Given that the acceleration vector of a particle is a(t)=-25cos(5t)i-25sin(5t)j+

Question

Given that the acceleration vector of a particle is a(t)=-25cos(5t)i-25sin(5t)j+3tk, the initial velocity is v(0)=i+k, and the initial position vectoris r(0)=i+j+k, compute: The velocity vector v(t)=___i+____+j____+k and the position vector r(t)=___i+___+j___k.

Explanation / Answer

A>acceleration=-25cos(5t)i-25sin(5t)j+3tk velocity=integrate acceleration =intg[-25cos(5t)i-25sin(5t)j+3tk] =intg[-25cos(5t)i] + intg [-25sin(5t)j] +intg [3tk] =(-5 sin5t + C1)i +( 5 cos5t+ C2)j+ (3t^2/2 + C3)k v(0)=i+k=C1 i+ (5+C2)j + C3 k =>C1=1, C2=-5, C3=1 SO V=(-5 sin5t + 1)i +( 5 cos5t-5)j+ (3t^2/2 + 1)k b>position=integrate velocity =intg [(-5 sin5t + 1)i +( 5 cos5t-5)j+ (3t^2/2 + 1)k] =intg[(-5 sin5t + 1)i] +intg[( 5 cos5t-5)j] +intg [(3t^2/2 + 1)k] =(cos 5t +t + C1)i + (sin 5t -5t +C2)j+ (t^3/2 +t +C3)k r(0)=i+j+k=(1+C1)i + C2j +C3 k so, C1=0, C2=1, c3=1 so r=(cos 5t +t )i + (sin 5t -5t +1)j+ (t^3/2 +t +1)k EXPLAINED PLZ RATE IT FIRST

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