How do I calculate the net velocity of an object given its speed and the speed of an external influence, such as ocean current?

Answer 1

To calculate the net velocity of an object influenced by another speed, like an ocean current, you can use vector addition. Here's how you can do it step by step:

1. Identify the velocities: You need to know the velocity of the object and the velocity of the external influence (like the ocean current). Remember that velocity is a vector quantity that has both magnitude (speed) and direction.

2. Represent the velocities as vectors:

   • If the object is moving at a speed \( V_o \) in a particular direction, you can represent this as a vector \( \vec{V_o} \).
   • Similarly, if the ocean current has a speed \( V_c \) in a different direction, you represent this as a vector \( \vec{V_c} \).

3. Break down the vectors into components: If the velocities are not in the same line or direction, you may need to break them down into their x (horizontal) and y (vertical) components using trigonometric functions, depending on the angle of movement. For example:

   • \( \vec{V_o} = V_o \cdot (\cos(\theta_o), \sin(\theta_o)) \)
   • \( \vec{V_c} = V_c \cdot (\cos(\theta_c), \sin(\theta_c)) \) Here, \( \theta_o \) and \( \theta_c \) are the angles at which the object and the current are moving, respectively.

4. Add the velocity vectors: To find the net velocity \( \vec{V_{net}} \), add the corresponding components of the velocity vectors: \[ \vec{V_{net}} = \vec{V_o} + \vec{V_c} \] This gives you: \[ V_{net,x} = V_{o,x} + V_{c,x} \] \[ V_{net,y} = V_{o,y} + V_{c,y} \]

5. Calculate the net velocity magnitude and direction: Once you have the components, you can find the magnitude of the net velocity using the Pythagorean theorem: \[ V_{net} = \sqrt{V_{net,x}^2 + V_{net,y}^2} \] For the direction (angle), you can use the arctangent function: \[ \theta_{net} = \tan^{-1}\left(\frac{V_{net,y}}{V_{net,x}}\right) \]

Example:

• If the object has a speed of 5 m/s at an angle of 30° from the positive x-axis (east) and the ocean current has a speed of 3 m/s coming from the north (90°).

1. Object Velocity Components: \[ V_{o,x} = 5 \cdot \cos(30°) \approx 4.33 \text{ m/s} \] \[ V_{o,y} = 5 \cdot \sin(30°) = 2.5 \text{ m/s} \]

2. Current Velocity Components: \[ V_{c,x} = 0 \text{ m/s} \] \[ V_{c,y} = -3 \text{ m/s} \text{ (since it’s going south)} \]

3. Net Velocity Components: \[ V_{net,x} = 4.33 + 0 = 4.33 \text{ m/s} \] \[ V_{net,y} = 2.5 - 3 = -0.5 \text{ m/s} \]

4. Magnitude of Net Velocity: \[ V_{net} = \sqrt{(4.33)^2 + (-0.5)^2} \approx 4.34 \text{ m/s} \]

5. Direction of Net Velocity: \[ \theta_{net} = \tan^{-1}\left(\frac{-0.5}{4.33}\right) \approx -6.7° \text{ (below the x-axis)} \]

This process will give you the net velocity of the object considering the influence of the external speed.