A plane is initially flying north with a speed v0 = 430 mph relative to the ground while the wind has a constant speed vw = 12 mph in the north-south direction. The plane performs a circular turn with radius of ρ = 0.45 mi. Assume that the airspeed indicator on the plane measures the absolute value

Question

A plane is initially flying north with a speed [latex]v_{ 0 }[/latex] = 430 mph relative to the ground while the wind has a constant speed [latex]v_{ w }[/latex] = 12 mph in the north-south direction. The plane performs a circular turn with radius of ρ = 0.45 mi. Assume that the airspeed indicator on the plane measures the absolute value of the component of the relative velocity of the plane with respect to the air in the direction of motion. Then determine the value of the tangential component of the airplane’s acceleration when the airplane is halfway through the turn, assuming that the airplane maintains constant the reading of the airspeed indicator.

Accepted Answer

Let the subscripts P and W denote quantities pertaining to the airplane and wind, respectively. Then, referring to the figure on the the right and using a normal tangential component system, the velocity of the airplane and wind are

[latex]\vec{v}_P=v_P \ \hat{u}_t \quad ext { and } \quad \vec{v}_W=v_W\left(-\cos heta \ \hat{u}_t+\sin heta \ \hat{u}_n ight).[/latex] (1)

Therefore the velocity of the airplane relative to the wind is

[latex]\vec{v}_{P / W}=\left(v_P+v_W \cos heta ight) \hat{u}_t-v_W \sin heta \ \hat{u}_n .[/latex] (2)

Now we recall that that the quantity measured by the airspeed indicator is the component of [latex]\vec{v}_{P / W}[/latex] that is in the direction of motion, i.e.,

[latex]v_{ ai }=\left|\vec{v}_{P / A} \cdot \hat{u}_t ight|=v_P+v_W \cos heta[/latex]

where the subscript ‘ai’ stands for airspeed indicator. Now, recalling that the measure of θ in radians is given by [latex] heta=s / ho[/latex], we can rewrite [latex]v_{ ai }[/latex] as follows:

[latex]v_{ ai }=v_P+v_W \cos \frac{s}{ ho}.[/latex] (3)

Observing that for θ = 0 we have [latex]v_r=v_0+v_W[/latex] and recalling that [latex]v_{ ai }[/latex] is maintained constant along the turn, we can solve Eq. (3) for vP as a function of s to obtain

[latex]v_P(s)=v_0+v_W\left(1-\cos \frac{s}{ ho} ight).[/latex] (4)

The tangential component of acceleration is the time derivative of Eq. (4). Hence, we can write

[latex]a_t=\frac{d v_P}{d s} \frac{d s}{d t}=v_P \frac{d v_P}{d s} \Rightarrow a_t=\left[v_0+v_W\left(1-\cos \frac{s}{ ho} ight) ight] \frac{v_W}{ ho} \sin \frac{s}{ ho}.[/latex] (5)

For [latex]s / ho=(\pi / 4) \ rad[/latex] , we have [latex]\sin (s / ho)=\cos (s / ho)=\sqrt{2} / 2[/latex] . Therefore, midway through the turn, we have

[latex]a_t=\left[v_0+v_W\left(1-\frac{\sqrt{2}}{2} ight) ight] \frac{v_W \sqrt{2}}{2 ho} .[/latex] (6)

Recalling that [latex]v_0=430 \ mph =630.7 \ ft / s , v_W=12 \ mph =17.60 \ ft / s ext {, and } ho=0.45 \ mi =2376 \ ft[/latex], we can evaluate the above expression to obtain

[latex]a_t=3.33 \ ft / s ^2 ext {. }[/latex]

Question 2.254: A plane is initially flying north with a speed v0 = 430 mph ...

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