Distance Covered in Arithmetic Progression

First speed $v_1 = 960\text{ m/min} = 57.6\text{ km/h}$. First time $t_1 = 30\text{ min} = 0.5\text{ h}$. Let the AP of times be $t_1, t_2, t_3, t_4$. Since total time = 3 hours, $\sum t_i = 4 t_1 + 6 d_t = 2 + 6 d_t = 3 \implies d_t = 1/6\text{ h} = 10\text{ min}$. So times are $0.5, 2/3, 5/6, 1.0$ hours. Let speeds be $v_1, v_1+d_v, v_1+2d_v, v_1+3d_v$. Total distance $D = v_1 t_1 + v_2 t_2 + v_3 t_3 + v_4 t_4 = 224\text{ km}$. $57.6(0.5) + (57.6+d_v)(2/3) + (57.6+2d_v)(5/6) + (57.6+3d_v)(1.0) = 224$. $28.8 + 38.4 + (2/3)d_v + 48 + (5/3)d_v + 57.6 + 3d_v = 224$. $172.8 + (16/3)d_v = 224 \implies (16/3)d_v = 51.2 \implies d_v = 9.6\text{ km/h}$. For the fourth part: $v_4 = 57.6 + 3(9.6) = 86.4\text{ km/h}$, $t_4 = 1\text{ h}$. Distance in fourth part = $86.4\text{ km} = 86400\text{ meters}$. Wait, let's re-verify options: Option 3 is 86400. Option 1 is 76800. Let's check $v_4 \times t_4$: $86.4 \times 1 = 86.4$ km = 86400 m, which is option 3.