NTA Abhyas JEE Main2020PhysicsMotion in One DimensionPractice
Two trains A and B of length 400 m each are moving on two parallel tracks with a uniform speed of 72 km h 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 and the original distance between them is x , then calculate value of x 10 ?
Correct answer
45.00
Step-by-step solution
Length of each train, l ⁡ A = l ⁡ B = 4 0 0  m u ⁡ A ⁡ = 7 2 × 5 1 8 m/s = 20 m/s Distance travelled by train A in 50 s s A = u A × t (As for unaccelerated motion, distance = Speed x Time) s A = 2 0 × 5 0 = 1000 m Distance travelled by train B in 50 s , s B = u B t + 1 2 a B t 2 (As motion of train B is an accelerated motion) s B = 2 0 × 5 0 + 1 2 × 1 × 5 0 2 = 1000 + 1250 = 2250 m relative distance bet