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 B. The driver of B decides to overtake A and accelerated by 1 m/s². If after 50 s, the guard of B just brushes past the driver of A, what was the original distance between them?
Doubt by Suhani
Solution :
Method 1
Length of each train (L) = 400 m
Initial velocity of each train (u) = 72 km/h
= 72×(5/18)
= 20 m/s
= 72×(5/18)
= 20 m/s
Acceleration of Train A
aA=1m/s²
aA=1m/s²
Acceleration of Train B
aB=0m/s²
aB=0m/s²
Time Taken (t) = 50 sec
Let the original distance between the two trains = x
Now, the distance covered by the train B in 50 sec
xB=ut+½at²
xB=20(50)+½(1)(50)²
xB=1000+½(2500)
xB=1000+1250
xB=2250 m
xB=20(50)+½(1)(50)²
xB=1000+½(2500)
xB=1000+1250
xB=2250 m
Similarly Distance covered by Train A in 50 sec
xA=ut+½at²
xA=20(50)+½(0)t²
xA=1000+0
xA=1000 m
xA=ut+½at²
xA=20(50)+½(0)t²
xA=1000+0
xA=1000 m
Original Distance between Train A and B is given by
x=xB-xA
x=2250-1000
x=1250 m
x=1250 m
Method 2
Using the concept of Relative Velocity
Relative initial Velocity of Train B w.r.t A
Relative initial Velocity of Train B w.r.t A
uBA=uB-uA
uBA=20-20
uBA=0 m/s
Relative Acceleration of Train B w.r.t. A
uBA=20-20
uBA=0 m/s
Relative Acceleration of Train B w.r.t. A
aBA=aB-aA
aBA=1-0
aBA=1-0
aBA=0 m/s²
Using Equation
s=ut+½at²
x=uBAt+½aBAt²
x=0(50)+½(1)(50)²
x=0+½(2500)
x=1250 m
x=uBAt+½aBAt²
x=0(50)+½(1)(50)²
x=0+½(2500)
x=1250 m
Hence, the original Distance between Train A and B is 1250 m