Question

5. I A position vector in the first quadrant has an x-component of 6m and a magnitude of 10m . What is the value of its y-component?

Short answer

y-component = 8 m

Step-by-step solution

Step 1. Given Information

A position vector in the first quadrant has an -component of $6\text{m}$and a magnitude of $10\text{m}$.

This can be represented as

The position vector $\overrightarrow{r}$whose magnitude $r$$10\text{m}$is has an x - component of 6 m. It makes an angle $\theta$with the positive x-axis in the first quadrant.

Here;

$\overrightarrow{r}$is the position vector

$r=10m$$r_x-\text{x}$-component of the position vector.

$r_{\text{x}}=6\text{m}$$r_y-\text{y}$-component of the position vector.

$\theta$- The angle that the position vector makes with the positive x- axis.

Formula used: x-component,$r_x=r\cos \theta$

y-component,$r_y=r\sin \theta$

Step 2. Explanation

x-component, $r_x=r\cos \theta$

Substituting the value of $\text{rand}{r}_x$in the relation.

$\begin{array}{l}{r}_{\text{x}}=r\cos \theta \\ 6\text{m}=10\text{m}\times \cos \theta \\ \cos \theta =\frac{6\text{m}}{10\text{m}}\\ \theta ={\cos }^{-1}\left[\frac{6}{10}\right]\\ \theta ={53.13}^{°}\end{array}$

Thus, the y-component is

$\begin{array}{l}{r}_y=r\sin \theta \\ =10\text{m}\times \sin \left({53.13}^{°}\right)\\ =7.99\text{m}\approx 8\text{m}\end{array}$