Kinematics-Please Help me out

A car starts from rest and accelerates at constant rate in a straight line.In the first second the car covers a distance of 2m.The velocity of the car at the end of second(sec)will be
(a)4.0ms-1 (b)8.0ms-1 (c)16ms-1 (d)None of these
(According to me,
since the body is at rest initially and has uniform acceleration thereafter,
v2-u2=2as
v2=4a (after t=1s)
a=v24
Also,v=u+at
v=v24
v=4ms-1
But the given correct answer is 8 ms-1
where did i go wrong??? )

10 Answers

1
sc ·

I think it's for confusing you, and to ask its velocity after the '2nd' second? I have no other idea, your solution is correct.

11
epsilon ·

yes..it is coming roughly 8ms-1.
use the formula of nth sec..

1057
Ketan Chandak ·

@anik....stop posting "Please Help me out" in every thread....we will help you out.... :P

158
Anik Chatterjee ·

even if we use formula of nth sec
s=u+an-12a
2=0+v24-12xv24
2=v24-v28
2=2v2-v28
v2=16
v=4ms-1
@epsilon...pleez explain how u got the solution..
@ketan...i knoe...just 2 signify that i m really unable to get the solution...wont repeat it again though...:P

1057
Ketan Chandak ·

aey bhai....question dekh le....after 2 seconds toh nahi hai na?

11
epsilon ·

s=u+an-1/2a

here a is acc. and n is the nth sec.
2=1/2a
a=4
now in 2nd sec. the distance traversed is
s=4*2-1/2*4
= 6
now
v2=2as
=48
roughly coming v coming 8

11
epsilon ·

I think I am wrong somewhere

158
Anik Chatterjee ·

I think 2 sec hi hai.....
if the question is 2 sec, fir 8 hi answer aata hai

11
epsilon ·

o acha at the end of 2 sec. too then correct..

158
Anik Chatterjee ·

haa....i think....2s ke baad hi hogaa

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