The position of a particle is given by `r=2.00t hati-1.00t^(2)hatj+3.00hatk` where t is in seconds and the coefficients have the proper unit for r to be in metres. Find the velocity and acceleration of a particle then what is the magnitude and direction of velocity of the particle at t = 2 s?

Explanation

Given position of the particle is
`r=2.00 hati-1.00t^(2)hatj+3.00 hatk`
Velocity : The rate of change of acceleration is called velocity
`"velocity v"=(dr)/(dt)=(d)/(dt)(2.00hati-1.00t^(2)hatj+3.00hatk)`
`=[2.00hati-2.00thatj+0]ms^(-1), v=[2.00hati-2.00thatj]ms^(-1)`
Acceleration: The rate of change of position of the particle is called acceleration
`"Acceleration a"=(dv)/(dt)=(d)/(dt)(2.00hati-2.00thatj)`
`=0-200hatj=-2.00hatjms^(-2)`
At `"time"_(1)t=2s`.
velocity `=[2.00hati-2.00thatj[ms^(-1),=[2.00hati-2.00xx2hatj]`
`=[2.00hati-4.00hatj]`
Magnitude and direction of the particles.
`v=sqrt(((2)^(4)+(-4)^(2)))=sqrt(4+16)=sqrt(20)=4.47ms^(-1)`
If `theta` is the angle which v makes with x-axis, then, `tan theta=(v_(y))/(v_(x))=(-4)/(2)=-2=-tan 63.5^(@)`
`theta=63.5^(@)` below the x-axis.