Westonci.ca is the ultimate Q&A platform, offering detailed and reliable answers from a knowledgeable community. Explore comprehensive solutions to your questions from knowledgeable professionals across various fields on our platform. Explore comprehensive solutions to your questions from a wide range of professionals on our user-friendly platform.
Sagot :
We know that the wind is blowing from north to south with a speed of 32 km per hour, let's call the wind vector W, then it can be written as
[tex]\vec{W}=-32\hat{j}[/tex]We also know that Diane wants the plane to fly north of east at an angle of 25° and a speed of 140 km/h. This means that the resultant should be
[tex]\vec{R}=140\cos 24\hat{i}+140\sin 24\hat{j}[/tex]Now, we would like to know at what angle Diane should fly tha plane. To find this we will introduce a vector T whose magnitud and direction are unknown.
Then
[tex]\vec{T}=T\cos \theta\hat{i}+T\sin \hat{j}[/tex]Now, we know that the resultant is
[tex]\vec{R}=\vec{T}+\vec{W}[/tex]then
[tex]\begin{gathered} 140\cos 25\hat{i}+140\sin 25\hat{j}=T\cos \theta\hat{i}+T\sin \theta\hat{j}+(-32\hat{j}) \\ 140\cos 25\hat{i}+140\sin 25\hat{j}=T\cos \theta\hat{i}+(T\sin \theta-32)\hat{j} \end{gathered}[/tex]Since the unit vector i and j are independent this gives us two equations
[tex]\begin{gathered} 140\cos 25=T\cos \theta \\ 140\sin 25=T\sin \theta-32 \end{gathered}[/tex]From the first equation we have that
[tex]T=\frac{140\cos 25}{\cos \theta}[/tex]Plugging the value of T in the second equation we have
[tex]140\sin 25=(\frac{140\cos25}{\cos\theta})\sin \theta-32[/tex]Now we need to solve this equation for theta.
[tex]\begin{gathered} 140\sin 25=(\frac{140\cos25}{\cos\theta})\sin \theta-32 \\ 140\sin 25+32=(140\cos 25)\frac{\sin \theta}{\cos \theta} \\ \frac{\sin\theta}{\cos\theta}=\frac{140\sin 25+32}{140\cos 25} \end{gathered}[/tex]Now we have to remember that
[tex]\frac{\sin\theta}{\cos\theta}=\tan \theta[/tex]hence
[tex]\begin{gathered} \tan \theta=\frac{145\sin 25+32}{145\cos 25} \\ \theta=\tan ^{-1}(\frac{145\sin 25+32}{145\cos 25}) \\ \theta=35.7 \end{gathered}[/tex]Therefore Diane has to direct the airplane at an angle of 35.7° North of east.
Thank you for your visit. We're committed to providing you with the best information available. Return anytime for more. Thank you for your visit. We're committed to providing you with the best information available. Return anytime for more. Thank you for visiting Westonci.ca, your go-to source for reliable answers. Come back soon for more expert insights.
