1. **State the problem:** We have a velocity-time graph for a motorcycle braking. We need to estimate: a) The deceleration at $t=4$ seconds. b) The distance travelled in the first 2 seconds. 2. **Recall formulas and concepts:** - Deceleration (acceleration) is the rate of change of velocity with respect to time: $$a = \frac{\Delta v}{\Delta t}$$ - Distance travelled is the area under the velocity-time graph. 3. **Part a: Estimate deceleration at $t=4$ seconds** - From the graph, approximate velocity at $t=2$ is about 16 m/s. - Velocity at $t=4$ is about 12 m/s. - Velocity at $t=6$ is about 0 m/s. Calculate average deceleration between $t=2$ and $t=4$: $$a_1 = \frac{12 - 16}{4 - 2} = \frac{-4}{2} = -2\ \text{m/s}^2$$ Calculate average deceleration between $t=4$ and $t=6$: $$a_2 = \frac{0 - 12}{6 - 4} = \frac{-12}{2} = -6\ \text{m/s}^2$$ Deceleration at $t=4$ is roughly the average of these two: $$a = \frac{-2 + (-6)}{2} = \frac{-8}{2} = -4\ \text{m/s}^2$$ 4. **Part b: Estimate distance travelled in first 2 seconds** - Distance is area under velocity-time graph from $t=0$ to $t=2$. - Approximate velocity at $t=0$ is 18 m/s, at $t=2$ is 16 m/s. - Approximate area as trapezium: $$\text{Distance} = \frac{(v_0 + v_2)}{2} \times (2 - 0) = \frac{18 + 16}{2} \times 2 = 17 \times 2 = 34\ \text{m}$$ **Final answers:** - a) Deceleration at $t=4$ is approximately $4$ m/s$^2$ (slowing down, so negative acceleration). - b) Distance travelled in first 2 seconds is approximately $34$ metres.