Question

Vector A has a magnitude of 35.1 m/sec and a direction of 215 degrees. What are its x- and y- components?

Answer 1

Answer on Question 64216, Physics, Astronomy, Astrophysics

Question:

Vector $A$ has a magnitude of 35.1 $m / s$ and a direction of 215 degrees. What are its $x$ - and $y$ - components?

Solution:

We can find $x$ - and $y$ - components of vector $A$ from the geometry:

$$A _ {x} = A \cos \theta ,$$ $$A _ {y} = A \sin \theta ,$$

here, $A$ is the magnitude of the vector, $\theta$ is the angle that vector $A$ makes with the positive $x$ -axis.

Then, we can calculate $A_{x}$ and $A_{y}$ (since vector $A$ lies in the third quadrant its $x$ - and $y$ - components will be negative as shown in figure above):

$$A _ {x} = A \cos \theta = 3 5. 1 \frac {m}{s} \cdot \cos 2 1 5 {}^ {\circ} = - 2 8. 7 5 \frac {m}{s}.$$ $$A _ {y} = A \sin \theta = 3 5. 1 \frac {m}{s} \cdot \sin 2 1 5 {}^ {\circ} = - 2 0. 1 3 \frac {m}{s}.$$

Answer:

$$A _ {x} = - 2 8. 7 5 \frac {m}{s}, A _ {y} = - 2 0. 1 3 \frac {m}{s}.$$

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