The position vector of a particle at time `t` is given by `vecr=2t hati+5t hatj+4sin omegat hatk` where `omega` is a constant.Answer the following questions
Acceleration of the particle is

Options

A. Constant in magnitude but variable with direction
B. constant
C. Constant in direction but variable with magnitude
D. Varying with magnitude as well as direction

Answer

D

Explanation

`vecr=2t hati+5thatj+4omegasinomegat hatk`
`rArr vecV=2hati+5hatj+4sinomegathatk rArr veca=-4omega^(2)sinomegathatk`
We can earily acc/ is varify with magnitude but along a constant direction. As velocity is not changing along `x` axis and `y` axis, we can early realise a vector parallel to `XY` plane which is `_|_^(r)` to velocity. say `x.hati+y hatj=vecA` Hence `vecV.vecA=0`(for `vecV _|_ vecA`)
The above equation gives are solution
`x=5 and y=-2 or x=10 and y=-4` and so on (Realise the above solution in `3D` and you will get a clear idea).