On a two-lane road,car $A$ is travelling with a speed of $36 \; km/h$. Two cars $B$ and $C$ approach car $A$ in opposite directions with a speed of $54 \; km/h$ each. At a certain instant,when the distance $AB = AC = 1 \; km$,car $B$ decides to overtake car $A$ before car $C$ does. What minimum acceleration (in $m/s^2$) of car $B$ is required to avoid an accident?

The velocities of $A$ and $B$ are $\vec{v}_A = 2 \hat{i} + 4 \hat{j}$ and $\vec{v}_B = 3 \hat{i} - 7 \hat{j}$. The velocity of $B$ as observed by $A$ is:

$A$ bird is tossing (flying to and fro) between two cars moving towards each other on a straight road. One car has a speed of $54 \text{ kmh}^{-1}$ while the other has a speed of $36 \text{ kmh}^{-1}$. The bird starts moving from the first car towards the other and is moving with a speed of $36 \text{ kmh}^{-1}$ when the two cars are separated by $36 \text{ km}$. The total distance covered by the bird before the cars meet each other is: (in $\text{ m}$)

$A$ car travels with a speed of $40 \text{ km h}^{-1}$. Rain drops are falling at a constant speed vertically. The traces of the rain on the side windows of the car make an angle of $30^{\circ}$ with the vertical. The magnitude of the velocity of the rain with respect to the car is

$A$ person walks up a stalled escalator in $90 \ s$. When standing on the same moving escalator,he reaches the top in $60 \ s$. The time it would take him to walk up the moving escalator will be: (in $s$)

$A$ particle $(A)$ moves due north at $3\,km/h$ and another particle $(B)$ moves due west at $4\,km/h$. The relative velocity of $A$ with respect to $B$ is $(\tan 37^{\circ} = 3/4)$.