Question

A vector$\stackrel{\mathbf{\rightharpoonup }}{\mathbf{B}}$, with a magnitude of$\mathbf{8}\mathbf{.}\mathbf{0}\mathbf{}\mathbf{m}$, is added to a vector $\stackrel{\mathbf{\rightharpoonup }}{\mathbf{A}}$, which lies along an x axis. The sum of these two vectors is a third vector that lies along the y axis and has a magnitude that is twice the magnitude of$\stackrel{\mathbf{\rightharpoonup }}{\mathbf{A}}$. What is the magnitude of $\stackrel{\mathbf{\rightharpoonup }}{\mathbf{A}}$?

Short answer

The magnitude of$\stackrel{\rightharpoonup }{A}$is$3.6\mathrm{m}$

Step-by-step solution

To understand the concept

This problem is based on the triangle law of vector addition. In this law, when two vectors are represented by sides of the triangle taken in the same order of magnitude and direction, the resultant sum of these two vectors is given by the third vector. This concept can be used here to find the magnitude of vector$\stackrel{\mathbf{\rightharpoonup }}{\mathbf{A}}$.

Using the above concept third vector C can be written as

$\stackrel{\rightharpoonup }{C}=\stackrel{\rightharpoonup }{A}+\stackrel{\rightharpoonup }{B}$

Given are,

$\stackrel{\rightharpoonup }{B}=8.0\mathrm{m}$

To find the magnitude of A⇀

Here, vector$\stackrel{\rightharpoonup }{A}$is along x ais, so the angle at $\stackrel{\rightharpoonup }{A}$is 0. The vector$\stackrel{\rightharpoonup }{C}$lies along y axis, so the angle for $\stackrel{\rightharpoonup }{C}$is $90°$.

Using Pythagorean theorem,

$$\begin{gathered} B=\sqrt{A^2+C^2} \\ 8.0\mathrm{m}=\sqrt{{\mathrm{A}}^2+4{\mathrm{A}}^2} \\ \mathrm{A}=\frac{8}{\sqrt{5}}=3.6\mathrm{m} \end{gathered}$$