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

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

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

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