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⁻¹ in the same direction, with A ahead of B. The driver of B dicides to overtake A and accelerates by 1 m s⁻². If after 50 s, the guard of B just moves past the driver of A. What was the original distance between them?

Answer 1

Hii dear,

# Answer- 450m

# Explaination-
For train A:
Initial velocity, u = 72 km/h = 20 m/s
Time, t = 50 s
Acceleration, a1 = 0
From second equation of motion, distance (s1)covered by train A can be obtained as:
s1 = ut + (1/2)a1t^2
s1 = 20 × 50 + 0
s1 = 1000 m

For train B:
Initial velocity, u = 72 km/h = 20 m/s
Acceleration, a2 = 1 m/s2
Time, t = 50 s
From second equation of motion, distance (s2) covered by train A can be obtained as:
s2 = ut + (1/2)a2t^2
s2 = 20 X 50 + (1/2) × 1 × (50)2
s2 = 2250 m

Length of both trains = 2 × 400 m = 800 m

Hence, the original distance between the driver of train A and the guard of train B is 2250 – 1000 – 800 = 450m.

Hope this is helpful.

Answer 2

Given :

For Train A :

Initial velocity =u= 72km/h=72x 5/18=20 m/s

Time=t= 50 sec

a=0 m/s 2²

s=sA

From second equation of motion :

S=ut+1/2at²

sA=20x50+1/2 x0x 50²

=1000m

For train B:

Initial speed=u= 72km/hr=72x5/18=20 m/s

a=1 m/s²

time=t=50 sec

s=sB

From second equation of motion :

S=ut+1/2at²

sB=20x50+1/2 x1 x50 ²=2250m

Length of both trains = 400+ 400 m = 800 m

Hence, the original distance between the driver of train A and the guard of train B is 2250 – 1000 – 800 = 450m.