1. **Problem statement:** We have two athletes, A and B, running in a relay race. Athlete A runs at a constant velocity of 10 m/s until he starts to decelerate at 5 m/s² and eventually stops. Athlete B starts from rest with constant acceleration until reaching 10 m/s. The baton exchange happens 2 seconds after B starts running, when both have velocity 8 m/s and B is 1 m ahead of A. We need to find: (i) The time $t$ when A starts to decelerate. (ii) The distance between A and B at the instant B starts running. --- 2. **Relevant formulas and rules:** - Velocity under constant acceleration: $v = u + at$ - Displacement under constant acceleration: $s = ut + \frac{1}{2}at^2$ - For constant velocity, displacement: $s = vt$ - Deceleration is negative acceleration. --- 3. **Part (i): Find when A starts to decelerate** - Athlete A runs at constant velocity 10 m/s until deceleration. - At deceleration start time $t = t_1$, velocity is still 10 m/s. - After $t_1$, A decelerates at $a = -5$ m/s² until velocity reaches 0. - From the graph, at $t=2$ s, velocity of A is 8 m/s (since B has 8 m/s and they exchange baton). Using velocity equation during deceleration: $$v = u + at$$ Here, $v=8$, $u=10$, $a=-5$, and $t = 2 - t_1$ (time since deceleration started). So, $$8 = 10 - 5(2 - t_1)$$ $$8 = 10 - 10 + 5t_1$$ $$8 = 5t_1$$ $$t_1 = \frac{8}{5} = 1.6 \text{ seconds}$$ **Answer (i):** A starts to decelerate at $t=1.6$ seconds. --- 4. **Part (ii): Calculate distance between A and B when B starts running (at $t=0$)** - Let’s define positions at $t=0$ as $x_A(0)$ and $x_B(0)$. - We want $d = x_B(0) - x_A(0)$. - At $t=2$ s, B is 1 m ahead of A: $$x_B(2) - x_A(2) = 1$$ - Find $x_A(2)$: From $t=0$ to $t=1.6$, A runs at 10 m/s: $$s_1 = 10 \times 1.6 = 16 \text{ m}$$ From $t=1.6$ to $t=2$, A decelerates from 10 m/s to 8 m/s: - Time interval: $0.4$ s - Displacement during deceleration: $$s_2 = ut + \frac{1}{2}at^2 = 10 \times 0.4 + \frac{1}{2}(-5)(0.4)^2 = 4 - 0.4 = 3.6 \text{ m}$$ Total displacement of A from $t=0$ to $t=2$: $$x_A(2) - x_A(0) = s_1 + s_2 = 16 + 3.6 = 19.6 \text{ m}$$ - Find $x_B(2)$: B starts from rest with constant acceleration $a_B$ until velocity reaches 8 m/s at $t=2$: Using $v = u + at$: $$8 = 0 + a_B \times 2 \Rightarrow a_B = 4 \text{ m/s}^2$$ Displacement of B from $t=0$ to $t=2$: $$x_B(2) - x_B(0) = ut + \frac{1}{2}at^2 = 0 + \frac{1}{2} \times 4 \times 2^2 = 8 \text{ m}$$ - Using the relative position at $t=2$: $$x_B(2) - x_A(2) = 1$$ Substitute: $$(x_B(0) + 8) - (x_A(0) + 19.6) = 1$$ $$x_B(0) - x_A(0) = 1 + 19.6 - 8 = 12.6 \text{ m}$$ **Answer (ii):** The distance between B and A at $t=0$ is 12.6 meters, with B ahead. --- **Final answers:** (i) $t = 1.6$ seconds (ii) Distance between A and B at $t=0$ is 12.6 meters