The position of a particle is given by ` vec r= 3.0 t hat I - 2 2- 0 t^@ hat j + 4.0 hat k m`, wher 9t) in seconds and the coefficits have the proper units for ` vec r` to be in metres. (a) Fing the ` vec v` and ` ve a` of the particle ? (b) What is the magnitude and direction fo velocity of the particle at ` t= 2 s`?

Explanation

(a) Velocity, ` vec v = (d vec r0/ (dt) = d/(dt0 ( 3.0 t hat I - 2.0 t hat j + 4.0 hat k) = [ 3. 0 hat I - 4.0 t hat j] ms^(-1)`
Acceleration , ` vec a = (d vec v)/(dt0 = d/(dt) ( 3.0 hat I - 4 .0 t hat j ) = 0 -4.0 hat =- 4.0 hat j ms^(-2)`
(b) At time ` t 2 s, vec v = 3.0 hat i - 4. 0 xx 2 hat j = 3. 0 hat i - 8.0 hat j ` :. ` vec = sqrt (93.0)^2+ (-8)^2) = sqrt 72 = 8. 54 ms^(-10`
If ` theta` is the angle which ` vec v` makes with x-axis, then ` atn = v_y/v_x - (-8)/3 =- 2.667 =- tan 69. 5^@`
` theta = 69.5^@` below teh x-axis`.