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Two trains a and b each of length 400 m are moving on two parallel tracks in the same direction (with a ahead ofb.With the same speed of 72 km/h, the driver of b decides to overtake a and accelerates by 1 m/s. After 50 s, the driver of b and the guard of a are in the same line. What is the original distance between a and b? (1) 750 m (2) 1000 m (3) 1250 m (4) 2250 m |
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Answer» Initial velocity, u = 72 km/h = 20 m/s Initial velocity, u = 72 km/h = 20 m/sTime, t = 50aI = 0 (Since it is movicity) city)From second equation of MOTION, distance (sI)covered by train A can be obtained as: city)From second equation of motion, distance (sI)covered by train A can be obtained as:s = ut + (1/2)a1t2 city)From second equation of motion, distance (sI)covered by train A can be obtained as:s = ut + (1/2)a1t2= 20 × 50 + 0 = 1000 m city)From second equation of motion, distance (sI)covered by train A can be obtained as:s = ut + (1/2)a1t2= 20 × 50 + 0 = 1000 mFor train B: city)From second equation of motion, distance (sI)covered by train A can be obtained as:s = ut + (1/2)a1t2= 20 × 50 + 0 = 1000 mFor train B:Initial velocity, u = 72 km/h = 20 m/s city)From second equation of motion, distance (sI)covered by train A can be obtained as:s = ut + (1/2)a1t2= 20 × 50 + 0 = 1000 mFor train B:Initial velocity, u = 72 km/h = 20 m/sAcceleration, a = 1 m/s2 city)From second equation of motion, distance (sI)covered by train A can be obtained as:s = ut + (1/2)a1t2= 20 × 50 + 0 = 1000 mFor train B:Initial velocity, u = 72 km/h = 20 m/sAcceleration, a = 1 m/s2Time, t = 50 s city)From second equation of motion, distance (sI)covered by train A can be obtained as:s = ut + (1/2)a1t2= 20 × 50 + 0 = 1000 mFor train B:Initial velocity, u = 72 km/h = 20 m/sAcceleration, a = 1 m/s2Time, t = 50 sFrom second equation of motion, distance (sII) covered by train A can be obtained as: city)From second equation of motion, distance (sI)covered by train A can be obtained as:s = ut + (1/2)a1t2= 20 × 50 + 0 = 1000 mFor train B:Initial velocity, u = 72 km/h = 20 m/sAcceleration, a = 1 m/s2Time, t = 50 sFrom second equation of motion, distance (sII) covered by train A can be obtained as:sII = ut + (1/2)at2 city)From second equation of motion, distance (sI)covered by train A can be obtained as:s = ut + (1/2)a1t2= 20 × 50 + 0 = 1000 mFor train B:Initial velocity, u = 72 km/h = 20 m/sAcceleration, a = 1 m/s2Time, t = 50 sFrom second equation of motion, distance (sII) covered by train A can be obtained as:sII = ut + (1/2)at2= 20 X 50 + (1/2) × 1 × (50)2 = 2250 m city)From second equation of motion, distance (sI)covered by train A can be obtained as:s = ut + (1/2)a1t2= 20 × 50 + 0 = 1000 mFor train B:Initial velocity, u = 72 km/h = 20 m/sAcceleration, a = 1 m/s2Time, t = 50 sFrom second equation of motion, distance (sII) covered by train A can be obtained as:sII = ut + (1/2)at2= 20 X 50 + (1/2) × 1 × (50)2 = 2250 mLength of both trains = 2 × 400 m = 800 m |
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