Question

# Position vector of a particle at time=t is given by ((3t2)^i+(4t)^j+12^k)m. Choose the correct option(s).

A
Initially the particle is in the x-y plane
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B
Initial velocity of the particle is (4^j)m/s
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C
at time=1s acceleration of the particle is 6m/s2 along x-axis
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D
Acceleration of the particle is constant
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Solution

## The correct options are A Initial velocity of the particle is (4^j)m/s B at time=1s acceleration of the particle is 6m/s2 along x-axis C Acceleration of the particle is constant→x=3t2→i+4t→j+12→k→v=dxdt=6t→i+4→j→a=dvdt=6→iClearly acceleration is constant, 6m/s2 as it is not dependent on any variable.Put t=0 in the above equations to get initial values of velocity i.e. 4→jm/s

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