Question

Acceleration of a particle which is at rest at x = 0 is a = (4 - 2x) i Select the correct alternative (s)​

Answer 1

Explanation:

a=dvdt=dvdx×dxdt=dvdx×va=dvdt=dvdx×dxdt=dvdx×v

∴Vdvdx=(4−2x)∴Vdvdx=(4-2x) or vdv=(4−2)dxvdv=(4-2)dx

Intefrating it whithin the limits (when x=−0x=-0 to x=4x=4, the velocity changes 00 to vv)

:. ∫v0vdv=∫x0(4−2x)dx∫0vvdv=∫0x(4-2x)dx

v22=4x−x2v22=4x-x2 or v2=8x−2x2v2=8x-2x2

or v=(8−2x2)1/2v=(8-2x2)1/2 ....(i)

viszeroatiszeroat x=0x-2 x^2]^(1//2)]=0oror1/2( 8 x- 2 x^2)^(-1//2) xx (8 - 4) =0oror 8-4 x= ooror x=2.

It means the velocity of particle is maximum at mean position. From (i),

v(max)=(×2−2×22)1/2=22–√unitsv(max)=(×2-2×22)1/2=2rooy2units.

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