Question

A motor boat whose speed is $24\text{ km/h}$ in still water takes $1\text{ hour}$ more to go \[32\text{ km}\] upstream than to return downstream to the same spot. Find the speed of the stream.

Answer 1

Hint: The main concept behind this type of question is that when an object moves in the direction of water current (downstream) then the net speed of object is the sum of its speed in still water and that of stream and when it flows in the direction opposite to that of water current (upstream) then the net speed of object is the difference of speed of object and that of stream. This concept is known as relative velocity.

Complete step-by-step solution -
Now, we have been given that the velocity of a motor boat in still water is $24\text{ km/h}$.
Let the speed of the stream is $u\text{ km/h}$.
Hence the speed of boat in upstream is $(24-u)\text{ km/h}$ and the speed of boat in downstream is \[(24+u)\text{ km/h}\].
Let the time taken by boat to go $32\text{ km}$ upstream be \[t\text{ hours}\]. Then as per the question the time taken by boat to cover the same distance downstream is $(t-1)\text{ hours}$ .
We know that;
\[\text{Distance travelled = speed }\!\!\times\!\!\text{ time taken}.............................\text{(1)}\]
From equation \[\left( 1 \right)\], for journey along upstream we have;
\[\begin{align}
  & 32=\left( 24-u \right)\times \left( t \right) \\
 & t=\dfrac{32}{\left( 24-u \right)}\ldots \ldots \ldots .............................................\ldots \ldots \ldots \ldots .\left( 2 \right) \\
\end{align}\]
From equation \[\left( 1 \right)\], for journey along downstream;
\[32=\left( 24+u \right)\times \left( t-1 \right)..........................................................(3)\]
Putting the value of $t$ from equation \[\left( 2 \right)\] in equation \[\left( 3 \right)\], we get;
\[32=\left( 24+u \right)\times \left( \dfrac{32}{\left( 24-u \right)}-1 \right)\]
Taking L.C.M. we get;
\[\begin{align}
  & 32=(24+u)\times \left( \dfrac{32-(24-u)}{\left( 24-u \right)} \right) \\
 & 32=(24+u)\times \left( \dfrac{8+u}{24-u} \right) \\
\end{align}\]
By cross multiplying we get;
\[\begin{align}
  & 32\times (24-u)=(24+u)\times (8+u) \\
 & 768-32u=192+24u+8u+{{u}^{2}} \\
 & {{u}^{2}}+64u-576=0 \\
 & {{u}^{2}}+72u-8u-576=0 \\
 & (u+72)(u-8)=0 \\
 & \therefore u=-72\text{ or }u=8. \\
\end{align}\]
The speed cannot be negative and hence, the speed of the stream is \[u\text{ km/h}=\text{ 8 km/h}\].

Note: We have to understand the concept of relative velocity, quadratic equation and speed-distance formula. For an object moving downstream the net speed of object is the sum of object in still water and that of stream and when the object moves upstream the net speed of object is the difference of speed of object in still water and that of stream.