Question

a pilot wants to fly on a bearing of 61.5°. by flying due east, she finds that a 47.7 - mph wind, blowing from the south, puts her on course. find the airspeed and the groundspeed of the plane. the airspeed is □□ mph. (round to the nearest integer as needed.) the groundspeed is □□ mph. (round to the nearest integer as needed.)

Explanation:

Step 1: Analyze the problem

We have a pilot flying with a wind blowing from the south. The bearing she wants is \(61.5^\circ\), and the wind speed is \(47.7\) mph. We can model this as a right triangle or use trigonometry to find the airspeed and groundspeed. Let's assume the airspeed is \(v\), the wind speed is \(w = 47.7\) mph, and the angle between the airspeed vector and the east direction is related to the bearing. The bearing of \(61.5^\circ\) means the angle from the north to the east direction is \(61.5^\circ\), so the angle between the airspeed (which we can consider as having a north - south and east - west component) and the east direction is \(90^\circ- 61.5^\circ=28.5^\circ\)? Wait, no. Let's think in terms of components. The wind is blowing from the south, so it is a northward wind? Wait, wind blowing from the south means it is moving towards the north. So the wind vector is in the north direction with speed \(47.7\) mph. The pilot wants to fly on a bearing of \(61.5^\circ\), which is \(61.5^\circ\) from the north towards the east. So the airspeed vector (the velocity of the plane relative to the air) and the wind vector (velocity of air relative to the ground) add up to the groundspeed vector (velocity of plane relative to the ground). Let's denote:
- Let the airspeed be \(v\) (magnitude of the plane's velocity relative to air).
- The wind speed \(w = 47.7\) mph (northward).
- The bearing of the plane's desired path (groundspeed) is \(61.5^\circ\) from north to east. So the groundspeed vector has components: \(V_{g,x}=V_g\sin(61.5^\circ)\) (eastward) and \(V_{g,y}=V_g\cos(61.5^\circ)\) (northward), where \(V_g\) is the groundspeed.
- The airspeed vector has components: \(V_{a,x}=v\sin(61.5^\circ)\) (eastward) and \(V_{a,y}=v\cos(61.5^\circ)- 47.7\) (northward, since the wind is adding \(47.7\) mph northward to the airspeed's northward component to get the groundspeed's northward component). But since the pilot is on course, the northward component of the groundspeed should be equal to the northward component of the airspeed plus the wind speed? Wait, no. If the wind is blowing from the south, it is a northward wind. So the velocity of the plane relative to the ground (\(\vec{V}_g\)) is equal to the velocity of the plane relative to the air (\(\vec{V}_a\)) plus the velocity of the air relative to the ground (\(\vec{V}_w\)). \(\vec{V}_w=(0, 47.7)\) (assuming north is the positive y - direction and east is the positive x - direction). \(\vec{V}_a=(v\sin(61.5^\circ),v\cos(61.5^\circ))\) (since the airspeed vector is at an angle of \(61.5^\circ\) from the north towards the east, so the x - component (east) is \(v\sin(61.5^\circ)\) and the y - component (north) is \(v\cos(61.5^\circ)\)). Then \(\vec{V}_g=\vec{V}_a+\vec{V}_w=(v\sin(61.5^\circ),v\cos(61.5^\circ)+ 47.7)\). But the direction of \(\vec{V}_g\) is also \(61.5^\circ\) from the north towards the east. So the ratio of the x - component to the y - component of \(\vec{V}_g\) should be \(\tan(61.5^\circ)\). So \(\frac{V_{g,x}}{V_{g,y}}=\tan(61.5^\circ)\). But \(V_{g,x}=v\sin(61.5^\circ)\) and \(V_{g,y}=v\cos(61.5^\circ)+ 47.7\). Wait, this seems complicated. Maybe a better approach: Since the wind is from the south (so northward wind), and the pilot is flying on a bearing of \(61.5^\circ\), the airspeed's north - south component must balance the wind? No, wait, the wind is adding to the north - south component. Wait, maybe the angle between the airspeed vector and the north direction is \(61.5^\circ\), and the wind is in the north direction. Let's consider the triangle formed by the airspeed (\(v\)), wind (\(w = 47.7\)), and groundspeed (\(V_g\)). The angle between the airspeed and the wind is \(61.5^\circ\)? Wait, no. Let's use trigonometry. If we consider that the wind is in the north direction, and the airspeed has a component in the north direction and a component in the east direction. Let the airspeed be \(v\), with north component \(v\cos(61.5^\circ)\) and east component \(v\sin(61.5^\circ)\). The wind is in the north direction with speed \(47.7\), so the north component of the groundspeed is \(v\cos(61.5^\circ)+ 47.7\), and the east component is \(v\sin(61.5^\circ)\). But the groundspeed should be in the direction of the bearing \(61.5^\circ\), so the ratio of east component to north component of the groundspeed should be \(\tan(61.5^\circ)\). So \(\frac{v\sin(61.5^\circ)}{v\cos(61.5^\circ)+ 47.7}=\tan(61.5^\circ)\). Wait, \(\tan(61.5^\circ)=\frac{\sin(61.5^\circ)}{\cos(61.5^\circ)}\), so:

\(\frac{v\sin(61.5^\circ)}{v\cos(61.5^\circ)+ 47.7}=\frac{\sin(61.5^\circ)}{\cos(61.5^\circ)}\)

Multiply both sides by \(v\cos(61.5^\circ)+ 47.7\) and by \(\cos(61.5^\circ)\):

\(v\sin(61.5^\circ)\cos(61.5^\circ)=\sin(61.5^\circ)(v\cos(61.5^\circ)+ 47.7)\)

Divide both sides by \(\sin(61.5^\circ)\) (assuming \(\sin(61.5^\circ) eq0\), which it isn't):

\(v\cos(61.5^\circ)=v\cos(61.5^\circ)+ 47.7\)

This gives \(0 = 47.7\), which is impossible. So my initial assumption about the components is wrong.

Wait, maybe the wind is blowing from the south, so it is a southward wind? Wait, wind blowing from the south means the wind is moving towards the north? No, wind direction is the direction it is coming from. So wind blowing from the south is moving towards the north. So the wind vector is \((0, 47.7)\) (northward). The pilot is flying due east (wait, the problem says "by flying due east, she finds that a 47.7 - mph wind, blowing from the south, puts her on course". Oh! Wait, the problem says: "A pilot wants to fly on a bearing of \(61.5^\circ\). By flying due east, she finds that a 47.7 - mph wind, blowing from the south, puts her on course." So when she flies due east (airspeed vector is eastward, so airspeed has magnitude \(v\), direction east), and the wind is blowing from the south (so wind vector is northward, magnitude \(47.7\) mph), the resultant (groundspeed) vector is on the bearing of \(61.5^\circ\). Ah! That's the key. So:

- Airspeed vector: \(\vec{V}_a=(v,0)\) (eastward, magnitude \(v\))
- Wind vector: \(\vec{V}_w=(0, 47.7)\) (northward, since wind blows from south to north)
- Groundspeed vector: \(\vec{V}_g=(v, 47.7)\) (east and north components)

The bearing of the groundspeed vector is the angle \(\theta\) from the north to the east, so \(\tan(\theta)=\frac{\text{east component}}{\text{north component}}=\frac{v}{47.7}\)

And we know that \(\theta = 61.5^\circ\)

So \(\tan(61.5^\circ)=\frac{v}{47.7}\)

Step 2: Solve for airspeed \(v\)

We know that \(\tan(61.5^\circ)=\frac{v}{47.7}\)

So \(v = 47.7\times\tan(61.5^\circ)\)

Calculate \(\tan(61.5^\circ)\approx1.847\)

\(v=47.7\times1.847\approx47.7\times1.847 = 47.7\times(1 + 0.8+0.04 + 0.007)=47.7+38.16 + 1.908+0.3339 = 88.0919\approx88\) mph? Wait, no, let's calculate \(\tan(61.5^\circ)\) more accurately. \(\tan(61.5^\circ)=\tan(60^\circ + 1.5^\circ)=\frac{\tan(60^\circ)+\tan(1.5^\circ)}{1 - \tan(60^\circ)\tan(1.5^\circ)}\). \(\tan(60^\circ)=\sqrt{3}\approx1.732\), \(\tan(1.5^\circ)\approx0.02618\)

\(\tan(61.5^\circ)=\frac{1.732 + 0.02618}{1-1.732\times0.02618}=\frac{1.75818}{1 - 0.0453}\approx\frac{1.75818}{0.9547}\approx1.841\)

Then \(v = 47.7\times1.841\approx47.7\times1.841\)

\(47\times1.841 = 86.527\), \(0.7\times1.841 = 1.2887\), total \(86.527+1.2887 = 87.8157\approx88\) mph. Wait, but let's use a calculator for \(\tan(61.5^\circ)\). \(61.5^\circ\) in radians is \(61.5\times\frac{\pi}{180}\approx1.073\) radians. \(\tan(1.073)\approx1.847\)

So \(v = 47.7\times1.847\approx47.7\times1.847\)

\(47\times1.847 = 86.809\), \(0.7\times1.847 = 1.2929\), total \(86.809 + 1.2929=88.1019\approx88\) mph (airspeed)

Step 3: Solve for groundspeed \(V_g\)

The groundspeed vector has components \(v\) (east) and \(47.7\) (north). So the magnitude of the groundspeed is \(V_g=\sqrt{v^{2}+47.7^{2}}\)

We found \(v\approx88\) mph, so \(V_g=\sqrt{88^{2}+47.7^{2}}=\sqrt{7744 + 2275.29}=\sqrt{10019.29}\approx100\) mph (wait, \(88^2 = 7744\), \(47.7^2=47.7\times47.7 = (48 - 0.3)^2=48^2-2\times48\times0.3 + 0.3^2=2304 - 28.8+0.09 = 2275.29\), sum is \(7744+2275.29 = 10019.29\), square root of \(10019.29\) is approximately \(100.1\approx100\) mph. Wait, but let's use the exact value of \(v\) from the first step. Since \(v = 47.7\times\tan(61.5^\circ)\), then \(V_g=\sqrt{(47.7\times\tan(61.5^\circ))^{2}+47.7^{2}}=47.7\sqrt{\tan^{2}(61.5^\circ)+1}=47.7\sec(61.5^\circ)\) (since \(\tan^{2}\theta + 1=\sec^{2}\theta\))

\(\sec(61.5^\circ)=\frac{1}{\cos(61.5^\circ)}\), \(\cos(61.5^\circ)\approx0.478\), so \(\sec(61.5^\circ)\approx2.092\)

\(V_g = 47.7\times2.092\approx47.7\times2.092\)

\(47\times2.092 = 98.324\), \(0.7\times2.092 = 1.4644\), total \(98.324+1.4644 = 99.7884\approx100\) mph.

Wait, let's re - do the airspeed calculation. Since \(\tan(\theta)=\frac{\text{opposite}}{\text{adjacent}}=\frac{v}{47.7}\), where \(\theta = 61.5^\circ\), so \(v = 47.7\times\tan(61.5^\circ)\)

\(\tan(61.5^\circ)=\tan(60^\circ+1.5^\circ)=\frac{\tan60+\tan1.5}{1 - \tan60\tan1.5}\)

\(\tan60=\sqrt{3}\approx1.732\), \(\tan1.5^\circ=\tan(\frac{1.5\pi}{180})=\tan(0.02618)\approx0.02618\)

\(\tan(61.5^\circ)=\frac{1.732 + 0.02618}{1-1.732\times0.02618}=\frac{1.75818}{1 - 0.0453}=\frac{1.75818}{0.9547}\approx1.841\)

\(v = 47.7\times1.841\approx47.7\times1.841 = 87.8157\approx88\) mph (airspeed)

Groundspeed: \(V_g=\sqrt{88^{2}+47.7^{2}}=\sqrt{7744 + 2275.29}=\sqrt{10019.29}\approx100\) mph

Answer:

The airspeed is \(\boxed{88}\) mph.
The groundspeed is \(\boxed{100}\) mph.