Distance between A and B is 72 km. Two men started walking from A and B at the same time towards each other. The person who started from A travelled uniformly with average speed of 4 km/hr. While the other man travelled with varying speed as follows: in the first hour his speed was 2 km/hr, in the second hour it was 2.5 km/hr, in the third hour it was 3 km/hr, and so on. When will they meet each other?
Explanation:
Since A and B are moving in opposite directions, we will add their speeds to calculate the effective speeds. In other words, in the first hour they would effectively cover a distance of (4 + 2) = 6 km towards each other. In the second hours, they would effectively cover a distance of (4 + 2.5) = 6.5 km towards each other. In the third hour, (4 + 3) = 7 km. In the fourth hour, (4 + 3.5) = 7.5 km and so on. We can see that the distances that they cover in each hour are in AP, viz. 6, 6.5, 7, 7.5 ... with a = 6 and d = 0.5. Since they have to effectively cover a distance of 72 km, the time taken to cover this much distance would be the time taken to meet each other. So the sum of the first n terms of our AP has to be 72. If we are to express this in our equation of sum of first n terms of the AP, we will get Sn = n2× [2a + (n – 1)d]. Substituting our values, we will get
72 = n2 × [12 + 0.5(n – 1)]
Solving this, we get n = 9 hr. In 9 hr A would have covered (9 × 4) = 36 km. So B would also have covered (72 – 36) = 36 km. Hence, they would meet mid-way between A and B.
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