Question

need part B please

1/2 points I Previous Answers A spaceship is traveling through deep space towards a space station and needs to make a course correction to go around a nebula. The captain orders the ship to travel 1.9 x 106 kilometers before turning 70° and traveling 1.7 x 106 kilometers before assuming the path towards the space station. If the captain had not ordered a course correction, what would have been the magnitude and direction of the path of the spaceship if it had traveled through the nebula? My Notes Ask Your Teacher space station nebula spaceship magnitude 2.9506 direction e counterclockwise from the +x-axis

Answer 1

$\vec{A}=1.9*10^{6}\,\hat{i}\,km$

$\vec{B}=1.7*10^{6}(\cos70\degree\hat{i}+\sin70\degree\hat{j})\,km$

Resultant displacement $\vec{R}=\vec{A}+\vec{B}=1.9*10^{6}\,\hat{i}+1.7*10^{6}(\cos70\degree\hat{i}+\sin70\degree\hat{j})$

$\vec{R}=(1.9*10^{6}+1.7*10^{6}*\cos70\degree)\,\hat{i}+1.7*10^{6}\sin70\degree\hat{j}$

Magnitude of displacement $|\vec{R}|=\sqrt{(1.9*10^{6}+1.7*10^{6}*\cos70\degree)^2+(1.7*10^{6}\sin70\degree)^2}$

$|\vec{R}|=2.95*10^{6}\,km$

Direction of resultant displacement makes an angle $\phi=\tan^{-1}{\left(\frac{1.7*10^{6}\sin70\degree}{1.9*10^{6}+1.7*10^{6}\cos70\degree} \right )}$

$\phi=\tan^{-1}{\left(\frac{1.7\sin70\degree}{1.9+1.7\cos70\degree} \right )}$

$\phi=\tan^{-1}{\left(\frac{1.597477}{2.481434} \right )}=32.77\degree$ counter clockwise with +x axis.