1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question

# A 150 m long train accelerates uniformly from rest. If the front of the train passes a railway worker 50 m away from the station at a speed of 25 m/s, what will be the speed of the back part of the train as it passes the worker?

A
30 m/s
No worries! We‘ve got your back. Try BYJU‘S free classes today!
B
40 m/s
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
50 m/s
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
D
60 m/s
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution

## The correct option is C 50 m/sApplying 3rd equation of motion v2=u2+2as 25×25=0+2×a×50 a=254 m/s2 Now for speed of the back part of the train as it passes the worker We have v′2=v2+2al where l is the length of the train or distance covered by the back part to pass the worker v′2=252+2×254×150 v′2=625+(25×75)=2500 v′=50 m/s

Suggest Corrections
21
Join BYJU'S Learning Program
Related Videos
Motion Under Constant Acceleration
PHYSICS
Watch in App
Explore more
Join BYJU'S Learning Program