New answers tagged

1

$\alpha$ the angle the air speed makes with the person $\omega$ the uniform angular speed $t = \alpha/\omega$ the time of the race $vt$ the length of the road, because air friction is considered negligible


4

There is one factor missing from your consideration: the velocity of the medium! Take the general case of two cars driving at different speeds towards each other while both blow their horns. You could take either car as being stationary, and wind up with the same relative velocity (car to car). But substituting this into the various equations you cite ...


6

Relative velocity for sound waves is not a "symmetric" situation. For example, in the extreme case of a fighter jet approaching a stationary observer at Mach 1, the jet will be traveling as fast as its sound waves. The observer will not hear anything until the jet gets to his position. On the other hand, for the case of a stationary sound source and an ...


0

As it is already answer by so many but i want to add one more point ,THAT IS -change in distance of B with repect to time will be equal to ym/s if you take your frame of reference as origin according to origin it velocity changes because it s distance with respect to time change equal to y=m/s but now yiu change frame of reference that is A so the velocity ...


0

given two vectors $\vec{v}_{01}$ and $\vec{v}_{02}$ thus: $$\vec{v}_{12}=\vec{v}_{10}+\vec{v}_{02}\,,\quad\text{("zero cancel")}$$ where $\vec{v}_{10}=-\vec{v}_{01}$ your example $\vec{v}_{01}=x$ $\vec{v}_{02}=-y$ $\Rightarrow$ $\vec{v}_{12}=-x-y=-(x+y)\quad \surd$ $\vec{v}_{21}=-\vec{v}_{12}=(x+y)$


2

The problem is that your spacetime diagram is wrong: you're using Euclidean geometry, when you should be using Minkowski. The actual diagram, if you'll forgive the low quality picture, looks something like this: where I've drawn the $t^2 - x^2 = \pm 1$ hyperbolae, and the angles obey $\tan \alpha = v_1$ and $\tan \beta = v_2$. Note that that unit vectors ...


1

Suppose body $A$ is going to the right with a speed $x$ and another body $B$ is going to the left at a speed $y$. The motion of the bodies can be represented as vector diagram 1. To both motions add a velocity to the left of equal magnitude to that of the velocity of body $A$, ie stopping body $A$, as shown in vector diagram 2. On adding the two ...


3

The top diagram shows the velocities in the lab frame. Particle $A$ is moving east at speed $x$ and particle $B$ is moving west at speed $y$. I'm taking the east direction to be positive, so the velocity of $A$ is positive and the velocity of $B$ is negative. To find the velocity of $B$ relative to $A$ we have to transform to the frame where $A$ is ...


0

Everything we currently know about physics, including quantum physics, implies that it is not possible to have a physical phenomenon which depends on a system's velocity relative to another system with which it is not interacting. This means, in particular, that the influence of the local vacuum near to any system is not detectable through velocity, but it ...


1

Your question asks about accounting for the Earth's, Sun's, and Milky Way's speed. The principle of relativity says these don't matter. That's because the laws of physics should be the same for every observer, no matter where they are or how fast they're moving. If we did have to take into account that the Earth is moving (for this experiment), then that's ...


0

My recommendation is that you calculate your answer twice, first by adopting the accelerating reference frame of the train, and then by adopting the inertial frame of reference of the ground. Then keep working at it till your two answers agree. If you adopt the train as your reference frame, then you can treat the effects of its acceleration as like a ...


-1

You can use the train as a frame of reference as long as you take into account that it is not an inertial frame of reference. To account for the acceleration of the train you will have to introduce a fictitious force of magnitude $mx$ N acting horizontally on an object of mass $m$ kg. This fictious force is similar to gravity, but it acts horizontally ...


0

That relative motion expression for your question was non relativistic motion. When consider non relativistic motion, people take it from the usual experience. The additive property was assumed. i.e. the so called "if you are traveling at 20 km/h, you add an extra 30 km/h in the same direction, your speed would be at 50 km/h". With velocity, it's the same ...


0

If I observe the motion of another car through my frame of reference i.e.car, will I be at rest with my frame of reference or in motion? You will always be "at rest" in your own frame of reference. You will not be at rest with respect to the frame of reference of the person in the other car. Nor will the person in the other car be "at rest" with respect ...


0

If we model you as a "disembodied spatial tangent vector" (i.e. a point particle with a preferred spatial direction to represent your line of sight), then you'll need to spin in place to keep an eye on the other guy, but that doesn't count as motion. If we think of you as three dimensional, then parts of your body will have to move, but there's no need ...


2

You're always at rest with respect to your own frame. It doesn't matter what else is going on.


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