Answer to Question #237523 in Mechanics | Relativity for hanifa

Question

An airplane takes off and flies 225 km on a course of 25.0° north of west and then changes direction and flies 135 km due north where it lands. Find the displacement from its starting point to its landing point

Answer 1

Explanations & Calculations

• Refer to the figure attached in which each displacement is shown.
• Displacement is a vector quantity thus should be defined with a magnitude & a direction.
• Here initial displacements are given & the total/net displacement is requested. For that, we need to draw a vector triangle like the attached one & using simple trigonometry & geometry concepts, the required parameter can be found.

$\qquad\qquad \begin{aligned} \small AC^2 &=\small AB^2+BC^2 \\ &=\small (225\cos 25)^2+(225\sin25 +135)^2\\ &=\small (203.92)^2+(95.09+135)^2\\ &=\small (204km)^2+(230km)^2\\ \small X &=\small \sqrt{94516\,km^2}\\ &=\small \bold{307.43\approx307km} \end{aligned}$

• Now the magnitude is found & the direction is needed which can be found as follows.

$\qquad\qquad \begin{aligned} \small \tan\theta&=\small \frac{BC}{AC}=\frac{230}{204}\\ \small \theta&=\small \tan^{-1}\Big(\frac{230}{204}\Big)\\ &=\small 48.4^0 \end{aligned}$

• Therefore, the total displacement is $\small 307km\, 48.4^0\,North\,of\,West$