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Motion in Straight Line

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Physics Part I

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Physics

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CBSE Gujarat Board Haryana Board

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Class 10 Class 12
Two trains A and B of length 400 m each are moving on two parallel tracks with a uniform speed of 72 km h–1 in the same direction, with A ahead of B. The driver of B decides to overtake A and accelerates by 1 m s–2. If after 50 s, the guard of B just brushes past the driver of A, what was the original distance between them ?

In case of train A: 

Initial velocity, u = 72 km/h = 20 m/s

Time, t = 50 s

Since the train is moving with uniform velocity, 

Acceleration, aI = 0

Now, using second equation of motion, 

Distance covered by train A is given by, 

straight s space equals space ut space plus space left parenthesis 1 divided by 2 right parenthesis straight a subscript 1 straight t squared space

space space space equals space 20 cross times 50 plus space 0 space equals space 1000 space straight m 

For train B, 

Initial velocity, uB72 km/h = 20 m/s

Acceleration, a = 1 m/s

Time, t = 50 s

Now, using second equation of motion, 

Distance travelled by second train B is, 

straight s subscript II space equals space ut space plus space 1 half at squared space
space space space space space equals space 20 space straight X space 50 space plus space left parenthesis 1 divided by 2 right parenthesis space cross times space 1 space cross times space left parenthesis 50 right parenthesis squared space

space space space space space equals space 2250 space straight m space

Length of both trains = 2 cross times 400 space straight m space equals space 800 space straight m
Therefore, the original distance between the driver of train A and the guard of train B = 2250 - 1000 - 800
= 450m.




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