a grizzly bear makes a displacement of 1563 m due west, followed by a displacenent of 3348 m in a direction 32.0 north of west. what are the magnitude and direction of the displacement needed for the bear to return to its starting point

The reverse of the sum of the two vectors that you mentioned. Show your work for further assistance, if needed.

Can you please give me an example of what the sum of the two vectors mean?

To find the magnitude and direction of the displacement needed for the bear to return to its starting point, we can use vector addition.

First, let's break down the two displacements into their x and y components.

For the first displacement of 1563 m due west, since it is in the west direction, the x-component would be -1563 m (negative because it's in the opposite direction of the positive x-axis) and the y-component would be 0 m (no displacement in the y direction).

For the second displacement of 3348 m in a direction 32.0° north of west, we need to find its x and y components. We can use trigonometry to do that.

The x-component would be -3348 m * cos(32.0°) since it is in the west direction. The y-component would be 3348 m * sin(32.0°) since it is in the north direction.

Now, let's add the x and y components separately to find the final x and y components of the displacement needed for the bear to return to its starting point.

Final x-component = -1563 m + (-3348 m * cos(32.0°))
Final y-component = 0 m + (3348 m * sin(32.0°))

Once we have the final x and y components, we can find the magnitude and direction of the displacement using the Pythagorean theorem and trigonometry.

Magnitude of the displacement = sqrt((Final x-component)^2 + (Final y-component)^2)

Direction of the displacement = atan(Final y-component / Final x-component)

By plugging in the values and evaluating the equations, you will get the magnitude and direction of the displacement needed for the bear to return to its starting point.