Civil Engineering - UPSC Civil Service Exam Questions - Discussion

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

[A]. 10 m
[B]. 20 m
[C]. 50 m
[D]. 100 m

Answer: Option D

Explanation:

No answer description available for this question.

Devbrat said: (Sep 10, 2015)  
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.

Devkate Goving said: (Oct 17, 2015)  
Not a=4, a-2.

Naresh said: (Nov 19, 2017)  
Here, a=2 is correct.

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