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Question

The position of a particle is given by r=3t^i+3t2^j4^k, where t is in seconds and r is in meters. Find out magnitude and direction of velocity v with horizontal at t=3 s.

A
35 m/s, θ=tan1(2)
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B
35 m/s, θ=tan1(23)
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C
32 m/s, θ=tan1(3)
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D
35 m/s, θ=tan1(12)
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Solution

The correct option is A 35 m/s, θ=tan1(2)
Given that , r=3t^i+3t2^j4^k.
We know that,velocity is derivative of position with respect to time,
v=drdt=3^i+23t^j

Now to find the magnitude and direction of velocity at t=3 s we put the value in the above equation

v(t=3 s)=3^i+6^j,
magnitude of the vector a^i+b^j is given by a2+b2 and direction is given by tanθ=ba.
Hence,
|v|=32+62=35 m/s

θ=tan1(63)=tan1(2)

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