The position vector of a particle is given by the relation `vecr=vecalpha(1-gammat+betat^(2))`, where `vecalpha` is a constant vector while, `beta` and `gamma` are positive constants. Which of the following statement is true ?

Options

A. Displacement in first two seconds is `vecalpha(1-2gamma+4beta)`
B. Velocity at `t=0` is `-vecalphagamma`.
C. Acceleration at `t=0` is `2betavecalpha`.
D. Speed is decreasing with time at `t=0`.

Explanation

`vecr_(1)=vecalpha(1-gammat+beta t^(2))`, as `vecalpha` is constant so magnitude of the position vector is increasing but the direction of position vector is constant. Hence motion is a straight line motion
Displacement `(vecx)=vecr_(1)-vecr_(0)=(betat^(2)-gammat)vecalpha`
Velocity , `vecv=(dvecx)/(dt)=vecalpha(2beta t-gamma)`
Acceleration `veca=2betavecalpha`
As the motion is straight line motion speed is same as the magnitude of velocity