1. **Problem statement:** We have the speed of a jet at 5-second intervals from 0 to 25 seconds and want to estimate the length of runway used for takeoff (part a). Then, we need to explain if this estimate is an underestimate or overestimate assuming smooth acceleration (part b). 2. **Given data:** Time (s): $0, 5, 10, 15, 20, 25$ Speed (m/s): $2, 5, 10, 18, 28, 42$ 3. **Formula and approach:** The length of runway is the distance traveled during acceleration. Distance can be estimated by integrating speed over time. Since we have discrete speed data, we use the trapezium rule to approximate distance: $$\text{Distance} \approx \sum_{i=1}^n \frac{(v_{i-1} + v_i)}{2} \times \Delta t$$ where $v_i$ is speed at time $t_i$ and $\Delta t = 5$ seconds. 4. **Calculate each trapezium area:** Between 0 and 5 s: $$\frac{2 + 5}{2} \times 5 = \frac{7}{2} \times 5 = 17.5$$ Between 5 and 10 s: $$\frac{5 + 10}{2} \times 5 = \frac{15}{2} \times 5 = 37.5$$ Between 10 and 15 s: $$\frac{10 + 18}{2} \times 5 = \frac{28}{2} \times 5 = 70$$ Between 15 and 20 s: $$\frac{18 + 28}{2} \times 5 = \frac{46}{2} \times 5 = 115$$ Between 20 and 25 s: $$\frac{28 + 42}{2} \times 5 = \frac{70}{2} \times 5 = 175$$ 5. **Sum all distances:** $$17.5 + 37.5 + 70 + 115 + 175 = 415$$ meters So, the estimated length of runway used is $415$ meters. 6. **Part (b) explanation:** Since the jet accelerated smoothly, the speed increased continuously, not abruptly. The trapezium rule uses straight lines between speed points, which slightly underestimates the area under a curve that is concave up (speed increasing at an increasing rate). Therefore, the estimate of $415$ meters is an **underestimate** of the actual runway length used. **Final answers:** (a) Length of runway used $= 415$ meters (b) This is an underestimate of the actual length because the jet's acceleration was smooth and increasing.