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A particle starts its motion from origin with a velocity $4$ m/s in positive $x$ direction. It's acceleration is related with position as $a = (2x + 2)$. Find magnitude of velocity of particle at $x=4$.

I've tried a lot to bring it with respect to time. But I could not proceed any further. Is this problem valid or there's a typo in my book.

John Rennie
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Shubhraneel Pal
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2 Answers2

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$$a = \frac{dv}{dt} = \frac{dv}{dt}.\frac{dx}{dx} = \frac{dv}{dx}.\frac{dx}{dt} = v.\frac{dv}{dx}$$

Abhijeet Melkani
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You should solve a differential equation: $\frac{d^2x}{dt^2}=2x+2$ Solve for x and than calculate the derivative of x that is the velocity.

user154508
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