Speed: 15.3 m/s
Direction: [tex]78.7^{\circ}[/tex] north of east
Explanation:
In order to find velocity, we have to find the components along the east-west direction and the north-south direction first.
Along the east-west direction, we have
[tex]v_e = 6 m/s[/tex] east
[tex]v_w = 3 m/s[/tex] west
So the net velocity in this direction is
[tex]v_x = 6 - 3 = 3 m/s[/tex] east
In the north-south direction, the net velocity is
[tex]v_y = 15 m/s[/tex] north
So, the speed (which is the magnitude of the velocity) is
[tex]v=\sqrt{v_x^2+v_y^2}=\sqrt{3^2+15^2}=15.3 m/s[/tex]
While the direction is given by
[tex]\theta=tan^{-1}(\frac{v_y}{v_x})=tan^{-1}(\frac{15}{3})=78.7^{\circ}[/tex] north of east.
Learn more about velocity:
brainly.com/question/5248528
#LearnwithBrainly
