Answer to Question #93753 in Mechanics | Relativity for lacey

Question

REMARKS Note that the negative solution has been disregarded, because it's not physically meaningful. In checking (step 8), note that the units are correct and that an approximate answer can be obtained by using the easier quantities, 500 km and 400 km. Doing so gives an answer of 300 km, which is approximately the same as our calculated answer of 270 km.

QUESTION What is the answer if the distance traveled due east and the direct return distance are both doubled?

km
PRACTICE IT
Use the worked example above to help you solve this problem. An airplane travels 4.07 102 km due east and then travels an unknown distance due north. Finally, it returns to its starting point by traveling a distance of 505 km. How far did the airplane travel in the northerly direction?

km

Answer 1

Distance traveled in east$(AB)=102 \space km$

Let distance traveled in north$(BC)=D$

Now the particle travels towards the starting point a distance of $505 \space Km(CA)$ making the total path a right angle triangle.

By using pythagorus theorm for the triangle,

$505^2=102^2+D^2$

$D=\sqrt{505^2-102^2}=\sqrt{244621}\approx494 Km$

By rounding off ,$D=500 Km$

This means $BC\approx CA$ or we can say it almost comes back through northern point to the starting point.