1. **Problem statement:** A sprinter runs 100m with two acceleration phases and then constant speed, finishing in time $T$. After the finish, he decelerates over 20m to stop. We find (a) total time $T$ to complete 100m, (b) deceleration during stopping. 2. **Given data:** - Initial speed $u_0=0$ m/s - First acceleration to $v_1=8$ m/s in $t_1=1.5$ s - Second acceleration to $v_2=12$ m/s in $t_2=3$ s - Constant speed $v_2=12$ m/s for remaining distance - Total race distance $d=100$ m - Deceleration distance after finish $d_{dec}=20$ m 3. **Step 1: Calculate accelerations and distances in acceleration phases.** - First acceleration $a_1=\frac{v_1-u_0}{t_1}=\frac{8-0}{1.5}=\frac{8}{1.5}=\frac{16}{3}$ m/s$^2$ - Distance in first phase: $$s_1 = u_0 t_1 + \frac{1}{2} a_1 t_1^2 = 0 + \frac{1}{2} \times \frac{16}{3} \times (1.5)^2 = \frac{8}{3} \times 2.25 = 6$$ m - Second acceleration $a_2=\frac{v_2 - v_1}{t_2} = \frac{12 - 8}{3} = \frac{4}{3}$ m/s$^2$ - Distance in second phase: $$s_2 = v_1 t_2 + \frac{1}{2} a_2 t_2^2 = 8 \times 3 + \frac{1}{2} \times \frac{4}{3} \times 9 = 24 + 6 = 30$$ m 4. **Step 2: Calculate remaining distance and time at constant speed.** - Distance remaining: $$s_3 = 100 - (s_1 + s_2) = 100 - (6 + 30) = 64$$ m - Time at constant speed: $$t_3 = \frac{s_3}{v_2} = \frac{64}{12} = \frac{16}{3} \approx 5.3333 \text{ s}$$ 5. **Step 3: Total time to complete race:** $$T = t_1 + t_2 + t_3 = 1.5 + 3 + \frac{16}{3} = 4.5 + 5.3333 = 9.8333 \text{ s}$$ 6. **Step 4: Deceleration after finish to stop over 20m.** - Initial speed for deceleration $v = 12$ m/s - Final speed $v_f = 0$ m/s - Distance $d_{dec} = 20$ m - Use formula: $$v_f^2 = v^2 + 2 a d_{dec} \implies 0 = 12^2 + 2 a \times 20$$ $$0 = 144 + 40 a \implies a = -\frac{144}{40} = -3.6 \text{ m/s}^2$$ **Final answers:** - (a) Time to complete race $T = 9.83$ seconds (approx) - (b) Deceleration $a = -3.6$ m/s$^2$