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.                                                                                         

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r=3ti^+2t2j^+5k^Velocity of particle isv=drdt=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

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