Civil Engineering - UPSC Civil Service Exam Questions - Discussion
Discussion Forum : UPSC Civil Service Exam Questions - Section 20 (Q.No. 41)
41.
A train starts from rest on a curved track of radius 800 m. Its speed increases uniformly and after 3 minutes it is 72 km/hr. Its normal acceleration after 2 minutes is
Discussion:
4 comments Page 1 of 1.
Rahul Kumar said:
4 years ago
V = 72km/hr = 72 * 5/18 =2 0m/sec. u = 0m/sec.
t = 3minutes = 3*60 = 180sec ,t = 2 * 60minutes=120sec r=800m.
V=u+at . 20=0+a*180. a=20/180=1/9m/sec^2.
Again V=u+at=0+1/9*120=40/3m/sec.
Normal centripetal acceleration (a)=V^2/r =(40/3)^2/800=1600/9/800=1600/9*800=4/18 m/sec^2.
t = 3minutes = 3*60 = 180sec ,t = 2 * 60minutes=120sec r=800m.
V=u+at . 20=0+a*180. a=20/180=1/9m/sec^2.
Again V=u+at=0+1/9*120=40/3m/sec.
Normal centripetal acceleration (a)=V^2/r =(40/3)^2/800=1600/9/800=1600/9*800=4/18 m/sec^2.
Kdm said:
4 years ago
After 3 minute = 72km/hr=20m/s.
3 x 60 = 20m/s.
By Unitary method;
1 second = 20/3x60(2) = 4/18.
3 x 60 = 20m/s.
By Unitary method;
1 second = 20/3x60(2) = 4/18.
Shivani rayishetty said:
3 years ago
Can anyone explain this answer in detail? please.
Bharat said:
8 years ago
Anybody solve this problem?
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