Question

A vector has a magnitude of 46.0 m and points in a direction 20.0° below the positive x-axis. A second vector, , has a magnitude of 86.0 m and points in a direction 42.0° above the negative x-axis. a) Sketch the vectors A⃗ , B⃗ , and C⃗=A⃗+B⃗ .
b) Using the component method of vector addition, find the magnitude of the vector C⃗ .
c) Using the component method of vector addition, find the direction of the vector C

Answer 1

Answer with Step-by -step explanation:

We are given that

b.$\mid A\mid=46 m$

$\theta=20^{\circ}$ below the positive x-axis

Therefore, the angle made by vector A in counter clockwise direction when measure from positive x-axis=$x=360-20=340^{\circ}$

x-component of vector A=$A_x=\mid A\mid cosx=46cos 340=46\times 0.94=43.24$

y-Component of vector A=$A_y=\mid A\mid sinx=46sin340=46(-0.34)=-15.64$

Magnitude of vector B=86 m

The vector B makes angle with positive x- axis=$x'=42^{\circ}$

x-component of vector B=$B_x=86cos42=63.64$

y-Component of vector B=$B_y=86sin42=57.62$

Vector A=$A_xi+A_yj=43.24i-15.64j$

Vector B=$B_xi+B_yj=63.64i+57.62j$

Vector C=A+B

Substitute the values

$C=43.24i-15.64j+63.64i+57.62j$

$C=106.88i+41.98j$

c.Direction=$\theta=tan^{-1}(\frac{y}{x})=tan^{-1}(\frac{41.98}{106.88})=21.5^{\circ}$

The direction of the vector C=21.5 degree