The position of a particle is given by ` vec r= 3.0 t hat i - 2.0 t^2 hat j + 4.0 hat k m`, wher (t) in seconds and the coefficients have the proper units for ` vec r` to be in metres. (a) Fing the ` vec v` and ` vec a` of the particle ? (b) What is the magnitude and direction fo velocity of the particle at ` t= 2 s`?

Explanation

Here `vec r =(3.0t hati -2.0t^(2) hatj +4.0 hatk)m`
(b) Velocity =`vecv=(vecdr)/(dt)=(3.9hati-4.0t hatj +0) m//s`
Acceleration `=veca=(d vecv)/(dt)=0-4.0hatj=-4.0 hatj m//s^(2)`
(b) At t =2 s
`vecv=(3.0hati-4.0xx2 hatj)`
`=(3.0hati-8.0hatj) =(3hati-8hatj) m//s`
`therefore v=sqrt(3^(2)+(-8)^(2))=sqrt(9+64)`
`v=sqrt73=8.54 ms^(-1)`
Let `theta` be the angle which `vecv` makes with the X-axis . Therefore , `tan theta =v_(y)//v_(x)= -(8)/(3)=-2.7`
`therefore theta =tan^(-1) (-2.7) =- 69.7^(@)=-70^(@)` with the X-axis (below).