The position of a particle is given by r =3t i +2t2J +5k , where t is in second and the coefficients have the proper units for r to be in metre. Find the magnitude and direction of the velocity of the particle at t=3s. Share with your friends Share 21 Sanjay Singh answered this Dear student r→=3ti^+2t2j^+5k^Velocity of particle isv⇀=dr→dt=d(3ti^+2t2 j^+5k^)dtv⇀=(3i^+4tj^)putting t=3 velocity of particle at 3 sec is v⇀=(3i^+12j^)magnitude of velocity is V=VX2+VY2=32+122=153 m/sX component of velocity of particle is VX=3m/sY component of velocity of particle is VY=12m/sLet angle made by velocity of particle with X axis is θ so tanθ=VYVX=123=4θ=tan-1(4)Regards 41 View Full Answer