Question

An ion's position vector is initially r with arrow = 7.0i hat â 7.0j + 1.0k, and 5.0 s later it is r with arrow = 7.0i hat + 7.0j â 8.0k, all in meters. What was its v with arrow avg during the 5.0 s?

Answer 1

Answer:

Shown in the explanation

Explanation:

Position vector of a  particle at a given instant is given by:

$\vec{r}=x\hat{i}+y\hat{j}+z\hat{k}$

On the other hand, the average velocity is the change in the particle’s position vector divided by time interval:

$\vec{v}=\frac{\Delta \vec{r}}{\Delta t}=\frac{\vec{r_{2}}-\vec{r_{1}} }{t_{2}-t_{1}} \\ \\ \\ Where: \\ \\ \\ \vec{r_{1}}: Initial \ position \\ \\ \vec{r_{2}}: Final \ position \\ \\ t_{1}:Initial \ time \\ \\ t_{2}:Final \ time$

So we have:

$\vec{r_{1}}=7\hat{i}+7\hat{j}+1\hat{k} \\ \\ \vec{r_{2}}=7\hat{i}+7\hat{j}+8\hat{k} \\ \\ \\ Then: \\ \\ \Delta \vec{r} = \vec{r_{2}}-\vec{r_{1}}=7\hat{i}+7\hat{j}+8\hat{k}-(7\hat{i}+7\hat{j}+1\hat{k}) \\ \\ \Delta \vec{r}=7\hat{k} \\ \\ \\ We \ also \ know: \\ \\ \Delta t=5s$

Finally, the average velocity is:

$\vec{v}=\frac{\Delta \vec{r}}{\Delta t} \\ \\ \\ \vec{v}=\frac{7\hat{k}}{5} \\ \\ \boxed{\vec{v}=\frac{7}{5}\hat{k} \ units/s}$