Answer:
< 2.11, 4.53 >, < -3.03, -1.75 >, <2.93 cos 108.26, 2.93 sin 108.26 >
Step-by-step explanation:
First, let's decompose Bruce's velocity along the x- and y- direction. Bruce is moving 5 m/s at 25 degrees east of north, so its angle with respect to the positive x-direction is actually 90 - 25 = 65 degrees. So its components are
[tex]b_x = (5 m/s) cos 65^{\circ} =2.11 m/s\\b_y = (5 m/s) sin 65^{\circ} =4.53 m/s[/tex]
So, Bruce's vector is
< 2.11, 4.53 >
The current is moving 3.5 m/s at an angle 60 degrees west of south, which means an overall angle of 210 degrees, measured counterclockwise from the positive x-axis. So, the components of the current's velocity are
[tex]c_x = (3.5 m/s) cos 210^{\circ}=-3.03 m/s\\c_y = (3.5 m/s) sin 210^{\circ}=-1.75 m/s[/tex]
So, the current's vector is
< -3.03, -1.75 >
Finally, we can add the components of the two vectors to find Bruce's actual velocity:
[tex]v_x = b_x + c_x = 2.11 + (-3.03)=-0.92 m/s\\v_y = b_y + c_y = 4.53+(-1.75)=2.78 m/s[/tex]
So, Bruce's actual velocity is
< -0.92, 2.78 >
The magnitude is
[tex]v=\sqrt{(-0.92)^2+(2.78)^2}=2.93 m/s[/tex]
And the direction is
[tex]\theta=180^{\circ} - tan^{-1} (\frac{v_y}{v_x})=180^{\circ} - tan^{-1}(\frac{2.78}{-0.92})=180^{\circ}-71.7^{\circ}=108.3^{\circ}[/tex]
