To find the magnitude and direction of the resultant velocity of the boat, we can use vector addition.
Let's assume that the north direction is positive and the east direction is also positive.
The velocity of the boat moving north can be represented as Vboat = 15 m/s (north).
The velocity of the current moving east can be represented as Vcurrent = 15 m/s (east).
To find the resultant velocity, we need to add these two velocities as vectors.
The magnitude of the resultant velocity can be calculated using the Pythagorean theorem:
Resultant velocity magnitude = √((Vboat)^2 + (Vcurrent)^2)
Substituting the values:
Resultant velocity magnitude = √((15 m/s)^2 + (15 m/s)^2) = √(225 m^2/s^2 + 225 m^2/s^2) = √(450 m^2/s^2) = 21.2 m/s (approximately)
The direction of the resultant velocity can be calculated using trigonometry.
The angle θ between the resultant velocity and the north direction can be calculated as:
θ = arctan(Vcurrent / Vboat)
Substituting the values:
θ = arctan(15 m/s / 15 m/s) = arctan(1) = 45°
Therefore, the magnitude of the resultant velocity is 21.2 m/s, and its direction is 45° east of north.