Question

The speed of the boat in the still water is 8 km/hr. It can go 15 km upstream and 22km downstream in 5 hours. Find the speed of the stream in km/hr.

Answer 1

Hint: When the boat is moving in the upstream it overcomes the speed of the stream. While the boat is moving in the downstream stream speed adds up to the boat speed.
Speed of upstream= speed of the boat in still water – speed of the stream.
Speed of downstream= speed of the boat in still water + speed of the stream
And speed \[=\dfrac{\text{distance}}{\text{time}}\]

Complete step by step answer:
In the question, it is given that, speed of the boat in the still water is 8km/hr and it covers 15km in upstream and 22km in downstream. The total time of the journey is given as 5 hrs.
Consider “t” hrs of journey the boat travels in upstream and (5-t)hrs of the journey the boat travels in downstream.
Let’s consider:
Velocity of the boat in the still water = \[{{v}_{b}}\]
Velocity of the boat in the upstream = \[{{v}_{u}}\]
Velocity of the boat in the downstream = \[{{v}_{d}}\]
Velocity of the steam = \[{{v}_{s}}\]
We know,
Speed of upstream= speed of the boat in still water – speed of the stream.
$\Rightarrow {{v}_{u}}={{v}_{b}}-{{v}_{s}}$
And, Speed of downstream= speed of the boat in still water + speed of the stream.
$\Rightarrow {{v}_{d}}={{v}_{b}}+{{v}_{s}}$
We also know that the time taken for a journey is the ratio of distance of the journey to the speed of the vehicle. So, time of travel in upstream is given as:
Time boat travel in the upstream =\[\dfrac{\text{distance travelled by the boat in the upstream}}{\text{velocity of the boat in upstream}}\]
\[\Rightarrow t=\dfrac{15}{8-{{v}_{s}}}\]\[.....(1)\]
And, the time of travel in downstream is given as :
Time boat travel in the downstream =\[\dfrac{\text{distance travelled by the boat in the downstream}}{\text{velocity of the boat in downstream}}\]
\[\Rightarrow 5-t=\dfrac{22}{8+{{v}_{s}}}\]\[.....(2)\]
Now, we will substitute equation 1 in 2. On substitution, we get:
\[5-\dfrac{15}{8-{{v}_{s}}}=\dfrac{22}{8+{{v}_{s}}}\]
\[\Rightarrow 5=\dfrac{15}{8-{{v}_{s}}}+\dfrac{22}{8+{{v}_{s}}}\]
\[\Rightarrow 5=\dfrac{15\left( 8+{{v}_{s}} \right)+22\left( 8-{{v}_{s}} \right)}{64-{{v}_{s}}^{2}}\]
\[\Rightarrow 5=\dfrac{8\left( 15+22 \right)+{{v}_{s}}\left( 15-22 \right)}{64-{{v}_{s}}^{2}}\]
\[\Rightarrow 5=\dfrac{8\left( 37 \right)-7{{v}_{s}}}{64-{{v}_{s}}^{2}}\]
\[\Rightarrow -5{{v}_{s}}^{2}+7{{v}_{s}}+320-296=0\]
\[\Rightarrow 5{{v}_{s}}^{2}-7{{v}_{s}}-24=0\]
We can see that the above equation is a quadratic in ${{v}_{s}}$ . We know the roots of the quadratic equation $a{{x}^{2}}+bx+c=0$ is given as $x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$ . So, the roots of the equation \[5{{v}_{s}}^{2}-7{{v}_{s}}-24=0\] is given as:
\[{{v}_{s}}=\dfrac{7\pm \sqrt{49+4(24)(5)}}{2(5)}\]
\[\Rightarrow {{v}_{s}}=\dfrac{7\pm 23}{10}=3\]
Here the negative value of \[{{v}_{s}}\] cannot be considered because speed is a scalar quantity which has only magnitude. So, the negative value of \[{{v}_{s}}\] is meaningless.

Therefore, the velocity of the stream is 3km/hr.

Note: While doing substitution, be careful of the sign. Do not make sign mistakes as they can lead to wrong answers. Sign mistakes are common, so students should be very careful. While doing substitution, be careful of the sign. Do not make sign mistakes as they can lead to wrong answers. Also remember, the direction along the stream is called downstream and direction against the stream is called upstream. If speed of boat is uniform in still water, then the formula is
$downstream=\left( u+v \right)km/hr$
$upstream=\left( u-v \right)km/hr$
But here, the speed of the boat is given, so directly use the formula and find accordingly values.
 Sign mistakes are common, so students should be very careful.