Question

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?

Answer 1

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

Answer 2

why isnt your answer right?

Answer 3

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

Answer 4

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

Answer 5

nope

Answer 6

hi @DLS

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

Answer 7

Just figured out a moment before u posted!

Answer 8

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

Answer 9

first derivate velocity then put values of x

Answer 10

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

Answer 11

ans is 30

Answer 12

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

Answer 13

ok muku you got it right :P