Question

# The position of a particle is given byr=3.0t^i−2.0t2^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)=d→r/dt =3.0^i−4.0t^j→a(t)=d→v(t)/dt =−4.0^j(b) →v(2)=3.0^i−8.0^jv(2)=8.54m/s∠v(2)=tan−1(−8/3)=−69.44∘with x axis

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