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An aircraft A is flying in a vertical plane containing two tracking stations P and Q which are 15km apart. At a cetain instant θ (measured anticlockqise from horizontal line PQ to the line PA) is 60° and the $ \dot \theta $ (angular velocity) is -0.025rad/s. At the same time, α (measured anticlockqise from extended horizontal line PQ to the line QA) is 150° and the $\dot \alpha$ is -0.02rad/s. Determine the magnitude and the direction of the velocity of the aircraft.

I used the radial and transverse components of the velocities relative to P and Q and calculated the resultant.

Relative to P, radial component = $$\frac{d(PA)}{dt}= \frac{d(15000Cos(\theta))}{dt} = -15000Sin(\theta) \dot \theta $$

The transverse component = $$r \dot \theta =-(PA)0.025$$

The components of velocity relative to Q could be obtained in the same way but I'm not sure about what to do next. Is taking the resultant of them correct? I took the resultant and got the following answers.

Magnitude - 15133m/s

Direction - 88.89° north of east

S.Dan
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  • Why would be the direction north of east when it is traveling in a vertical plane? I am typing out the solution, it will take some time. – Satish Ramanathan Jul 11 '14 at 08:44
  • @BeaumontTaz, 18.35*3600 = 66060 Km/Hr does not fit the normal range, do you have rough idea as to how you approached the problem in a descriptive manner. – Satish Ramanathan Jul 11 '14 at 10:04
  • @satishramanathan, I made a mistake by using the angular velocities in terms of degrees per second rather than in terms of radians per second. I posted my updated answer below. The result is much much more realistic. – BeaumontTaz Jul 11 '14 at 10:06
  • This has been posted twice to physics, where it was closed. That shows the difference in homework policies. – Ross Millikan Jul 11 '14 at 23:11
  • @Ross Millikan, Sorry, Ross. I did not mean to feed the OP with the answer. I like to solve problems like this to keep my gray cells active. I will tone down. – Satish Ramanathan Jul 12 '14 at 04:41
  • @satishramanathan: I like math's homework policy better than physics'. I like encouraging work by OP, but I think the immediate closing on physics makes the site poorer. As is often true, somewhere in between would be better. We do see many full answers here. The activity level here is much higher than physics-maybe this is part of it. It was more effective for OP here. – Ross Millikan Jul 12 '14 at 04:49

2 Answers2

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Consider the triangle $\Delta PQA$. Denote the measure of angle $\angle APQ$ as $\theta$ and note $\theta=\pi/3\;\text{rad}$, the measure of angle $\angle AQP$ as $\beta$ and note $\beta=\pi-\alpha$ where $\alpha$ is the measure of the angle between the extended line $QP$ and segment $AQ$. Since $\alpha=5\pi/6\;\text{rad}$ we know $\beta=\pi/6\;\text{rad}$. We know that $\theta' = -0.025\;\text{rad/s}$ and that $\alpha' = -0.02\;\text{rad/s}$. This implies that $\beta' = 0.02\;\text{rad/s}$.

Define our coordinates centered at point $P$ with the $x$ direction along segment $PQ$ and the $y$ direction perpendicular to segment $PQ$.

Let point $A=(x,y)$. The velocity of point $A$ (denoted $V_A$) is $(x',y')$. The magnitude of the velocity is $\sqrt{x'^2 + y'^2}$ and the direction (relative to the horizontal $PQ$) is $\tan{y'/x'}$. It's easy to show that $A=(x,y)=(3.75\;\text{km},6.495\;\text{km})$.

Denote the point $B$ as the vertical projection (perpendicular to $PQ$) of point $A$ onto segment $PQ$. Consider the triangle $\Delta PBA$. The triangle is a right triangle with a vertical side of length $y$ and a horizontal side of length $x$. Note that

$$\tan\theta = \frac{y}{x}$$

which can be derived to produce the line

$$y'=(\tan\theta) x' + x(\sec^2{\theta})\theta'$$

where all but $x'$ and $y'$ are known. Substituting all known values reduces the line to

$$y'=1.732x'-0.375\;\text{km/s}$$

Now consider triangle $\Delta QBA$. Again note that this is a right triangle with a vertical side of length $y$ and a horizontal side of length $15-x$. Again note that

$$\tan{\beta}=\frac{y}{15-x}$$

which can be derived to produce the line

$$y'=-(\tan{\beta})x'+(15-x)(\sec^2{\beta})\beta'$$

where again, all is known except for $x'$ and $y'$. Substituting in the known values reduces the line to

$$y' = -0.5774x'+0.3\;\text{km/s}$$

The intersection of these two lines will provide us with the velocity vector

$$(x',y')=(0.292\;\text{km/s},0.131\;\text{km/s})$$

The magnitude, as described above is, $320\;\text{m/s}$ and the direction is $0.418\;\text{rad}=23.9^o$ above the horizontal. To be even more clear we will say that the ground projection of the plane is heading toward $Q$ and away from $P$.

BeaumontTaz
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  • It seems like you have added the derivative of the horizontal distance which I did not do. Although, by that addition, your velocity has increased by 5 m/s. I agree conceptually yours is better. The question could have been worded better. Also, if you could draw a diagram, the OP will be ecstatic. Nice answer and superb conceptually. – Satish Ramanathan Jul 11 '14 at 14:04
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enter image description here

$$OA = OPtan(\theta)$$ $$d(OA) = d(OP)tan(\theta) + OPsec^{2}(\theta)d\theta ......(1)$$

Similarly,

$$OA = OQtan(\alpha)$$ $$d(OA) = d(OQ)tan(\alpha) + OQsec^{2}(\alpha)d\alpha ........(2)$$

OQ = 15000-OP

$$\frac{OA}{OP} = tan(60)$$

$$\frac{OA}{OQ} = tan(30) = \frac{OA}{15-OP} = tan(30)$$

Solving these two : You will find that $OP = 3.75$$

$$ \frac{d(OA)}{dt} = \frac{d(OP)}{dt}tan(\theta) +OPsec^{2}(\theta)\frac{d\theta}{dt}...(3)$$ $$ \frac{d(OA)}{dt} = \frac{d(OQ)}{dt}tan(\alpha)+ OQsec^{2}(\alpha)\frac{d\alpha}{dt}.....(4)$$

$$\frac{d\theta}{dt} = \theta^{o}$$ $$\frac{d\alpha}{dt} = \alpha^{o}$$

Substituting the values of $\theta, \alpha,\theta^{o},\alpha^{o}, OP, OQ$,

we get, and solving (3) and (4) we get the y component $\frac{d(OA)}{dt} $ and the x component $\frac{d(OP)}{dt}$ and take the magnitude $\sqrt{x^2+y^2}$ou will get

$$ v = 320$$m/sec and the angle will be the $tan^{-}(\frac{y}{x}) = 23.9$ degrees.

Satish Ramanathan
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  • That is the reason why I asked the OP the question about vertical plane. If it is vertical there is no horizontal component. Am I mistaken? – Satish Ramanathan Jul 11 '14 at 09:17
  • The plane moves in the vertical plane, meaning in the y-z plane if the ground is the x-y plane. Basically, the ground projection of the plane should be colinear with station P and Q. – BeaumontTaz Jul 11 '14 at 09:19
  • That could be a twist, still you would call it an horizontal plane with repect to the ground. I am wondering if the OP meant it that way. Further, on hindsight, plane would not move vertically in a commonsense point of view. Let me retry it, till then I will delete it. – Satish Ramanathan Jul 11 '14 at 09:22
  • Also, you solution is not compatible with the restriction on $\theta'$ because your solution has $\theta$ increasing whereas it is actually decreasing. – BeaumontTaz Jul 11 '14 at 09:24
  • The negative sign is not meant to be negative, I got to understand that when he posted magnitude and the direction with a hyphen. Am I mistaken again? – Satish Ramanathan Jul 11 '14 at 09:25
  • I interpret "x is -5" to be $x = -5$. And he words it as "$\theta'$ is -0.025rad/s". I would say that segment AQ is rotating clockwise around point Q (and changing in length) and segment AP is also rotating clockwise but around point P (and also changing in length). I'm not comparing your answer to his. I'm comparing the direction that the angles are changing relative to what he states the directions to be changing. – BeaumontTaz Jul 11 '14 at 09:27
  • This solution can be obtained by simplifying the components of the velocities in the vertical and horizontal directions. And it does not need to be this long and complex does it? Velocity relative to each P and Q could be obtained using $$ v = r \omega $$ and the resultant of the two velocities give the above magnitude and the direction. I tried it earlier but then again I tried to consider the radial and transverse components of the motion. Relative to P, the components would be $$ \frac{d(PA)}{dt} \ and \ r \omega $$ – S.Dan Jul 11 '14 at 22:37
  • Added the information to the question – S.Dan Jul 11 '14 at 22:48