Hello Aspirants, \r\nRelative motion problems are pretty common in all competative exams
\r\nSuch problems seems to be difficult for people with non-physics background. These type of problems are pretty\r\nsimple to solve if you know the concept of relative frame of reference. In this article, We are going to share\r\n some tricks and direct results for the people with non-science background. Also, It can be helpful in time saving for\r\npeople relating with science stream.
\r\n\r\nSome important tricks and formulae
\r\nIf a vehicle A moves with velocity a & other vehicle B moves with velocity b in opposite direction of A, then
\r\nthe relative velocity of A assuming velocity of B as 0 becomes a+b.
\r\nThis concept is pretty important in solving such problems.
\r\n$ a $ km/hr = $ a * frac{5}{18} $ m/s
\r\n$ a $ m/s = $ a * frac{18}{5} $ km/hr
\r\nIf the two bodies are moving in same direction with velocities u & v, then relative velocities of two bodies \r\nwith respect to each other will be (u-v).
\r\nThe -ve sign indicates that the body lagging behind to the other body by velocity u-v.
\r\nAnd +ve sign indicates that the body is moving forward w.r.t other body by velocity (u+v).
\r\nIf two trains of length a metres and b metres are moving in same direction at u m/s and v m/s, then the time \r\ntaken by faster train to cross slower train = $ frac{a+b}{u-v} $ sec.
\r\n\r\nIf two trains of length a metres and b metres are moving in same direction at um/s and v m/s, then the time \r\ntaken by trains to cross each other = $ frac{a+b}{u+v} $ sec.
\r\nIf two trains start at the same time from points A and B towards each other and after crossing they \r\ntake a and b secs. in reaching B and A respectively, then (A\'s speed) : (B\'s Speed) = ($ b^ frac{1}{2} $ : $ a^ frac{1}{2} $) .
\r\nLet\'s solve a problem:
\r\nEX. A train 100m long takes 6 seconds to cross a man walking at 5kmph in a direction opposite to that of train\r\n.Find the speed of the train.
\r\nSolution:
\r\nLet the speed of the train be x kmph.
\r\nSpeed of the train relative to man = (x+5)kmph = $ (x+5) * frac{5}{18} $ m/sec
\r\n$ frac{Distance}{speed} = time $ .
\r\nHence, $ frac{100}{(x+5) * frac{5}{18}} =6. $
\r\nAfter solving, the speed of the train is 55 kmph.
\r\n\r\n\r\n\r\nThank you for reading. Solve some more examples by yourself for practice. Keep visiting.\r\n