Question

The position of a particle is given by $\overrightarrow{r}=(3t\hat{i}-t^2\hat{j}+4\hat{k})$ , what will be the velocity and acceleration?

Answer 1

Given that,

Position vector $\overrightarrow{r}=(3t\hat{i}-t^2\hat{j}+4\hat{k})$

Now, the velocity is

$\overrightarrow{v}=\frac{d\overrightarrow{r}}{dt}=\frac{d}{dt}(3t\hat{i}-t^2\hat{j}+4\hat{k})$

$v=(3\hat{i}-2t\hat{j})m/s$

Now, the acceleration is

$\overrightarrow{a}=\frac{d\overrightarrow{v}}{dt}$

$\overrightarrow{a}=\frac{d}{dt}(3\hat{i}-2t\hat{j})$

$\overrightarrow{a}=(-2\hat{j})m/s^2$

Hence, this is the required solution