A particle travels with speed 50 m/s from the point (3 m, -7 m) in a direction $7\widehat{i}-24\widehat{j}$. Find its position vector after 3 seconds.

$\text{ Given speed }\mathrm{v}=50\mathrm{m}/\mathrm{s}\text{ in the direction }\overrightarrow{\mathrm{a}}=7\widehat{\mathrm{i}}-24\widehat{\mathrm{j}}$

$\overrightarrow{v}=v\widehat{a};\widehat{a}\text{ is the unit vector in the direction }7\widehat{i}-24\widehat{j}$

$\begin{array}{l}\widehat{a}=\frac{(7\widehat{i}-24\widehat{j})}{\sqrt{(24{)}^2+(7{)}^2}}=\frac{(7\widehat{i}-24\widehat{j})}{25}\\ \overrightarrow{v}=v\widehat{a}=50\cdot \frac{(7\widehat{i}-24\widehat{j})}{25}=2(7\widehat{i}-24\widehat{j})\mathrm{m}/\mathrm{s}\end{array}$

$\text{ Initially the particle is at }{\overrightarrow{\mathrm{r}}}_0=(3\widehat{\mathrm{i}}-7\widehat{\mathrm{j}})\mathrm{m}$

$\text{ Position of the particle after }3\sec ,\overrightarrow{r}={\overrightarrow{r}}_0+\overrightarrow{v}t$

$\Rightarrow \overrightarrow{\mathrm{r}}=(3\widehat{\mathrm{i}}-7\widehat{\mathrm{j}})+3\times 2(7\widehat{\mathrm{i}}-24\widehat{\mathrm{j}})=(45\widehat{\mathrm{i}}-151\widehat{\mathrm{j}})\mathrm{m}$