1. **Problem statement:** A particle decelerates from 40 m/s to a speed $V$ over 10 seconds, then decelerates to rest in 16 seconds. Initial deceleration is 2 m/s². 2. **Given:** - Initial velocity $u_1 = 40$ m/s - Time for first deceleration $t_1 = 10$ s - Deceleration $a_1 = -2$ m/s² (negative because it's deceleration) - Time for second deceleration $t_2 = 16$ s - Final velocity $v_2 = 0$ m/s 3. **Find:** (a) Sketch velocity-time graph (described in explanation) (b) Value of $V$ after first deceleration (c) Average velocity for whole journey 4. **Step (b) Find $V$:** Use formula for velocity under constant acceleration: $$v = u + at$$ For first deceleration: $$V = 40 + (-2) \times 10 = 40 - 20 = 20 \text{ m/s}$$ 5. **Step (c) Find average velocity:** First find deceleration $a_2$ for second phase: $$0 = V + a_2 \times 16 \Rightarrow a_2 = \frac{0 - V}{16} = \frac{-20}{16} = -1.25 \text{ m/s}^2$$ Calculate displacement in each phase: - First phase displacement $s_1$: $$s_1 = ut + \frac{1}{2}at^2 = 40 \times 10 + \frac{1}{2}(-2)(10)^2 = 400 - 100 = 300 \text{ m}$$ - Second phase displacement $s_2$: $$s_2 = Vt + \frac{1}{2}a_2 t^2 = 20 \times 16 + \frac{1}{2}(-1.25)(16)^2 = 320 - 160 = 160 \text{ m}$$ Total displacement: $$s = s_1 + s_2 = 300 + 160 = 460 \text{ m}$$ Total time: $$t = 10 + 16 = 26 \text{ s}$$ Average velocity: $$\bar{v} = \frac{s}{t} = \frac{460}{26} \approx 17.69 \text{ m/s}$$ 6. **Summary:** - (a) Velocity-time graph starts at 40 m/s, decreases linearly to 20 m/s over 10 s, then decreases linearly to 0 m/s over 16 s. - (b) $V = 20$ m/s - (c) Average velocity $\approx 17.69$ m/s