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Velocity of the river is?

Since it covers 1km in 15 mins, the relative velocity of the boat with respect to the river will be $$\frac 14 V_{br}=1$$ $$V_{br}=4km/h$$ So $$V_b=V_{br} + V_r$$ $$V_r=1km/h$$ The right answer is 3km/h, so what am I doing wrong?

Aditya
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    The flowing water will give lateral movement to the boat. Think of Pythagoras. Your initial calculation $V_{br}$ will reveal the velocity of water. The equality $V_b = V_{br} + V_r$ is equality of vectors. – AlvinL Sep 08 '19 at 15:18
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    You need to take into account, that velocities are vectors! In this case, it is sufficient to take them as 2-dimensional vectors. The Velocity is the absolute value of the velocity vector. Therefore: $v_b = \sqrt{v_{br} ^2 + v_r ^2}$. Given $v_{br}=4$ and $v_b = 5$, you find $v_r=3$. (recall: $3^2+4^2=5^2$). Try to work out the details of the addition (what are the directions of the velocities!) – Caroline Sep 08 '19 at 15:19
  • Yeah, thought that, but since the angles weren’t given, I wasn’t sure about it. – Aditya Sep 08 '19 at 15:19

2 Answers2

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The only way to go along the shortest path is by tilting the boat against the river current. A good picture helps:
enter image description here

You're given $V_{BW} = 5kmph$ and $V_{BS} = 1km ~in~15min$ .
You can find $V_{WS}$

AgentS
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You should use the equation

$$V_b^2=V_{br}^2+ V_r^2$$

instead of

$$V_b=V_{br} + V_r$$

because the three velocity vectors are not in the same direction. Rather, they form a right triangle.

Then, you get

$$V_r = \sqrt{ V_b^2 - V_{br}^2} = \sqrt{ 5^2 - 4^2}= 3 \text{km/h}$$

Quanto
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