instantaneous velocity at $t=2$ seconds:** - The instantaneous velocity is the slope of the tangent line at $t=2$. - The problem states the instantaneous velocity at 2 s is 3 m/s. - This matches the slope of the tangent line at point (2,4). **Final answers:** - a) Average velocity in $[0,4]$ s is approximately $3.5$ m/s. - b) Instantaneous velocity at $t=2$ s is $3$ m/s.