Harshit Singh
Last Activity: 4 Years ago
Dear Student
Initial speed of the train, u= 90 km/h = 25 m/s (1km/hr = 5/18 m/s)
Final speed of the train, v = 0 (finally the train comes to rest and its velocity becomes 0)
Acceleration = - 0.5 m s^-2
According to third equation of motion:
v^2= u^2+ 2as
(0)^2= (25)^2+ 2( - 0.5) s
Where, s is the distance covered by the train
S = 25^2
2(0.5)
= 625
The train will cover a distance of 625 m before coming to rest.
Thanks