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1

From your diagram you have $x = y \tan \theta$. Taking the time derivative, and using the speed of the plane is $\dot{x} = v$ (taken as constant) you have $v = y \, \sec^2 \theta \, \dot{\theta}$ assuming that the height of the plane, $y$, is constant. So, for constant height (and taking the speed constant) you have $\dot{\theta} = C \cos^2 \theta$ where ...

1

I'm too lazy to do the detailed math, but it's clear that, for constant velocity and constant rate of climb, it is possible to distinguish between a level path and a rising or falling path. Take point A as the intersection point of a level path and a rising path. At some time the aircraft both occupied A. Point B is any aircraft location on the level path ...

5

Artists learn to draw using perspective - the further away an object, the smaller it appears. The series of identical boxes are pictured above, with perspective used to show their apparent size from the artists eye. A runner passing one box per second would appear small in the distance, but would be running at a constant speed. The situation described in ...

7

What you are measuring is the angular speed $\dot{\theta}$ of the plane and the geese. The speed of the plane or geese with respect to the ground will be $L \times (\cos \theta) \dot{\theta}$ where $L$ is their distance (if you look at the triangle formed by the solid lines some simple trigonometry gives $d = L \sin \theta$). The important thing is that an ...

1

I do not buy into the view that spacelike geodesics are not physical entities or, at best, can only be understood in tachyon terms. Mathematically spacelike geodesic paths are well understood in 3+1 spacetime. They can be computed in principle given a metric. Given 2 events in a local (but not infinitesimal) part of manifold they can be joined by a unique ...

1

You need to find a relation for x and y which holds true along the hypotenuse of the blue triangle. We have two data points: at x = (b-a), y = 0 at x = 0, y = h (taking the green dot as the origin) the hypotenuse is a line with the equation y = mx + b Here, b = h when x = 0 and m = rise / run = (h - 0) / (0 - (b-a)) therefore, m = h / (a - b) so the ...

0

The formula is $$v = \varphi\cdot r/t$$ Where v is the velocity, φ the traversed angle in radians, r the distance and t the time, if the object is moving in a transversal direction. If it is approaching or receding you have to split the radial and transversal components with Pythagoras. In your example I get around 1140 ft/sec.

1

The question is rather incomplete and confusing. By the way, it is used to consider surfaces as vectors when needed for computing surface integrals, like flux integrals, where the scalar product between a vector field $\vec A$ and a infinitesimal surface $\mathrm d\vec S$ is considered: $\vec A\cdot\mathrm d\vec S$. To this aim, the differential surface is ...

2

About second part of your question, I should say that I couldn't understand it because it may the polygon like a star has no side contacted with the ground. About first part of your question, I should say "It is not possible". “Let us say a polygon shaped object is stable on a side when the center of mass "falls" inside the base”. If you accept this phrase ...

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