if a variable accleration of -(1/2x^2) is working then at what time the particle will reach at x =0.25 m.if the velocity is 0 at x=1m at t=0.pls give me the analysed solution.

Dear Vijay,

Ans:- v dv/dx= -(1/2x^2)

integrating within the limits x=1 to x=x && v=0 to v=v we get,

v²=(1/x - 1)

or v=√(1/x - 1)

Again v= - dx/dt ( negatine sign comes, as with the increase of time x decreases) 

hence   - √x/√(1 - x) dx=dt

 Then integrating from x=1 to .25 and t=0 to t we get

t=∏/3 + √3/4 sec (ans)

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