The displacement of a particle at time t is given by
`vecx=ahati+bthatj+(C )/(2)t^(2)hatk`
where a, band care positive constants. Then the particle is

Options

A. accelerated along `hatk` direction
B. decelerated along `hatk` direction
C. decelerated along `hatj` direction
D. accelerated along `hatj` direction

Answer

A

Explanation

`X_("coordinate")=a`
`Y_("coordinate")=bt`
`Z_("coordinate")=(C)/(2)t^(2)`
Velocity along x-axis i.e. `(dx)/(dt)` is zero
velocity along y-axis i.e. `(dy)/(dt)` is constant
velocity along z-axis i.e. `(dz)/(dt)` is changing
`(dz)/(dt)=(d(c)/(2)t^(2))/(dt)=2(c)/(2)t=ct`