If it takes twice as long for a passgr train to pass a freight train after it first over takes it as it takes?

the q should be phrased "after it first reaches it"

let the lengths of the passenger/freight trains be P/F resply

& their speeds V and v, then

(P+F)/(V-v) = 2(P+F)/(V+v)

(V+v)/(V-v) = 2, then by componendo/dividendo,

V/v = (2+1)/(2-1) = 3 <-------

let the length of the trains be 'l' each., velocity of the passenger train be 'v' and the velocity of the freight train be 'v1'

when the passenger train and the freight train are travelling in the opposite directions, then the relative velocity of the passenger train with respect to the freight train(i.e, the velocity of the passenger train as observed by a man sitting in the freight train) =v+v1

and the distance which the passenger train covers to pass the freight train is 'l'+'l' =2l

so time taken =t = 2l/(v+v1)

whent he trains are travelling in the same direction, then the realtive velocity of the passenger train is

=v-v1

again the length is=2l (to overtake the freight train)

so time taken now is= t1 = 2l/(v-v1)

as given in the question, t1 = 2*(t)

so, 2*2l/(v+v1) = 2l/(v-v1)

so, v+v1 =2(v-v1)

so, v= 3*v1

so the passenger train is 3 times faster than the freight train.

hope you know relative velocity.

according to doppler's effect the speed of fright train=x+y.,

where x is the speed of passenger train and y is the speed of freight train..

hiii please can anyone solve this:

a car A starts from a point P towards another point Q.Another car B starts (also from p) 1 h after the first car and overtakes it after covering 30% of the distance PQ .after that the cars continue and on reaching Q, Car B reverses and meet car A at 23(1/3) of the distance QP.find the time taken by car B to cover the distance PQ(in hours)

arccot(sqrt(sina))- arctan(sqrt(sina)=x then sina=? please sir help