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(D) For two particles to collide,their position vectors at time $t$ must be equal: $r_1(t) = r_2(t)$.Given $r_1(t) = r_1 + v_1 t$ and $r_2(t) = r_2 +

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$8$

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(D) For two particles to collide,their position vectors at time $t$ must be equal: $r_1(t) = r_2(t)$.Given $r_1(t) = r_1 + v_1 t$ and $r_2(t) = r_2 + v_2 t$,the condition for collision is $r_1 + v_1 t = r_2 + v_2 t$.Rearranging,we get $r_1 - r_2 = (v_2 - v_1) t$.Substitute the given values:$r_1 - r_2 = (3 \hat{i} + 5 \hat{j}) - (-5 \hat{i} - 3 \hat{j}) = 8 \hat{i} + 8 \hat{j}$.$v_2 - v_1 = (a \hat{i} + 7 \hat{j}) - (4 \hat{i} + 3 \hat{j}) = (a - 4) \hat{i} + 4 \hat{j}$.Given $t = 2 \ s$,substitute into the equation:$8 \hat{i} + 8 \hat{j} = ((a - 4) \hat{i} + 4 \hat{j}) 	imes 2$.Dividing by $2$:$4 \hat{i} + 4 \hat{j} = (a - 4) \hat{i} + 4 \hat{j}$.Comparing the coefficients of $\hat{i}$:$a - 4 = 4 \implies a = 8$.

Two particles having position vectors $r_1 = (3 \hat{i} + 5 \hat{j}) \ m$ and $r_2 = (-5 \hat{i} - 3 \hat{j}) \ m$ are moving with velocities $v_1 = (4 \hat{i} + 3 \hat{j}) \ m/s$ and $v_2 = (a \hat{i} + 7 \hat{j}) \ m/s$. If they collide after $2 \ s$,then the value of $a$ is:

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