Question

A particle is projected with velocity along x-axis. The deceleration on the particle is proportional to the square of the distance from the origin ie, \( a=-a x^{2} \). The (distance) \( t \) which the particle stons is: \( u \equiv x \) चाiिe

Answer 1

(3⁰v²/2a(alpha))⅓ .Ans

solⁿ→ WE KHOW THAT DECLARATION IS GIVEN IN THE FORM OF X SO V ALSO GIVEN IN X TYPE FUNCTION.... 

THE FORMULA WHEN *VELOCITY* FUNCTION GIVEN IN TERM OF X .... 

a(declaration)=VdV/dX

a=—@x²

-@x²dx=VdV both side doing INTEGRATION 

NOW U CAN ALSO SOLVE IT THANX

         

Answer 2

\(a=\frac{dv}{dt}=\frac{dv\,dx}{dx\,dt}=v\frac{dv}{dx}dx^ 2\)