Physics Help Forum A question on physics kinematics

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 Jul 1st 2018, 02:18 AM #1 Junior Member   Join Date: Jul 2018 Posts: 4 A question on physics kinematics The velocity of a particle moving in the positive direction I of the x axis varies as v=a*root of x,where a is a +'ve constant .assuming that at moment t=0 the particle was located at the point x=0.find: A)the time dependence of the velocity and acceleration of the particle
Jul 1st 2018, 03:58 AM   #2
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 Originally Posted by Mithilaa The velocity of a particle moving in the positive direction I of the x axis varies as v=a*root of x,where a is a +'ve constant .assuming that at moment t=0 the particle was located at the point x=0.find: A)the time dependence of the velocity and acceleration of the particle
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 Jul 1st 2018, 05:12 AM #3 Senior Member   Join Date: Aug 2010 Posts: 369 Assuming that by "root of x" you mean the square root, $\displaystyle x^{1/2}$, "v= a root of x" is $\displaystyle \frac{dx}{dt}= ax^{1/2}$. That can be written as $\displaystyle x^{-1/2}dx= a dt$. Integrate. topsquark likes this.
 Jul 1st 2018, 06:21 AM #4 Junior Member   Join Date: Jul 2018 Posts: 4 My doubt is how to integrate Is it 2* root of x equals to at
 Jul 1st 2018, 06:22 AM #5 Physics Team   Join Date: Apr 2009 Location: Boston's North Shore Posts: 1,570 I guess hoi thinks the rules don't apply to him.
 Jul 1st 2018, 06:25 AM #6 Junior Member   Join Date: Jul 2018 Posts: 4 I actually have the solution of this question but I have some doubts on the question.thats why I asked it Last edited by Mithilaa; Jul 1st 2018 at 06:29 AM.
 Jul 1st 2018, 06:26 AM #7 Junior Member   Join Date: Jul 2018 Posts: 4 I am sorry .it was my mistake
Jul 1st 2018, 06:23 PM   #8
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 Originally Posted by Mithilaa I actually have the solution of this question but I have some doubts on the question.thats why I asked it
In a situation like that, show us, not only what solution you have, but how you got that solution.

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