Civil Engineering - UPSC Civil Service Exam Questions - Discussion
Discussion Forum : UPSC Civil Service Exam Questions - Section 1 (Q.No. 31)
31.
A particle of mass 3 kg moving in a straight line decelerates uniformly from a speed of 40 m/s to 20 m/s in a distance of 300 m. Before it comes to rest, it will travel a further distance pf
Discussion:
5 comments Page 1 of 1.
Priya said:
5 years ago
V2 = u2 + 2 as,
(40^2-20^2)=2 * a * 300.
a=2 m/s^2.
V2=U2+2 as
V=0.
U2=2 as.
20^2 = 2ax2xs,
S = 100m.
(40^2-20^2)=2 * a * 300.
a=2 m/s^2.
V2=U2+2 as
V=0.
U2=2 as.
20^2 = 2ax2xs,
S = 100m.
(2)
Devbrat said:
10 years ago
First find out acceleration.
v^2 = u^2+2as.
40^2-20^2 = 2*a*300.
a = 4 m/s^2.
Now find s by v^2 = u^2+2as.
Taking v=0, u=20 & a=4.
We get S = 100 m.
v^2 = u^2+2as.
40^2-20^2 = 2*a*300.
a = 4 m/s^2.
Now find s by v^2 = u^2+2as.
Taking v=0, u=20 & a=4.
We get S = 100 m.
(1)
Devkate goving said:
10 years ago
Not a=4, a-2.
Naresh said:
8 years ago
Here, a=2 is correct.
Badhi said:
4 years ago
Whats meaning U^2? Explain please.
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