A starts at 11:00AM and travels at a speed of 4km/hr. B starts at 1:00PM and travels at
1km/hr for the first 1hr and 2km/hr for the next hr and so on. At what time
they will meet each other.
Answers
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Let them meet at N hours in P. M. Obviously, T is after 1 PM. We assume that A and B are traveling in the same direction and they start at the same point.
Distance traveled by A by time N hours = 4 * (N+1) km
Distance traveled by B by time N hours = 1 km + 2 km + ... + (N-1) km
So 4 N + 4 = (N-1) N /2
8 N + 8 = N² - N
N² - 9 N - 8 = 0
Δ = 81 + 32 = 113
N = [9 +- √113 ]/2 = 9.815 hours PM
that is 9 PM and 48 minutes 54.26 sec
Distance traveled by A by time N hours = 4 * (N+1) km
Distance traveled by B by time N hours = 1 km + 2 km + ... + (N-1) km
So 4 N + 4 = (N-1) N /2
8 N + 8 = N² - N
N² - 9 N - 8 = 0
Δ = 81 + 32 = 113
N = [9 +- √113 ]/2 = 9.815 hours PM
that is 9 PM and 48 minutes 54.26 sec
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