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Question

The position of a particle is given byr=3.0t^i2.0t2^j+4.0^kmwhere t is in seconds and the coefficients have the proper units for r to be in metres(a) Find the v and a of the particle?(b) What is the magnitude and direction of velocity of the particle at t = 2.0 s ?

Solution
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(a)
v(t)=dr/dt
=3.0^i4.0t^j

a(t)=dv(t)/dt
=4.0^j

(b)
v(2)=3.0^i8.0^j
v(2)=8.54m/s
v(2)=tan1(8/3)=69.44with x axis

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