The distance between $A$ and $B$ is $30$ km. A bus left $A$ and first travelled at a constant speed. Ten minutes later a helicopter left $A$ and flew along the highroad to $B$. It overtook the bus in five minutes and continued on its way to $B$. Without landing at $B$, the helicopter turned back and again encountered the bus $20$ minutes after it left point $A$. Determine the speeds of the bus and the helicopter.
Given: Distance $AB = d\left(AB\right) = 30$ km
Let speed of bus $= u$ kmph
Helicopter starts after $10$ minutes after the bus
Therefore, time taken by bus $= \text{time taken by helicopter} + 10 \text{ min}$
Helicopter overtook the bus in $5$ minutes
i.e. $\;$ the bus has been traveling for $5 + 10 = 15$ min
Distance covered by the bus in $15$ min i.e. $\left(\dfrac{1}{4} \text{hr}\right)= \dfrac{u}{4}$ km
The helicopter covers a distance $\left(\dfrac{u}{4}\right)$ km in $5$ min i.e. $\dfrac{1}{12}$ hr
Therefore, speed of helicopter $= s_h = \dfrac{u/4}{1/12} = 3u$ kmph
The helicopter meets the bus in $20$ min
i.e. $\;$ the bus has been traveling for $20 + 10 = 30$ min
Let the bus reach point $A_1$ in $30$ min
Distance covered by the bus in $30$ min i.e. $\left(\dfrac{1}{2} \text{hr}\right) = d \left(AA_1\right) = \dfrac{u}{2}$ km
Remaining distance $= d\left(A_1B\right) = 30 - \dfrac{u}{2} km$
As per sum, distance covered by helicopter in $20$ min $= 30 + 30 - \dfrac{u}{2} = 60 - \dfrac{u}{2}$ km $\;\;\; \cdots \; (1)$
Also, distance covered by helicopter in $20$ min i.e. $\left(\dfrac{1}{3} \text{hr}\right) = \dfrac{3u}{3} = u$ km $\;\;\; \cdots \; (2)$
Therefore, we have from equations $\left(1\right)$ and $\left(2\right)$
$60 - \dfrac{u}{2} = u$
i.e. $\;$ $60 = u + u/2 = 3u /2$ $\implies$ $u = 40$
i.e. $\;$ speed of bus $= 40$ kmph
Therefore, speed of helicopter $= 3u = 3 \times 40 = 120$ kmph