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OpenStudy (dls):

The velocity of a particle moving on the X -axis is given by v=x^2+x where v is in m/s and x in m.Find its Acc. in m/s^2 when passing through the point x=2m?

OpenStudy (dls):

On differentiating,I got acceleration as 2x+1 On substituting, I got 5m/s^2,whats wrong?

OpenStudy (anonymous):

why isnt your answer right?

OpenStudy (dls):

Don't know..that's what im here for

OpenStudy (anonymous):

sorry I meant, is your answer not the right one?

OpenStudy (dls):

nope

OpenStudy (anonymous):

hi @DLS answer is incorrect because \[a=\frac{dv}{dt}\\not\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \frac{dv}{dx}\]

OpenStudy (dls):

Just figured out a moment before u posted!

OpenStudy (anonymous):

use this hint to find correct answer \[a=\frac{dv}{dt}=\frac{dv}{dx}\frac{dx}{dt}=v \frac{dv}{dx} \\ since \ \ \ \frac{dx}{dt}=v\]

OpenStudy (anonymous):

first derivate velocity then put values of x

OpenStudy (anonymous):

acc=2x that is a=4m :P is that your answer??

OpenStudy (dls):

ans is 30

OpenStudy (anonymous):

a=(x^2+x)(2x+1) x=2 a=6*5=30

OpenStudy (anonymous):

ok muku you got it right :P

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