Find the position vector of a partical

Question

Find the position vector for a particle with acceleration, initial velocity, and initial position given below.

ä(t) =(3t, 6 sin(t), cos(4t))

v(0) (-2,4, 2)

r(0) = (1,4, -3)

r(t) =

Answer 1

V(t) = ∫ a(t) dt =∫<3t,6sint,cos4t>dt = < 3t²/2 +κ, - 6cost + λ , ((sin4t)/4) +μ >

and v(0)= < − 2 , 4, 2 > then

1. 3t²/2 +κ = −2 , for t=0 hence κ = − 2
2. 6cost + λ = 4 , for t=0 then λ = − 2
3. ((sin4t)/4) +μ =2 , for t=0 therefore μ =2

This means V(t) = ∫ a(t) dt = < 3t²/2 +κ, - 6cost + λ , ((sin4t)/4) +μ > = < 3t²/2 −2, - 6cost −2 , ((sin4t)/4) +2 >

Or equivalently V(t) = < 3t²/2 −2, - 6cost −2 , ((sin4t)/4) +2 >

Now r(t) = ∫ V(t) dt = ∫ < 3t²/2 −2, - 6cost −2 , ((sin4t)/4) +2 > dt =

= < t³/2 −2t +φ , −6 sint t − 2t ÷ ξ , - ((cos4t)/16) +2t ÷ψ > .

Then r(t)= < t³/2 −2t +φ , −6 sint t − 2t ÷ ξ , - ((cos4t)/16) +2t ÷ψ > . Also r(0) = <1 ,4 , −3 > .That means

1. t³/2 −2t +φ = 1 for t=0 ⇒ φ = 1
2. −6 sint t − 2t ÷ ξ=4 for t=0 ⇒ ξ=4 and
3. - ((cos4t)/16) +2t ÷ψ = −3 for t=0 ⇒ -1/16 ÷ψ = −3 ⇒ ÷ψ = −3 + 1/16 ⇒ ψ = −47/16

Finally r(t)= < t³/2 −2t +1 , −6 sint t − 2t ÷ 4, - ((cos4t)/16) +2t − 47/16 >