Question

Vector has magnitude of 5.0 m and it is directed east. Vector has magnitude of 4.0 m and is directed 30o west of north. What are the magnitude and direction of the vector ?

Answer 1

Answer:

4.58 m, θ = 49 degree

Explanation:

Vector 1

A = 5 m East

Vector 2

B = 4 m towards 30 degree west of north

write down the vectors in vector form

$\overrightarrow{A}=5\widehat{i}$

$\overrightarrow{B}=4\left (-Sin30\widehat{i}+Cos30\widehat{j}  \right )$$\overrightarrow{B}=\left (-2\widehat{i}+3.464\widehat{j}  \right )$

The resultant of these two vectors is

$\overrightarrow{A}+\overrightarrow{B}=5\widehat{i}+\left (-2\widehat{i}+3.464\widehat{j}  \right )$

$\overrightarrow{A}+\overrightarrow{B}=3\widehat{i}+3.464\widehat{j}$

The magnitude of the resultant is given by

$=\sqrt{3^{2}+3.464^{2}}=4.58 m$

Let the resultant vector makes an angle θ from X axis

So,

$tan\theta =\frac{3.464}{3}=1.155$

θ = 49 degree