The x-component of vector A is -25.0 m and the y-component is +40.0 m.
a) what is the magnitude of vector A?
b) What is the angle between the direction of vector A and the positive direction of vector x?
can you please help me with the steps to accomplishing these?
The magnitude is obtained by taking the square root of the sum of squares of the components.
A vector with 3 and 4 units as the x- and y-components will have a magnitude of √(3²+4²)=5 units.
The angle of the vector measured from the x-axis is given by tan-1(y-component/x-component).
Be careful with the quadrant, as the tan-1 function usually ranges only from -π/2 to π/2.
Of course! I'd be happy to help you with these questions.
a) To find the magnitude of vector A, you can use the Pythagorean theorem. The magnitude of a vector is the square root of the sum of the squares of its components. In this case, since vector A has an x-component of -25.0 m and a y-component of +40.0 m, we can calculate the magnitude as follows:
Magnitude of vector A = √((-25.0 m)^2 + (40.0 m)^2)
Step 1: Square the x-component: (-25.0 m)^2 = 625.0 m^2
Step 2: Square the y-component: (40.0 m)^2 = 1600.0 m^2
Step 3: Add the squared components: 625.0 m^2 + 1600.0 m^2 = 2225.0 m^2
Step 4: Take the square root of the sum: √(2225.0 m^2) = 47.2 m (rounded to one decimal place)
Therefore, the magnitude of vector A is approximately 47.2 m.
b) To find the angle between the direction of vector A and the positive direction of vector x, you can use trigonometry. We can utilize the inverse tangent (arctan) function to determine this angle. The angle, represented as θ, can be calculated using the equation:
θ = arctan(y-component / x-component)
Step 1: Substitute the values: θ = arctan(40.0 m / -25.0 m)
Step 2: Use a calculator or table to find the arctan: θ ≈ -59.0 degrees
Therefore, the angle between the direction of vector A and the positive direction of vector x is approximately -59.0 degrees. Note that the negative sign indicates the angle is measured in the opposite direction of the positive x-axis.