Question #165776

A particle moves along ay-plane with a position ofx (t) = sin(2t) + 3 cos(4t)y (t) = sin(t) – 6 cos(3t)a. Determine the position vector of the particle.b. Determine the velocity vector of the particle.c. Determine the acceleration vector of the particle.d. Determine the position vector of the particle when t = 4s.e. Determine the average acceleration from t = 0s to t= 5s.f. Determine the instantaneous acceleration when të Is.g. Determine the average velocity from t = ls to t= 3s.​

1
Expert's answer
2021-02-22T16:03:34-0500

a)


r=(sin(2t)+3cos(4t)sin(t)6cos(3t))\vec{r}=\begin{pmatrix} \sin(2t) + 3 \cos(4t) \\ \sin(t) – 6 \cos(3t) \end{pmatrix}

b)


v=(2cos(2t)12sin(4t)cos(t)+18sin(3t))\vec{v}=\begin{pmatrix} 2 \cos(2t) -12\sin(4t) \\ \cos(t) +18 \sin(3t) \end{pmatrix}

c)


a=(4sin(2t)48cos(4t)sin(t)+54cos(3t))\vec{a}=\begin{pmatrix} -4 \sin(2t) -48\cos(4t) \\ - \sin(t) +54 \cos(3t) \end{pmatrix}

d)


r=(1.885.82)\vec{r}=\begin{pmatrix} -1.88 \\ - 5.82 \end{pmatrix}

e)


aav=(2.932.20)\vec{a}_{av}=\begin{pmatrix} -2.93 \\ 2.20 \end{pmatrix}

f)


a(5)=(17.440.1)\vec{a}(5)=\begin{pmatrix} -17.4\\ -40.1 \end{pmatrix}

g)


vav=(1.650.59)\vec{v}_{av}=\begin{pmatrix} 1.65 \\ -0.59 \end{pmatrix}


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