Find a vector of magnitude 18, which is perpendicular to both the vectors 4i^-j^+3^k and -2i^+j^-2k^ . Question no 2.The position of a particle is given by r = 3.0ti^-2.0t*ti^+4.0k^m where t is in seconds and the coefficiants have proper units for r to be in meter. A)find the velocity and acceleration of the particle and B)what is the magnitude and direction of velocity of the particle at t=2.0 sec

Asked by sharmistha.panda | 8th Aug, 2012, 02:20: PM

Expert Answer:

1. A vector perpendicular to the given two  vectors will be in the direction of their cross product.
    the unit vector in the direction of their cross product will be( cross product can be found by using
    the method of determinants):
   
and the required vector will be 18(unit vector) = 18/3(-i+2j+2k)=  -6i+12j+12k
 
2. position of particle: r= (3t i - 2t2 j + 4 k) m
    (A) velocity, v = dr/dt = (3 i - 4t j) m/s
         acceleration, a = dv/dt = -4 j m/s2
    (B) velocity at t=2:  v=  (3i - 8j) m/s
         magnitude: (32+82)1/2= (73)1/2
         direction: it makes an angle of tan-1(-8/3) from x axis.

Answered by  | 9th Aug, 2012, 07:51: AM

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