The displacement, force, velocity, and acceleration vectors’ components are given. Calculate the magnitude and direction of each vector. (a) x = +15.0 cm, y = -12.3 cm.

I am completely lost.

The magnitude |<x,y>| of a vector given as <x,y> can be obtained by a calculation similar to Pythagoras theorem. For example,

|<3,4>| = sqrt(3²+4²)=5
The direction θ with respect to the x-axis can be calculated using a calculator by:
θ=tan-1(y/x)
but add 180° if x is negative.
For example, for <-12,5>
θ=tan-1(-12/5)
=tan-1-67.4°
But since x<0, we add 180° to give
θ=112.6°

You can use the same procedure to calculate the magnitude and direction of all vectors when the x,y components are given.

To calculate the magnitude and direction of the given vector, we can use the Pythagorean theorem and trigonometric functions. Here's how you can approach the problem:

1. Draw a coordinate system with the x and y-axes.
2. Plot the displacement vector using the given x and y components. In this case, x = +15.0 cm and y = -12.3 cm. From the origin (0,0), move 15.0 cm to the right along the positive x-axis and 12.3 cm downwards along the negative y-axis.
3. Now, to find the magnitude of the vector, use the Pythagorean theorem:
magnitude = sqrt(x^2 + y^2)
Substitute the given values: magnitude = sqrt((15.0 cm)^2 + (-12.3 cm)^2)
Calculate the magnitude using a calculator or by hand: magnitude = 19.2 cm (rounded to one decimal place).

The magnitude of the vector is 19.2 cm.

4. To find the direction of the vector, you can use trigonometric functions. Calculate the angle between the vector and the positive x-axis.

Let θ be the angle between the vector and the positive x-axis. To find θ, use the inverse tangent function (tan^-1) of the ratio of the y-component to the x-component:
θ = tan^(-1)(y / x)
Substitute the given values: θ = tan^(-1)(-12.3 cm / 15.0 cm)
Calculate the angle using a calculator or by hand: θ ≈ -41.2° (rounded to one decimal place).

Note: The negative sign indicates that the vector is pointing downwards from the positive x-axis.

The direction of the vector is approximately -41.2° (clockwise from the positive x-axis).

Therefore, the magnitude of the vector is 19.2 cm and its direction is approximately -41.2° (downwards from the positive x-axis).