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

For the two given particles $A$ and $B$ as shown in the figure,how much time will it take for their horizontal separation to become zero?

$A$ swimmer can swim in still water with speed $v$ and the river is flowing with velocity $\frac{v}{2}$. To cross the river in the shortest time,he should swim making an angle $\theta$ with the upstream. What is the ratio of the time taken to swim across in the shortest time to that in swimming across over the shortest distance?

Airplanes $A$ and $B$ are flying with constant velocity in the same vertical plane at angles $30^{\circ}$ and $60^{\circ}$ with respect to the horizontal,respectively,as shown in the figure. The speed of $A$ is $100 \sqrt{3} \ m/s$. At time $t=0$,an observer in $A$ finds $B$ at a distance of $500 \ m$. This observer sees $B$ moving with a constant velocity perpendicular to the line of motion of $A$. If at $t = t_0$,$A$ just escapes being hit by $B$,then $t_0$ in seconds is:

Two cars $A$ and $B$ are moving with the same speed of $45\,km/h$ along the same direction. If a third car $C$ coming from the opposite direction with a speed of $36\,km/h$ meets the two cars in an interval of $5\,min$,the distance of separation between the two cars $A$ and $B$ should be.......$km$.

$A$ swimmer can swim with speed '$v$' with respect to still water in a river which is flowing with speed '$u$'. There is a float moving with the river. Now the swimmer overtakes the float,gets a lead of '$l$',and returns back to the float. The time taken by the swimmer in this process will be: