The position of a particle is given byr=3.0ti^−2.0t2j^+4.0k^mwhere 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 ?
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Solution
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(a)
v(t)=dr/dt
=3.0i^−4.0tj^
a(t)=dv(t)/dt
=−4.0j^
(b)
v(2)=3.0i^−8.0j^
v(2)=8.54m/s
∠v(2)=tan−1(−8/3)=−69.44∘withxaxis
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