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Question
Wed August 08, 2012

# 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

Thu August 09, 2012
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.
Related Questions
Mon February 20, 2017

# A ball  thrown up is caught by the thrower after 4 seconds . How high did it go and with what velocity was it thrown? How far below its highest point was in 3 second after start? acceleration due to gravity is 9.8.

Mon February 20, 2017

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