Question

Suppose that the acceleration vector of a particle is given by a(t)=⟨−16cos(4t),−16sin(4t),−4t⟩, the paricle's initial velocity is v(0)=⟨1,0,1⟩, and the particle's initial position vector is r(0)=⟨1,1,1⟩. Compute the following: A. The velocity vector of the particle at time t: v(t)= -4sin(4t)+1 i+ 4cos(4t)-4 j+ 4t^2/2-1 k B. The position vector of the particle at time t: r(t)= cos(4t)+t i+ sin4t-t+1 j+ 1 k

Answer 1

Answer with Explanation:

The acceleration vector is given by

`\overrightarrow{a}=-16cos(4t)\widehat{i}-16sin(4t)\widehat{j}-4t\widehat{k}`

Now by definition of acceleration we have

`\frac{d\overrightarrow{v}}{dt}=\overrightarrow{a}\\\\\int \overrightarrow{dv} =\int \overrightarrow{a}dt\\\\\overrightarrow{v}=\int \overrightarrow{a}dt\\\\\overrightarrow{v}-\overrightarrow{v_o}=\int _(0)^(t)(-16cos(4t)\widehat{i}-16sin(4t)\widehat{j}-4t\widehat{k})dt\\\\\overrightarrow{v}-\overrightarrow{v_o}=(-16sin(4t))/(4)\widehat{i}+(16cos(4t))/(4)\widehat{k}-(4t^2)/(2)\\\\\overrightarrow{v(t)}=-4sin(4t)+1)\widehat{i}+(4cos(4t)-4)\widehat{j}-(2t^2+1)\widehat{k}`

Similarly the position vector can be obtained by integrating the velociry vector as

`\frac{d\overrightarrow{r}}{dt}=\overrightarrow{v}\\\\\int \overrightarrow{dr} =\int \overrightarrow{v}dt\\\\\overrightarrow{r}=\int \overrightarrow{v}dt\\\\\overrightarrow{r}=\int _(0)^(t)(-4sin(4t)+1)\widehat{i}+(4cos(4t)-4)\widehat{j}-(2t^2+1)\widehat{k})dt\\\\\overrightarrow{r}=((4cos(4t))/(4)+t)\widehat{i}+((4sin(4t))/(4)-4t)\widehat{k}-((2t^3)/(3)+t)\widehat{k}\\\\\overrightarrow{r}=(cos(4t+t)\widehat{i}+(sin(4t)-4t+1)\widehat{j}+((2t^3)/(3)+t+1)\widehat{k}`