$$
begin{align*}
&v_{max}=5hspace{1mm}frac{m}{s}text{maximum speed of boat, direction of boat is the same as river’s current}
\
&v_{min}=1hspace{1mm}frac{m}{s}
text{minimum speed of boat, direction of boat is the opposite from river’s current}
end{align*}
$$

We are to find the magnitude and direction of the powerboat relative to the shore. First, let us review the given values.

$$
begin{align*}
v_b = 13 ,mathrm{frac{m}{s}}\
v_r = 5.0 ,mathrm{frac{m}{s}}
end{align*}
$$

where $v_b$ is the velocity of the powerboat due $textit{northwest}$ relative to the water and $v_r$ is the velocity of water flow in the river that is due $textit{north}$.

Given the equation:
begin{align}
v &= sqrt{v_{x}^{2} + v_{y}^{2}}
intertext{such that,}
v_x &= v_{b} cos{theta} + v_r\
v_y &= v_{b} sin{theta}
end{align}
where $v_x$ is the sum of all the velocity with respect to the $x$-axis and $v_y$ is the sum of the components with respect to $y$-axis

Since the velocity of the powerboat is due northwest, we can say that $theta = 45^{circ}$. We can now use the equation and plug in the values to get the magnitude of the velocity $v$.

$$
begin{align*}
v &= sqrt{v_{x}^{2} + v_{y}^{2}}\
&= sqrt{(v_{b} cos{theta} + v_r)^{2} + (v_{b} sin{theta})^{2}}\
&= sqrt{((13 ,mathrm{frac{m}{s}})cos{(45^{circ})} + 5.0,mathrm{frac{m}{s}})^{2} + ((13 ,mathrm{frac{m}{s}}) sin{(45^{circ})})^{2}}\
&= boxed{17 ,mathrm{frac{m}{s}}}
end{align*}
$$

Now, to get the direction, we can use the equation:

$$
begin{align}
tan{theta} = frac{v_y}{v_x}
end{align}
$$

We can isolate the $theta$ an calculate for its value.

$$
begin{align*}
tan{theta} &= frac{v_y}{v_x}\
theta &= tan^{-1} left(frac{v_y}{v_x} right)\
&= tan^{-1} left(frac{v_{b} sin{theta}}{v_{b} cos{theta} + v_r} right)\
&= tan^{-1} left(frac{(13 ,mathrm{frac{m}{s}}) sin{(45^{circ})}}{(13 ,mathrm{frac{m}{s}}) cos{(45^{circ})} + 5.0,mathrm{frac{m}{s}}} right)\
&= boxed{33^{circ} text{ West of North}}
end{align*}
$$

$$
v = 17 ,mathrm{frac{m}{s}} text{ due } 33^{circ} text{ West of North}
$$

Given the values:

$$
begin{align*}
v_p &= 175 ,mathrm{frac{km}{h}}\
v_w &= 85 ,mathrm{frac{km}{h}}
end{align*}
$$

such that the velocity of the airplane is denoted by $v_p$, which is due $textit{south}$ relative to the air, and $v_w$ is the velocity of the wind that is due $textit{east}$ relative to the ground.

Given the equation:

$$
begin{align}
v_R = sqrt{v_{p}^2 +v_{w}^2}
end{align}
$$

We can plug in the given values and solve for the magnitude of relative velocity $v_R$.

$$
begin{align*}
v_R &= sqrt{v_{p}^2 +v_{w}^2}\
&= sqrt{(175 ,mathrm{frac{km}{h}})^2 + (85 ,mathrm{frac{km}{h}})^2}\
&= 194 ,mathrm{frac{km}{h}}\
&= boxed{1.9 times 10^2 ,mathrm{frac{km}{h}}}
end{align*}
$$

Now, we will look for the angle $theta$ of the relative velocity. Given the equation:

$$
begin{align}
tan{theta} = frac{v_y}{v_x}
end{align}
$$

We can isolate $theta$ and substitute $v_y$ for $v_p$ and $v_x$ for $v_w$, plug in the given values, then solve for the value of $theta$.

$$
begin{align*}
tan{theta} &= frac{v_y}{v_x}\
theta &= tan^{-1} left(frac{v_p}{v_x}right)\
&= tan^{-1} left(frac{175 ,mathrm{frac{km}{h}}}{85 ,mathrm{frac{km}{h}}}right)\
&= boxed{64 ^{circ}}
end{align*}
$$

$$
v_R = 1.9 times 10^2 ,mathrm{frac{km}{h}} text{, } 64 ^{circ} text{ south of east}
$$

begin{align*}
intertext{We can write velocity of a wind, relative to the ground as sum of it’s components. We will use vectors because velocity is vector:}
&vec v_w=v_{wh}hat{i}+v_{wv}hat{j}\
intertext{$hat{i}$ and $hat{j}$ are unit vectors for x and y axis, we use them only to show direction of components, their value is equal to 1. }
&vec v_w=v_{w}cos 45 hat{i}+v_{w}sin 45hat{j}\
&vec v_b=(45.95 hat{i}+45.95hat{j})hspace{1mm}frac{m}{s}\
intertext{Velocity of airplane only has vertical component (north ):}
&vec v_a=235hat{j}hspace{1mm}frac{m}{s}
intertext{Now we can sum velocity of wind and airplane in order to find velocity of airplane relative to the ground.}
&vec v=vec v_w +vec v_w\
&vec v=(45.95 hat{i}+45.95hat{j})+235hat{j}\
&vec v=(45.95 hat{i}+280.95hat{j})hspace{1mm}frac{m}{s}\
intertext{Magnitude is:}
&v^2=v_h^2+v_v^2\
&v=sqrt{45.95^2+280.95^2}\
&v=285hspace{1mm}frac{m}{s}
end{align*}

$$
begin{align*}
&v=285hspace{1mm}frac{m}{s}\
&theta=80^{circ}hspace{1mm}text{Nort of east}
end{align*}
$$

begin{align*}
intertext{Data known from, the problem:}
&vec v_a=285hspace{1mm}frac{km}{h}hspace{1mm}text{Airplane’s velocity, relative to the air}\
&vec v_w=95hspace{1mm}frac{km}{h}hspace{1mm}text{Wind’s velocity, relative to the ground, directed $30^{circ}$ north of east.}
intertext{We are looking for airplane’s direction while flying, in order to
land at an airport due north, and we are looking for airplane’s speed relative to the ground. Speed is scalar, it is magnitude of velocity. But we are also looking for direction.}
intertext{North has direction of y axis, and east has direction of x axis, so velocity of wind actually has horizontal (to the east) component and vertical (to the north) component.}
end{align*}

begin{align*}
intertext{We can write velocity of a wind, relative to the ground as sum of it’s components. We will use vectors because velocity is vector:}
&vec v_w=v_{wh}hat{i}+v_{wv}hat{j}\
intertext{$hat{i}$ and $hat{j}$ are unit vectors for x and y axis, we use them only to show direction of components, their value is equal to 1. }
&vec v_w=v_{w}cos 30 hat{i}+v_{w}sin 30hat{j}\
&vec v_b=(82.27 hat{i}+47.5hat{j})hspace{1mm}frac{m}{s}\
intertext{If we wand airplane to land with only north component, east component, which is horizontal component of a wind, must cancel horizontal component of airplane. That can be happened if it has magnitude of horizontal of wind, but opposite direction.}
intertext{We can find direction by using cos of an angle. Cos of angle is length of the adjacent side divided by length of the hypotenuse. Hypotenuse is airplane’s velocity, and adjacent side is horizontal side, that we already said that it has to has magnitude of horizontal of wind, so:}
&costheta=frac{v_{wh}}{v_a}\
&costheta=frac{82.27}{285}\
&costheta=0.288\
&theta=cos^{-1}(0.288)\
&theta=73^{circ}hspace{1mm}text{north of west}
end{align*}

begin{align*}
intertext{Now we can sum velocity of wind and airplane in order to find velocity of airplane relative to the ground.We already said that horizontal components cancel each other, so we only have to sum vertical components.}
intertext{We calculated vertical component of wind:}
&v_{wv}=47.5hspace{1mm}frac{m}{s}
intertext{We need to calculate vertical component of airplane. It is flying with speed 285 $frac{km}{h}$ 73 $^circ$ north of west, so north component is:}
&v_{av}=v_asin 73\
&v_{av}=285cdot0.96\
&v_{av}=273.6hspace{1mm}frac{m}{s}\
intertext{Now we can find velocity of airplane, relative to the ground:}
&v=v_{av}+v_{wv}\
&v=273.6+47.5\
&v=321hspace{1mm}frac{m}{s}
end{align*}

$$
begin{align*}
&theta=73^{circ}hspace{1mm}text{north of west}\
&v=321hspace{1mm}frac{m}{s}
end{align*}
$$

Let $u$ be the velocity of the river;
Let $v_{x}$ be the x component, and,
$v_{y}$ be the y component of your velocity

$v_{x}$+u = 0;
Therefore, $v_{x}$= -u

Since the final point is directly opposite to initial starting point, the total displacement along x direction should be zero. The resultant velocity along x direction is $v_{x}$+u and this should equal to zero. The y component $v_{y}$ can have any non-zero value.

Hence, on moving with uniform velocity, the speed v=$sqrt{v_{x}^{2}+v_{y}^{2}}$ at an angle $theta$ from the x axis, where $tantheta$ = $left(dfrac{v_{y}}{v_{x}} right)$, you are able to move straight across the river.