Answer to Question #90205 in Physics for Paris Phoebe

Question

Find the magnitude and direction of the following vector represented by the following pairs of components
a) Ax=8.60cm. Ay=5.20cm
b) Ax=9.70m. Ay=2.45m
c) Ax= 7.75km. Ay=2.70km

Answer 1

a) We can find the magnitude of the vector $\overrightarrow{A}$ from the formula:

$$A = \lvert \overrightarrow{A} \rvert = \sqrt{A_x^2 + A_y^2} = \sqrt{(8.60 cm)^2 + (5.20 cm)^2} = 10 cm.$$

We can find the angle between the vector $\overrightarrow{A}$ and the $+x$-axis from the formula:

$$\theta = arctan(\dfrac{A_y}{A_x}) = arctan(\dfrac{5.20 cm}{8.60 cm}) = 31^{\circ}.$$

b) We can find the magnitude of the vector $\overrightarrow{A}$ from the formula:

$$A = \lvert \overrightarrow{A} \rvert = \sqrt{A_x^2 + A_y^2} = \sqrt{(9.70 m)^2 + (2.45 m)^2} = 10 m.$$

We can find the angle between the vector $\overrightarrow{A}$ and the $+x$-axis from the formula:

$$\theta = arctan(\dfrac{A_y}{A_x}) = arctan(\dfrac{2.45 m}{9.70 m}) = 14^{\circ}.$$

c) We can find the magnitude of the vector $\overrightarrow{A}$ from the formula:

$$A = \lvert \overrightarrow{A} \rvert = \sqrt{A_x^2 + A_y^2} = \sqrt{(7.75 km)^2 + (2.70 km)^2} = 8.21 km.$$

We can find the angle between the vector $\overrightarrow{A}$ and the $+x$-axis from the formula:

$$\theta = arctan(\dfrac{A_y}{A_x}) = arctan(\dfrac{2.70 km}{7.75 km}) = 19.21^{\circ}.$$

Answer:

a) $A = 10 cm$, $\theta = 31^{\circ}.$

b) $A = 10 m$, $\theta = 14^{\circ}.$

c) $A = 8.21 km$, $\theta = 19.21^{\circ}.$