Two cars are moving in the same direction with the same speed $30 \,km/hr$. They are separated by a distance of $5\, km$, the speed of a car moving in the opposite direction if it meets these two cars at an interval of $4$ minutes, will be.......$km/hr$
$40$
$45$
$30$
$15$
Two cars are moving in the same direction with the same speed $30 \,km/hr$. They are separated by a distance of $5\, km$, the speed of a car moving in the opposite direction if it meets these two cars at an interval of $4$ minutes, will be.......$km/hr$
The two car (say $A$ and $B$) are moving with same velocity, the relative velocity of one (say $B$) with respect to the other $A,\,{\overrightarrow v _{BA}} = {\overrightarrow v _B} - {\overrightarrow v _A} = v - v = 0$
So the relative separation between them ($= 5\, km$) always remains the same.
Now if the velocity of car (say $C$) moving in opposite direction to $A$ and $B$, is ${\overrightarrow v _C}$ relative to ground then the velocity of car $C$ relative to $A$ and $B$ will be
${\overrightarrow v _{rel.}} = {\overrightarrow v _C} - \overrightarrow v $
But as $\overrightarrow v $ is opposite to $v_C$
So $\,{v_{rel}} = {v_C} - ( - 30) = ({v_C} + 30)\,km/hr.$
So, the time taken by it to cross the cars $A$ and $B$
$t = \frac{d}{{{v_{rel}}}}\,\,\,\, \Rightarrow \,\,\,\frac{4}{{60}} = \frac{5}{{{v_C} + 30}}\,\,$
$ \Rightarrow \,\,\,{v_C} = 45\,km/hr.$