Chapter 3 Problem 64

Problem

An airplane is heading due south at a speed of 580 km/h. If a wind begins blowing from the southwest at a speed of 90.0 km/h (average), calculate

(a) the velocity (magnitude and direction) of the plane, relative to the ground, and

(b) how far from its intended position it will be after 11.0 min if the pilot takes no corrective action.

[Hint: First draw a diagram.]

Solution

(a)

${\vec {v}}_{\textrm {plane/ground}}={\vec {v}}_{\textrm {plane/air}}+{\vec {v}}_{\textrm {air/ground}}$

$-580{\hat {j}}{\textrm {km/h}}+\left(90.0\cos 45.0^{\circ }{\hat {i}}+90.0\sin 45.0^{\circ }{\hat {j}}\right){\textrm {km/h}}$

$\left(63.6{\hat {i}}-516{\hat {j}}\right){\textrm {km/h}}$

$\left|{\vec {v}}_{\textrm {plane/ground}}\right|=v_{\textrm {plane/ground}}={\sqrt {\left(63.6{\textrm {km/h}}\right)^{2}+\left(-516{\textrm {km/h}}\right)^{2}}}=520{\textrm {km/h}}$

$\theta =\tan ^{-1}{\frac {63.6}{-516}}=-7.0^{\circ }$ (7 degrees east of south)

(b)

The wind speed is 90.0 km/h, so after 11.0 min the plane is off course by:

$\Delta x=v_{x}t=\left(90{\textrm {km/h}}\right)\left(11{\textrm {min}}\right)\left({\frac {1{\textrm {h}}}{60{\textrm {min}}}}\right)=16.5{\textrm {km}}$