Answer:
7.07 miles/hour W 22° S.
Step-by-step explanation:
Given that the resultant velocity of the runner, [tex]v_r = 8[/tex] miles/hour due west.
Wind speed, [tex]v_w= 3[/tex] miles/hour N 28° W.
Assuming that the absolute velocity of the runner, [tex]v_a= x[/tex] miles/ hour W \alpha S as shown in the figure.
By using vector addition, [tex]\vec{v_r} = \vec {v_w} + \vec {v_a}[/tex] .
Now, from the figure,
[tex]|\vec {v_r}|=|\vec {v_w} |\sin{28°} + |\vec {v_a}| \cos \alpha \\\\\Rightarrow 8 = 3\sin{28°} + |\vec {v_a}| \cos \alpha \\\\\Rightarrow |\vec {v_a}| \cos \alpha =8-3\sin{28°} \cdots(i)[/tex]
And [tex]AC= |\vec {v_w} |\cos{28°} = |\vec {v_a}| \sin \alpha \\\\[/tex]
[tex]\Rightarrow |\vec {v_a}| \sin \alpha= 3\cos{28°} \cdots(ii)[/tex]
Dividing equation (ii) by equation (i), we have
[tex]\frac {|\vec {v_a}| \sin \alpha}{|\vec {v_a}| \cos \alpha}=\frac {3\cos{28°}}{8-3\sin{28°}} \\\\\Rightarrow \tan \alpha = 0.402 \\\\\Rightarrow \alpha = \tan ^{-1} (0.402)=22^{\circ} \\\\[/tex]
Putting the value of \alpha in equation (ii), we have
[tex]|\vec {v_a}| \sin {22^{\circ}}= 3\cos{28^{\circ}} \\\\\Rightarrow |\vec {v_a}| =(3\cos{28^{\circ}})/\sin {22^{\circ}} \\\\[/tex]
[tex]\Rightarrow |\vec {v_a}| = 7.07[/tex] miles/hour
Hence, the runner speed, without the effect of the wind, is 7.07 miles/hour W 22° S.
