Math Problem Statement

Ax𝐴π‘₯, Ay𝐴𝑦 =

Solution

Given that Aβƒ—\vec{A} has a magnitude of 8.0 m8.0 \, \text{m} and is directed along the negative yy-axis (at 270∘270^\circ or βˆ’90∘-90^\circ with respect to the positive xx-axis):

  • Ax=8.0Γ—cos⁑(270∘)=0 mA_x = 8.0 \times \cos(270^\circ) = 0 \, \text{m}
  • Ay=8.0Γ—sin⁑(270∘)=βˆ’8.0 mA_y = 8.0 \times \sin(270^\circ) = -8.0 \, \text{m}

So, the components are:

Ax=0 m,Ay=βˆ’8.0 mA_x = 0 \, \text{m}, \quad A_y = -8.0 \, \text{m}

Would you like any further details or have additional questions?

Here are some related questions:

  1. How would you find the components of a vector if the angle was 45Β° instead of 90Β°?
  2. What is the significance of the negative sign in AyA_y?
  3. How can the direction of a vector be verified using its components?
  4. How does changing the magnitude affect the components of a vector?
  5. How would the calculation change if the vector was not aligned with one of the axes?

Tip: Always ensure that the angle used in the trigonometric functions is correctly referenced from the positive xx-axis, adjusting for any necessary quadrant changes.

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Math Problem Analysis

Mathematical Concepts

Vectors
Trigonometry

Formulas

Vector components using trigonometric functions

Theorems

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Suitable Grade Level

High School