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?
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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
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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??
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