Acceleration dependent on displacement

Question

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.

Hi and welcome to the Physics SE! Please note that we don't answer homework or worked example type questions. Please see this Meta post on asking homework/exercise questions and this Meta post for "check my work" problems. — John Rennie, May 09 '17 at 14:05
Possible duplicate of How do I get the position at time $t$, if acceleration depends on position? — Yashas, May 09 '17 at 14:28

score -1 · Answer 1 · answered May 09 '17 at 14:14

-1

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

answered May 09 '17 at 14:14

Abhijeet Melkani

1,713

1

Please don't post complete solutions to homework questions. Please delete this answer. – garyp May 09 '17 at 15:05
It's a very common trick that every new student should know of and there's no other way I could have put it across. And it's not a complete answer anyway. – Abhijeet Melkani Jul 04 '17 at 13:57

score -1 · Answer 2 · answered May 09 '17 at 14:15

-1

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.

answered May 09 '17 at 14:15

user154508

95

Acceleration dependent on displacement

2 Answers2