Question

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 speed B. The driver of B decides to overtake A and accelerates by 1 m/s². If after 50s, the guard of B just brushes past the driver of A, what was the original distance between them?

• A. 100 m
• B. 1150 m
• C. 1300 m
• D. 1250 m

Hint:

When solving problems involving relative motion, use kinematic equations to calculate the distance covered by the moving objects and subtract their lengths to find the original distance between them.

Answer 1

The Correct Option is D

Let the initial speed of both trains be \( v_0 = 72 \, \text{km/h} = 20 \, \text{m/s} \).
Let the acceleration of train B be \( a = 1 \, \text{m/s}^2 \).
The time \( t = 50 \, \text{s} \) is given.
The distance covered by train B during 50 seconds, with initial speed \( v_0 \), can be calculated using the equation for motion: \[ s_B = v_0 t + \frac{1}{2} a t^2 \] Substituting the known values: \[ s_B = 20 \times 50 + \frac{1}{2} \times 1 \times 50^2 = 1000 + 1250 = 2250 \, \text{m} \] Since both trains have a length of 400 m, the original distance between the two trains will be the distance travelled by B minus the lengths of both trains: \[ \text{Distance} = s_B - 400 - 400 = 2250 - 800 = 1250 \, \text{m} \]
Thus, the correct answer is (d).