#### he position of a particle is given by r = 3.0 t i - 2.0 t^2 j + 4.0 k m Where t is in seconds and coefficients have the proper units for r to be in meters. (a) Find the v and a of the particle? (b) What is the magnitude and direction of velocity of the particle at t = 2.0 s?

a.

$\\ \text{Position of the particle is given by} \\ r=3.0 t i -2.0 t^{2} j +4.0 k m \\ Velocity v is given as follows: \\ v=\frac{d r}{d t}=\frac{d}{d t}\left(3.0 t i -2.0 t^{2} j +4.0 k \right) \\ So, v=3.0 t i -4.0 t j \\ Acceleration a of the particle is given by \\ a=\frac{d v}{d t}=\frac{d}{d t}(3.0 i-4.0 t j) \\ Or, a=-4.0 j$

b.

$\\ \text{We have velocity vector} v =3.0 i -4.0 t j \\ At t=2.0 s \\ v =3.0 i -8.0 j \\ The magnitude of velocity can be calculated as follows: \\ | v |=\sqrt{3^{2}+(-8) 2} \\ =\sqrt{73}=8.54 m / s \\ Direction \theta=\tan ^{-1} \frac{v_{y}}{v_{x}} \\ =\tan ^{-1} \frac{-8}{3} \\ =-\tan ^{-1} 2.667=-69.45^{\circ} \\ The negative sign shows that the direction of velocity is below x -axis$