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

Options

A. Maximum speed of the particle is `4` units
B. Particle further comes to rest at ` x =4`
C. Particle oscillates about ` x=2`
D. Particle will continuously acceleration along the x-axis.

Answer

B, C

Explanation

` a= (dv)/(dt) = (dv)/(dx) xx (dx)/(dt) = (dv)/(dx) xx v`
` :. V (dv)/(dx) = ( 4- 2 x) ` or ` v dv = (4-2) dx`
Intefrating it whithin the limits (when ` x=-0` to `x = 4`, the velocity changes ` 0` to `v`)
:. ` int_0^v v dv = int_0^x (4-2 x) dx`
` v^2/2 = 4 x-x^2` or ` v^2 = 8 x- 2 x^2`
or ` v= (8 - 2 x^2)^(1//2)` ....(i)
` `v` is zero at ` x=0` or ` x=4`
Acceleration of the particles is zero , at ` x=2`. Hence the particcle will oscillate about ` x=2`
Velocity will be maximum, when ` (dv)/(dx) =0`
or ` d/(dx) [[8 x-2 x^2]^(1//2)]=0`
or `1/2( 8 x- 2 x^2)^(-1//2) xx (8 - 4) =0`
or ` 8-4 x= o` or ` x=2.
It means the velocity of particle is maximum at mean position. From (i),
` v(max) = (xx 2 -2 xx 2^2) ^(1//2) =2 sqrt 2 units`.