A dog that can swim at 1.0 m/s jumps in a river that is flowing 3.0 m/s at an angle of N60*E. Solve the resultant velocity vector of the dog if he heads north. What direction is the dog acutally heading?
can someone please help thanks
To solve for the resultant velocity vector of the dog, we need to consider the vector addition of the dog's swimming velocity and the river's current velocity.
1. First, let's break down both vectors into their horizontal and vertical components.
The dog's swimming velocity (v_dog) heading north can be represented as:
v_dog = 1.0 m/s [N]
Since it is heading north, the horizontal component (v_dog_h) is zero, and the vertical component (v_dog_v) is 1.0 m/s.
The river's current velocity (v_river) is given as 3.0 m/s at an angle of N60*E. To find the horizontal and vertical components, we need to use basic trigonometry.
The horizontal component (v_river_h) can be found by taking the product of the river's speed (3.0 m/s) and the cosine of the angle (60 degrees), given by:
v_river_h = 3.0 m/s * cos(60 degrees)
The vertical component (v_river_v) can be found by taking the product of the river's speed (3.0 m/s) and the sine of the angle (60 degrees), given by:
v_river_v = 3.0 m/s * sin(60 degrees)
2. Calculate the horizontal and vertical components separately.
v_river_h = 3.0 m/s * cos(60 degrees) = 3.0 m/s * 0.5 = 1.5 m/s [E]
v_river_v = 3.0 m/s * sin(60 degrees) = 3.0 m/s * √(3)/2 = 2.598 m/s [N]
3. Add the horizontal and vertical components separately to find the resultant velocity vector.
The horizontal component of the resultant velocity (v_result_h) is the sum of the dog's horizontal velocity and the river's horizontal velocity:
v_result_h = v_dog_h + v_river_h = 0 m/s + 1.5 m/s = 1.5 m/s [E]
The vertical component of the resultant velocity (v_result_v) is the sum of the dog's vertical velocity and the river's vertical velocity:
v_result_v = v_dog_v + v_river_v = 1.0 m/s + 2.598 m/s = 3.598 m/s [N]
4. Determine the magnitude and direction of the resultant velocity vector.
The magnitude (speed) of the resultant velocity vector is given by the Pythagorean theorem:
|v_result| = √(v_result_h^2 + v_result_v^2) = √((1.5 m/s)^2 + (3.598 m/s)^2) ≈ 3.996 m/s
To find the direction, we can use inverse tangent:
θ = arctan(v_result_v / v_result_h) = arctan(3.598 m/s / 1.5 m/s) ≈ 67.27 degrees [N]
Therefore, the resultant velocity vector of the dog, if he heads north, has a magnitude of approximately 3.996 m/s and is pointed at an angle of approximately N67.27 degrees [E].