Civil Engineering - Applied Mechanics - Discussion
Discussion Forum : Applied Mechanics - Section 3 (Q.No. 13)
13.
The acceleration of a train starting from rest at any instant is 
where V is the velocity of the train in m/sec. The train will attain a velocity of 36 km/hour after travelling a distance of
where V is the velocity of the train in m/sec. The train will attain a velocity of 36 km/hour after travelling a distance ofDiscussion:
7 comments Page 1 of 1.
Lasith said:
7 years ago
v2=u2+2as
v=36km/hr = 10m/sec
a=1/66.
By putting values in the above formula
v2=u2(zero because of train starts from rest)+2as.
s=3300.
v=36km/hr = 10m/sec
a=1/66.
By putting values in the above formula
v2=u2(zero because of train starts from rest)+2as.
s=3300.
(1)
Vanathi said:
4 years ago
@Muhammad Hashim.
1 horu has 3600 sec put V in acceleration then we get;
V = (36*1000)/(60*60) = 10m/sec.
a = 1/6(10+1),
a = 1/66.
1 horu has 3600 sec put V in acceleration then we get;
V = (36*1000)/(60*60) = 10m/sec.
a = 1/6(10+1),
a = 1/66.
Manthan jain said:
7 years ago
You are right @Sachin.
The correct answer is 3300m.
The correct answer is 3300m.
Aparna said:
9 years ago
How to solve this? Explain the solution.
Engr M Danish said:
8 months ago
I think the Correct answer is 2000.
Sachin said:
8 years ago
The Correct answer is 3300m.
Muhammad Hashim said:
6 years ago
How you get a=1/66 ?
(1)
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