How Do You Calculate Train Velocities When Moving in Opposite Directions?

AI Thread Summary
Two trains, A and B, are moving in opposite directions, with lengths of 125m and 100m, respectively, and train B's velocity is three times that of train A. The trains take 4 seconds to completely pass each other, leading to the calculation of their velocities using the formula for relative velocity. The total distance covered during the passing is the sum of both trains' lengths, which is 225m. The calculations yield train A's velocity at 14.1 m/s and train B's at 42.3 m/s. The discussion clarifies that each train must travel its own length plus the length of the other train to fully pass.
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Homework Statement


Two trains A and B, 125m and 100m long respectively are moving in opposite directions on parallel tracks. the velocity of the train B is three times that of train A. The train takes 4s to pass each other, calculate the velocity of each train?

Homework Equations


Velocity=\frac {Distance}{Time}

The Attempt at a Solution


Let the velocity of train A be 'v', hence the velocity of the train B would be '3v'.

The relative velocity of train A w.r.t train B = v_A- v_B
=v-(-3v)=4v

The distance to be covered= 125+100= 225m

Velocity=\frac{Distance}{Time}

Hence, velocity, 4v= \frac{225}{4}
Therefore, 16v=225

Thus, v=14.1

Hence, v_A=14.1 m/s and v_B= 42.3 m/sI've reached the final answer, by using the textual methodology. But the thing which is bothering me is the distance covered, which is given by the addition of the length of both the trains (statement highlighted). Please explain me as to why do we add these lenghts, when the entire train (considering it as a whole, either A or B) moves across the length of the other, and not its own!
 
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Get two trains and try it...

The moment before they start passing each other:

Train 1 ------------------ Train 2 (total distance = sum of distances)
_______________/////////////////////////
////////////////////// ________________

The moment after they have passed each other:

Train2 ------------------ Train 1

//////////////////////// __________________
________________////////////////////////////


Notice (look at the tails) that each train has to travel its own length, and the other trains length. Hope this helps :)
 
oh, yes! got it! actually i was considering one train, and observing it going past the others!
 
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Since ##px^9+q## is the factor, then ##x^9=\frac{-q}{p}## will be one of the roots. Let ##f(x)=27x^{18}+bx^9+70##, then: $$27\left(\frac{-q}{p}\right)^2+b\left(\frac{-q}{p}\right)+70=0$$ $$b=27 \frac{q}{p}+70 \frac{p}{q}$$ $$b=\frac{27q^2+70p^2}{pq}$$ From this expression, it looks like there is no greatest value of ##b## because increasing the value of ##p## and ##q## will also increase the value of ##b##. How to find the greatest value of ##b##? Thanks
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