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?

Answers (1)

a. 

\ 	extPosition 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: \ $v=fracd rd t=fracdd tleft(3.0 t i -2.0 t^2 j +4.0 k 
ight)$ \ So, $v=3.0 t i -4.0 t j$ \ Acceleration a of the particle is given by \ $a=fracd vd t=fracdd t(3.0 i-4.0 t j)$ \ Or, $a=-4.0 j

 

b.

 

\ 	extWe 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: \ $| v |=sqrt3^2+(-8) 2$ \ $=sqrt73=8.54 m / s$ \ Direction $	heta=	an ^-1 fracv_yv_x$ \ $=	an ^-1 frac-83$ \ $=-	an ^-1 2.667=-69.45^circ$ \ The negative sign shows that the direction of velocity is below $x -axis

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