Relative Motion in Two Dimensions — Question 3

Question 3

A crosswind changes heading and ground speed

An aircraft has airspeed 100 m/s. Wind blows due east at 60 m/s relative to the ground. The desired ground course is due north over a distance of 100 km. Treat all velocities as constant. Use fixed east-positive x and north-positive y axes. Vector velocities satisfy $\vec v_{A/G}=\vec v_{A/B}+\vec v_{B/G}$ when expressed in the same axes.

1. Find the required air-relative heading and northward ground speed.

2. Find the travel time.

3. Could it maintain positive due-north ground motion if the eastward wind speed were 100 m/s or greater?

Question 3 - Solution

A crosswind changes heading and ground speed

(a) Cancel wind with the westward air-velocity component.

$$v_{A/{\mathrm{air}},x}=-60,\quad v_{A/G,y}=\sqrt{100^2-60^2}=80\quad\mathrm{m/s},\quad\boxed{\theta=\sin^{-1}(.6)=36.870^\circ}$$ The air-relative velocity must have x component $-60$ m/s. Its north component is $+\sqrt{100^2-60^2}=80$ m/s. Heading is $\sin^{-1}(60/100)\approx36.870^\circ$ west of north, and ground velocity is $(0,80)$ m/s.

(b) Use ground speed for a ground distance.

$$T=\frac{100\times10^3\,\mathrm m}{80\,\mathrm{m/s}}=1250\,\mathrm s=20.83\,\mathrm{min}$$ $100$ km is $100000$ m. The trip time is $100000/80=1250$ s, or approximately 20.83 min. Using the 100 m/s airspeed would ignore the portion used to oppose the crosswind.

(c) Check the residual north component.

$$v_{A/G,y}=\sqrt{V_{\mathrm{air}}^2-u^2}>0\ \Leftrightarrow\ |u|<V_{\mathrm{air}}=100\,\mathrm{m/s}$$ At 100 m/s wind, canceling eastward motion consumes the entire airspeed, leaving zero northward ground velocity. A stronger wind cannot be fully canceled at all. Positive due-north travel requires crosswind magnitude strictly less than the available airspeed in this model.