Question

A train 200 m long enters at 5 m/s and exits at 3 m/s from a bridge 300 m long.The time taken by it is

Answer 1

Answer:

2 minutes 5 second

Explanation:

Total distance S = 200 + 300 = 500m

Initial velocity u = 3m/s

Final velocity v = 5 m/s

We know,  

v​​​​​² ​​​​​- u​​​² = 2as

=> 25 - 9 = 1000a

=> a = (16/1000)m/s²

Again  

v = u + at

=> 5 = 3 + (16/1000)t

=> t = 2 × (1000/16) = 125 s = 2 minutes 5 second

Answer 2

Answer:

125 sec

Explanation:

Initial velocity (u) = 5m/s

Final velocity (v) = 3m/s

Total distance= distance of train + path distance

=200m + 300m = 500m

Applying the formula

v²-u²=2as

3²-5² = 2×a×500

-16 = 1000a

a = -16/1000

Applying

v = u + at

3 = 5 + (-16/1000) t

-2 = -16/1000 t

1 = 8/1000 t

t = 1000/8

t = 125 sec