Question

The position vector of a particle is determined by the expression $\overset{ \rightarrow }{r}=2t^{2}\hat{i}+3t\hat{j}+9\hat{k}$ , where $t$ represents time in seconds. The magnitude of its displacement (in $m$ ) in first two seconds is

Answer 1

$\overset{ \rightarrow }{r}=2t^{2}\hat{i}+3t\hat{j}+9\hat{k}$
At $t=0,\overset{ \rightarrow }{r_{1}}=9\hat{k}$
At $\text{t}=2\text{s},\overset{ \rightarrow }{\text{r}_{2}}=8\hat{\text{i}}+6\hat{\text{j}}+9\hat{\text{k}}$
$\therefore \, \Delta \overset{ \rightarrow }{\text{r}}=\overset{ \rightarrow }{\text{r}_{2}}-\overset{ \rightarrow }{\text{r}_{1}}=8\hat{\text{i}}+6\hat{\text{j}}$
$\therefore\left|\Delta \overrightarrow{r_{r}}\right|=\left|\overrightarrow{r_{2}}-\overrightarrow{r_{1}}\right|=\sqrt{(8)^{2}+(6)^{2}}=10 m$