Step 1: Convert Units
$v_1 = 960$ m/min $= \frac{960 \times 60}{1000}$ km/hr $= 57.6$ km/hr.
$t_1 = 30$ min $= 0.5$ hours.

Step 2: Find the Times ($t_1, t_2, t_3, t_4$)
Total time $= t_1 + t_2 + t_3 + t_4 = 3$ hours.
Since they are in AP: $0.5 + (0.5+d) + (0.5+2d) + (0.5+3d) = 3$.
$2 + 6d = 3 \implies 6d = 1 \implies d = 1/6$ hours (10 mins).
Times: $t_1=30, t_2=40, t_3=50, t_4=60$ mins.

Step 3: Solve for Speeds
Let common difference for speeds be $x$. Speeds are $57.6, 57.6+x, 57.6+2x, 57.6+3x$.
$\sum d_i = 224 \implies \sum v_i t_i = 224$.
$57.6(0.5) + (57.6+x)\frac{2}{3} + (57.6+2x)\frac{5}{6} + (57.6+3x)(1) = 224$.
$28.8 + 38.4 + \frac{2}{3}x + 48 + \frac{5}{3}x + 57.6 + 3x = 224$.
$172.8 + \frac{16}{3}x = 224 \implies \frac{16}{3}x = 51.2 \implies x = 9.6$ km/hr.

Step 4: Calculate Fourth Distance
$v_4 = v_1 + 3x = 57.6 + 3(9.6) = 86.4$ km/hr.
$t_4 = 1$ hour.
Distance $d_4 = 86.4$ km $= \mathbf{86400}$ meters.