\nLet the time taken for them to meet be T hours. The distance between A and B is 72 km.
\nMan 1 starts from A at a constant speed of 4 kmph. In T hours, Man 1 covers a distance of 4T km.
\nMan 2 starts from B. In the first hour, he covers 2 km. In the second hour, 2.5 km. In the third, 3 km, and so on. This is an arithmetic progression of distances covered per hour with first term a = 2 and common difference d = 0.5.
\nThe distance covered by Man 2 in T hours is the sum of the first T terms of this AP:
\nSum = (T\/2) * [2a + (T-1)d] = (T\/2) * [2*2 + (T-1)*0.5] = (T\/2) * [4 + 0.5T – 0.5] = (T\/2) * [3.5 + 0.5T].
\nWhen they meet, the sum of the distances covered by both men equals the total distance:
\nDistance by Man 1 + Distance by Man 2 = 72
\n4T + (T\/2) * (3.5 + 0.5T) = 72
\n4T + 1.75T + 0.25T^2 = 72
\n0.25T^2 + 5.75T – 72 = 0
\nMultiplying by 4 to clear decimals:
\nT^2 + 23T – 288 = 0
\nUsing the quadratic formula T = [-b \u00b1 sqrt(b^2 – 4ac)] \/ 2a:
\nT = [-23 \u00b1 sqrt(23^2 – 4*1*(-288))] \/ 2*1
\nT = [-23 \u00b1 sqrt(529 + 1152)] \/ 2
\nT = [-23 \u00b1 sqrt(1681)] \/ 2
\nSince sqrt(1681) = 41 (as 40^2=1600, 41^2=1681), and time must be positive:
\nT = (-23 + 41) \/ 2 = 18 \/ 2 = 9 hours.
\nThey meet after 9 hours.
\nDistance covered by Man 1 in 9 hours = 4 kmph * 9 hours = 36 km from A.
\nDistance covered by Man 2 in 9 hours = Sum of 9 terms of AP (a=2, d=0.5) = (9\/2) * [2*2 + (9-1)*0.5] = 4.5 * [4 + 8*0.5] = 4.5 * [4 + 4] = 4.5 * 8 = 36 km from B.
\nSince they meet after covering 36 km from A and 36 km from B, which adds up to the total distance of 72 km, they meet exactly midway between A and B.
\n<\/section>\n