How to find the equivalent speed of vector?

Question

I've been given this formula:

Suppose c is the current’s velocity vector, s is the velocity vector the object would have if the water was still, and f = c + s is the object's resultant velocity vector.

Below is a question I am trying to figure out.

A boat needs to travel south at a speed of 20 kmh-1. However a constant current of 6 kmh-1 is flowing from the south-east. Use vectors to find the equivalent speed in still water for the boat to achieve the actual speed of 20 km h-1

Is it wrong to apply the formula like so:

20^2 = 6^2 + x^2
400 = 36 + x^2
364 = x^2
x = √364

However, the answer says it is 24.6 km h-1. I am trying to figure out what I am doing wrong?

Answer 1

In vector notation, using a coordinate system oriented W-E ( $x-$axis) and S-N ($y-$axis), the current velocity vector is: $$ \vec u=(-3\sqrt{2},3\sqrt{2})^T $$

the final velocity vector of the boat is $$ \vec v=(0,-20)^T $$ and you want a vector $\vec x=(x,y)^T$ such that: $$ \vec u +\vec x= \vec v $$

so we must have: $$ \begin{cases} -3\sqrt{2}+x=0\\ 3\sqrt{2}+y=-20 \end{cases} $$ solve this system and find: $|\vec x|=\sqrt{x^2+y^2}= 24.611087..$.

Answer 2

The angle between south and south east directions is $45$ degrees.

We use the generalized Pythagore's Theorem.

$x^2=20^2+6^2-2.20.6.\cos(180-45)$

or

$x^2=436+240.\frac{\sqrt{2}}{2}$

$\approx 436+120.1,414$

$\approx 605,7$

and

$x\approx 24,6109 \; \; km.h{-1} $.