Answer:
The resultant velocity is 169.71 km/h at angle of 45° measured clockwise with the x-axis or the east-west line.
Explanation:
Considering west direction along negative x-axis and north direction along positive y-axis
Given:
The car travels at a speed of 120 km/h in the west direction.
The car then travels at the same speed in the north direction.
Now, considering the given directions, the velocities are given as:
Velocity in west direction is, [tex]\overrightarrow{v_1}=-120\ \vec{i}[/tex]
Velocity in north direction is, [tex]\overrightarrow{v_2}=120\ \vec{j}[/tex]
Now, since [tex]v_1\ and\ v_2[/tex] are perpendicular to each other, their resultant magnitude is given as:
[tex]|\overrightarrow{v_{res}}|=\sqrt{|\overrightarrow{v_1}|^2+|\overrightarrow{v_2}|^2}[/tex]
Plug in the given values and solve for the magnitude of the resultant.This gives,
[tex]|\overrightarrow{v_{res}}|=\sqrt{(120)^2+(120)^2}\\\\|\overrightarrow{v_{res}}|=120\sqrt{2} = 169.71\ km/h[/tex]
Let the angle made by the resultant be 'x' degree with the east-west line or the x-axis.
So, the direction is given as:
[tex]x=\tan^{-1}(\frac{|v_2|}{|v_1|})\\\\x=\tan^{-1}(\frac{120}{-120})=\tan^{-1}(-1)=-45\ deg(clockwise\ angle\ with\ the\ x-axis)[/tex]
Therefore, the resultant velocity is 169.71 km/h at angle of 45° measured clockwise with the x-axis or the east-west line.
