
The speed of a boat in still water is 15 km per hour. It can go 30 km upstream and return downstream to the original point in 4 hours 30 minutes. Find the speed of the stream.
Answer
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Hint: For solving this problem, let the speed of the stream be x. By using the definition of upstream and downstream, calculate the speed in both the cases. By using the distance and the time given in the problem, we can easily obtain the speed of the stream.
Complete step-by-step solution -
According to the problem statement, we are given the speed of a boat in still water as 15 km/hr. The distance travelled by the boat in upstream and downstream is 30 km in both cases. The total time taken to cover 60 kilometre is 4 hours 30 minutes.
Let the speed of the stream be x. When the boat is upstream, the relative velocity of the boat is given as (v - u). Therefore, the speed of the boat in upstream is (15 - x). When the boat is downstream, the relative velocity of the boat is given by (v + u). Therefore, the speed of the boat in upstream is (15 + x).
The relationship between speed, time and distance can be expressed as:
$\begin{align}
& \text{speed = }\dfrac{\text{distance}}{\text{time}} \\
& \text{time = }\dfrac{\text{distance}}{\text{speed}} \\
\end{align}$
From the above obtained expression, time taken by the boat to go 30 km upstream will be $\dfrac{30}{15-x}$.
Time taken by the boat to go 30 km downstream will be $\dfrac{30}{15+x}$.
As per the question, total time taken to reach the original position is 4 hours 30 minutes. Hence, we get
\[\begin{align}
& \dfrac{30}{15-x}+\dfrac{30}{15+x}=4+\dfrac{1}{2} \\
& 30\times \left( \dfrac{1}{15-x}+\dfrac{1}{15+x} \right)=\dfrac{9}{2} \\
& \dfrac{15+x+15-x}{{{15}^{2}}-{{x}^{2}}}=\dfrac{9}{2\times 30} \\
& \dfrac{30}{225-{{x}^{2}}}=\dfrac{3}{20} \\
& 225-{{x}^{2}}=\dfrac{30\times 20}{3} \\
& 225-{{x}^{2}}=200 \\
& {{x}^{2}}=25 \\
& x=\pm 5km/hr \\
\end{align}\]
Since the speed of the stream can never, so neglecting the negative value (-5).
Therefore, the speed of the stream is 5 km/hr.
Note: Students must be careful while forming the expression of speed of upstream and downstream respectively. Also, the final answer which the student obtained must possess only positive value after discarding the negative value.
Complete step-by-step solution -
According to the problem statement, we are given the speed of a boat in still water as 15 km/hr. The distance travelled by the boat in upstream and downstream is 30 km in both cases. The total time taken to cover 60 kilometre is 4 hours 30 minutes.
Let the speed of the stream be x. When the boat is upstream, the relative velocity of the boat is given as (v - u). Therefore, the speed of the boat in upstream is (15 - x). When the boat is downstream, the relative velocity of the boat is given by (v + u). Therefore, the speed of the boat in upstream is (15 + x).
The relationship between speed, time and distance can be expressed as:
$\begin{align}
& \text{speed = }\dfrac{\text{distance}}{\text{time}} \\
& \text{time = }\dfrac{\text{distance}}{\text{speed}} \\
\end{align}$
From the above obtained expression, time taken by the boat to go 30 km upstream will be $\dfrac{30}{15-x}$.
Time taken by the boat to go 30 km downstream will be $\dfrac{30}{15+x}$.
As per the question, total time taken to reach the original position is 4 hours 30 minutes. Hence, we get
\[\begin{align}
& \dfrac{30}{15-x}+\dfrac{30}{15+x}=4+\dfrac{1}{2} \\
& 30\times \left( \dfrac{1}{15-x}+\dfrac{1}{15+x} \right)=\dfrac{9}{2} \\
& \dfrac{15+x+15-x}{{{15}^{2}}-{{x}^{2}}}=\dfrac{9}{2\times 30} \\
& \dfrac{30}{225-{{x}^{2}}}=\dfrac{3}{20} \\
& 225-{{x}^{2}}=\dfrac{30\times 20}{3} \\
& 225-{{x}^{2}}=200 \\
& {{x}^{2}}=25 \\
& x=\pm 5km/hr \\
\end{align}\]
Since the speed of the stream can never, so neglecting the negative value (-5).
Therefore, the speed of the stream is 5 km/hr.
Note: Students must be careful while forming the expression of speed of upstream and downstream respectively. Also, the final answer which the student obtained must possess only positive value after discarding the negative value.
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