Question

Two particles having position vectors ${\vec{r}}_1=(3\hat{i}+5\hat{j})m$ and ${\vec{r}}_2=(-5\hat{i}-3\hat{j})m$ at $t=0$ are moving with velocities ${\vec{v}}_1=(4\hat{i}+3\hat{j})$ $m{s}^{-1}$ and ${\vec{v}}_2=(a\hat{i}+7\hat{j})m{s}^{-1}$. For what value of ' $a^{\prime }$ they collide at $=2$ $\sec$

Answer 1

Answer: 8

When, they collide then the final position of the particles will same it means,
${\vec{r}}_1+{\vec{v}}_{1}t={\vec{r}}_2+{\vec{v}}_{2}t$
$(3\hat{i}+5\hat{j})+(8\hat{i}+6\hat{j})=(-5\hat{i}-3\hat{j})+(2a\hat{i}+14\hat{j})$

$11=-5+2a$
$a=8$