A powerboat heads due northwest at 11 m/s relative to the water across a river that flows due north at 5.4 m/s. What is the velocity (both magnitude and direction) of the motorboat relative to the shore?

Vb = 11m/s @ 135o + 5.4m/s @ 90o.

X=11*cos135+5.4*cos90=-7.78 + 0=-7.78m/s
Y=11*sin135+5.4*sin90 =7.78+5.4=13.18 m/s.

tanAr = Y/X = 13.18/-7.78 = 1.69385.
Ar = -59.4o = Reference angle.
A = --59.4 + 180 = 120.6o = Direction.

Vb = X/cosA = -7.78/cos120.6=15.28 m/s=
Velocity of boat.

To find the velocity of the powerboat relative to the shore, we need to consider the velocity of the boat relative to the water and the velocity of the water itself.

Let's break down the given velocities:
- The velocity of the powerboat relative to the water is 11 m/s in the northwest direction.
- The velocity of the river relative to the shore is 5.4 m/s due north.

To find the velocity of the powerboat relative to the shore, we can add the two vectors together.

Step 1: Convert the northwest velocity into its components:
Since the northwest direction is a combination of north and west, we can break it down using trigonometry. Let's consider the right triangle formed by the north and west components of the powerboat's velocity.

The northwest velocity forms a 45-degree angle with the north direction (90 degrees divided by 2).
Using trigonometry, we can find the north and west components of the northwest velocity:

north component = velocity * sin(angle)
north component = 11 m/s * sin(45°)
north component ≈ 7.78 m/s (rounded to two decimal places)

west component ≈ north component (due to the 45-degree angle)

Step 2: Find the resultant velocity:
To find the resultant velocity, we need to add the north component of the powerboat's velocity to the north velocity of the river.

Resultant north velocity = north component of powerboat's velocity + north velocity of the river
Resultant north velocity = 7.78 m/s + 5.4 m/s
Resultant north velocity ≈ 13.18 m/s (rounded to two decimal places)

Since the west component of the powerboat's velocity is equal to the north component, the resultant west velocity will be the same.

Step 3: Calculate the magnitude and direction of the resultant velocity:
Using the magnitude (resultant speed) and direction (resultant angle) of the resultant velocity, we can apply the Pythagorean theorem and inverse tangent function to find the answers.

Magnitude of resultant velocity = √(resultant north velocity^2 + resultant west velocity^2)
Magnitude of resultant velocity = √(13.18 m/s)^2 + (7.78 m/s)^2
Magnitude of resultant velocity ≈ 15.05 m/s (rounded to two decimal places)

Direction of resultant velocity = arctan(resultant west velocity / resultant north velocity)
Direction of resultant velocity = arctan(7.78 m/s / 13.18 m/s)
Direction of resultant velocity ≈ 30.1° (rounded to one decimal place)

Therefore, the velocity of the powerboat relative to the shore is approximately 15.05 m/s, at an angle of about 30.1° north of west.

To find the velocity of the motorboat relative to the shore, we can use vector addition.

Let's assume that the north direction is positive and east direction is negative.

The velocity of the motorboat relative to the water is given as 11 m/s due northwest. To break this velocity into its north and east components, we can use trigonometry.

The northwest direction forms a right triangle with the north and west directions. The angle between the north direction and the motorboat's velocity is 45 degrees (since northwest is halfway between north and west).

Using trigonometry, we can find the north component of the velocity:
North component = Velocity * sin(angle)
North component = 11 m/s * sin(45 degrees)
North component ≈ 7.78 m/s (rounded to two decimal places)

Similarly, we can find the west component of the velocity:
West component = Velocity * cos(angle)
West component = 11 m/s * cos(45 degrees)
West component ≈ -7.78 m/s (rounded to two decimal places)

Now, since the river flows due north at 5.4 m/s, the velocity of the river can be considered as the north component of the boat's velocity relative to the shore. So, the north component is 5.4 m/s.

To find the resultant velocity, we can add the north and west components:
Resultant velocity = √(North component^2 + West component^2)

Plugging in the values, we get:
Resultant velocity = √((7.78 m/s)^2 + (-7.78 m/s)^2)
Resultant velocity = √(60.40 m^2/s^2 + 60.40 m^2/s^2)
Resultant velocity ≈ √(120.80 m^2/s^2)
Resultant velocity ≈ 10.99 m/s (rounded to two decimal places)

Finally, to determine the direction of the resultant velocity, we can use inverse trigonometry:
Direction = tan^(-1) (West component / North component)

Plugging in the values, we get:
Direction = tan^(-1) (-7.78 m/s / 5.4 m/s)
Direction = tan^(-1) (-1.44)
Direction ≈ -55.99 degrees (rounded to two decimal places)

So, the velocity (both magnitude and direction) of the motorboat relative to the shore is approximately 10.99 m/s at an angle of -55.99 degrees (counterclockwise from the north direction).