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4

It's a vector. Instantaneous velocity $\vec v$ is defined as $$\vec v \equiv \lim_{\Delta t \rightarrow 0}\frac{\Delta \vec r}{\Delta t}.$$ In that equation, $\Delta \vec r$ is the displacement that occurs during time interval $\Delta t$. Putting on my math hat for physicists, the numerator is a vector, and the denominator is a scalar, so the resulting ...


0

OK... once more: let's assume we are looking at a map-coordinate system where the x-axis points East, the y-Axis points North and the z-axis points towards the zenith. In such a system there are three acceleration components $\{a_E, a_N, a_Z - g\}$. Since the surface of the Earth is not an inertial system, the gravity of Earth always acts on the Zenith ...


0

 Why doesn't anything need the second derivative (acceleration)? Only Newton's gravity law does not use acceleration in the expression for force. In electromagnetic theory with retarded fields, forces are functions of past positions, velocities and accelerations of the charged particles.


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"Could we come up with a theory that only requires a snapshot of the positions? ...Does modern physics (e.g. relativity) have something to say about this curious thing?" Yes. The Theory of Special Relativity posits that there is no favored rest frame, and therefore there is no such thing as absolute velocity. In other words, velocity can only be measured ...


0

because we want to find the trajectory of the particle and if we know it's position in different times we can calculate it's motion. Also, the acceleration can be obtained when we have the X(t). then acceleration is not a new information to use and it is embedded in the X(t) (acceleration=$d^{2}X/dt^{2}$)


0

I am trying to answer your questions one by one. By the way, I saw your comment. 1. You say in the question: "Why doesn't anything need the second derivative (acceleration)?" Yes the acceleration is needed for obtaining the velocity. I some situation we are given the force and the mass, as in an electrical or gravitational field, not the velocities. Then ...


1

The reason that you only need to specify initial position and velocity to exactly solve the equations of motion for a system is simply because Newton's Second Law (which is the equation governing motion in Classical Mechanics) is a second-order differential equation. The upshot is that to solve a 2nd-order ODE, you basically need to take 2 integrals. Each ...


0

HINT: At any time,your resistance acts in horizontal and veritcal directions both. The $v$ is for total velocity, and not along any particular axis.


1

Looking strictly at the graph you provided, it shows that at time=0, distance=0. And at time=6s, distance=60m. That graph shows a uniform velocity of 10m/s. Generally, velocity is not an undefined quantity, it is defined as the rate of change in unit displacement per unit of time, define your units and you have the definition of velocity.


1

Let's take an instance of race during Olympics; Say we have a man $A$ at position $a$. Assume, we whistle to start the race but he doesn't start to run and and when the race is finished, after some time intervals, we calculate his average running velocity, we get $0/t$ so the velocity is found to be $zero$ as well. But what happens when again on starting ...


12

what if distance and time both become zero as at origin in the graph is It appears that you're trying to say that the velocity is equal to the position of the particle divided by the clock time at that position: $$v= \frac{x(t)}{t}\; ? $$ But this isn't correct. Average velocity $\bar v$ is defined as displacement $\Delta x = x(t_f) - x(t_i)$ ...


2

At (0,0), before the clock has started, you can say that there is no displacement, and no time measured. It is only with hindsight, by looking at the rest of the chart after the clock as started, can you see the slope of the line, and hence work out its velocity. Any single point on the line, without any other data, shows only an average speed, again, we ...


2

Velocity is defined as how fast the body changes its position with respect to time. Change of position with respect to a frame is called displacement. Velocity measures how fast the body changes its position. $$ v = \lim_{\Delta t \to 0} \dfrac{\Delta x}{\Delta t}$$. At $t = 0$ , the body was at a distance of $0$ from the frame. It was not moving and hence ...


3

Actually you have plotted the graph of displacement $x(t)$, And here the $x(t)$=$vt$ ,$v$ is some constant. now lets take the slope of the graph dx/dt =v The slope is constant (equal to v and we physically call it velocity) now what is the slope at (0,0)? since the slope is constant it will be still $v$,right? hence at the origin the velocity is $v$ and ...


3

$$\text{Velocity }= \dfrac{\text{change in postion}}{\text{time taken for that change}} \neq \dfrac{\rm distance}{\rm time}.$$ If we draw a graph for the change in position vs time difference, the case you are talking does not exist. Time difference has to be there when we talking about velocity or speed as they are a measure of rate of change of ...


2

Force was defined to be proportional to acceleration because that definition makes description of classical physics simple. For example on Earth we have a downward pointing constant force - gravity. If we define another quantity, one which is proportional to velocity, lets call it push, it would be quite useless in describing gravitational interactions. An ...


5

You mix the relations between the things. A force produces changes in the linear momentum. The acceleration $a = F/m$ produces changes in velocity $v = p/m$. So, your question should be either why don't we take the force proportional to the linear momentum, or why don't we take the acceleration proportional to the velocity. Now, the second thing is ...


3

Well, technically we do: $$ F_{net}=m\frac{dv}{dt} $$ The net force is proportional to the rate of change of velocity, which we call acceleration. Note also that it's not $F\propto \Delta v$ (force proportional to the changed velocity because the changed velocity occurs over a period of time, $\Delta t$, that is also important--consider the difference in ...


0

If you mean force proportional to velocity, that restricts the second law to only the specific case when the force is proportional to the velocity (the object will feel a drag or will accelerate exponentialy with time, depending on the sign of the proportionality constant). Such a law will not decribe the dynamics of any other objects.


5

You need to integrate acceleration to get the velocity. $$v(t) = \int_{t=0}^{t} a. dt$$ There are a number of ways of doing this numerically. I assume that you get these readings regularly with a spacing of $\delta t$, for example $\delta t = 100 ms$ or something like that. About the simplest way to do it is $$v(t) = v(0) + \sum a \times \delta t$$ ...


2

The only factor is the capacity to return to the original shape when it is deformed by an external force. An elastic object recovers its shape after it has been compressed and deformed, if you crush a plastic bottle and remove the cap it will partially return to his original length, a lead ball is almost completely irreversibly deformed. The Coefficient ...


1

Let's make a few simplifying assumptions. We'll use only one ball, and assume that when the ball returns to your hand you immediately throw it back up - we'll assume the time taken for the throw is short enough to be ignored. We'll also assume your arm isn't elastically storing energy, so for each throw you have to put in fresh energy from your muscles. ...


0

Most of us will have discovered that a glass rings if you tap it, and the sound of the ring will persist for several seconds after hitting the glass. This shows that elastic energy stored in the glass is not dissipated fast. I would guess that fewer people have tried the experiment with a lead beaker, but I have and I can report that the lead does not ring. ...


1

I would guess that as long as they are both constantly being juggled that it would be the same, because throwing a ball in the air at twice the speed will result in twice the time to fall, but also twice the work to achieve the speed. I'm just using deductive reasoning and some of the basics i have learned. I hope this helps!


2

The freespace dispersion equation is $\omega^2 = k^2\,c^2$ and this cannot change: this simply follows from considering plane wave components of propagating fields, which all fulfil the Helmholtz equation $$\nabla^2 A_j + \frac{\omega^2}{c^2} A_j = 0\tag{1}$$ which is fulfilled by all Cartesian components of the moncrhomatic EM field vectors and, for a ...


0

Wavenumber k is the number of waves per metre. Frequency w is number of waves per second. The number w/k is the speed of the wave.


3

Technically, $\omega^2/1^2-k^2/c^2=0$ is a degenerate hyperbola if that counts. But I don't think you can derive an equation of the form $\omega^2/a^2 - k^2/b^2 = 1$ for waves propagating in free space. You may however find something of the kind if you consider materials with fancier dispersion relations than $\omega = ck$, like e.g. plasmas.


2

Conservation of momentum works here like everywhere else. When A (with mass $m_A$) throws a ball with mass $m_b$ with velocity $v$, then $v_A=-v_b\frac{m_b}{m_A}$ so that after the ball is thrown, the net momentum is zero; note that the ball will not be moving towards $B$ at velocity $v$ but instead at $v-v_A$ since $A$ started moving backwards... When B ...


1

Would A move backwards at the same speed? no.. Momentum will be conserved,not speed and since A has greater mass than the ball so his speed will be less.. if A throws the ball with speed 'u1',then his speed(in the backward direction) will be m1u1/M1 where m1 is the mass of the ball and M1 is the mass of the person A Now, what would happen when B catches the ...


4

As far as we know and can test, space is continuous, not discrete. Since space is continuous, then the labels we associate with it (i.e., positions) are also continuous. Calculus requires continuous functions to do the derivative and integral, so this implies that velocities and accelerations are also continuous because they are derivatives of positions: $$ ...


1

Careful between $\hat{r}$ and $\vec{r}$. The trajectory $\vec{r}(t)$ is a function that maps $t$ to a position vector $\vec{r}$. But the tangent to the trajectory is not the same as $\vec{r}$. Thus, $\vec{v}\cdot \vec{r}$ tells you nothing about the tangent vector. To be tangent to something means to be going in the same direction at exactly one point. The ...


6

Related questions Why does your car lurch toward an oncoming truck as it passes you? Observer stationary A vehicle passing a stationary vehicle can produce a complex pressure wave From MEASUREMENT OF THE AERODYNAMIC PRESSURES PRODUCED BY PASSING TRAINS In this you can see that the stationary vehicle is first pushed away and then sucked back ...


2

The relative speed of your car and another car does not matter since your car is affected only by air. Probable explanation. A moving car is producing a wind, blowing in nearly perpendicular direction These streams of are are similar to rain drops. And it is a subject for Galileo transform. If you stay and rain is vertical, it falls on your top. If you ...


-2

Mayby the moving mass of your car makes it more stabile. The same as a giroscope. The mass of a ciclist is lower so he can experiënce more swaying. Large motorhomes and caravans also sway more when a truck passes. And those are riding at about 50 m/h. So learning while writing , its a combination of surface and streamline, and the stability the moving mass ...


1

You need to measure the sound level in decibels. This is a logarithmic scale where zero decibels corresponds to a root mean square pressure of 20 micropascals, and every 20 decibels corresponds to a tenfold increase in the pressure. Once you have measured the sound level and calculated the pressure you can use the equation for the particle velocity: $$ v = ...


0

You are actually completely right, and then at the last minute you look the wrong way!!! Lets examine the last bit. If you were to continue applying a net force on an object for an infinite time it would indeed accelerate infinitely. There is nothing wrong here. The key point is the NET force, that is, the net amount of force, and its direction, that is ...


2

The electron does not move - it has no well-defined position in the orbital state, and hence no well-defined momentum. Neither does it "teleport" around - as long as it is not interacting with something that forces it to be at a definite position, its state is "smeared" all over the electron as an electron cloud. Yes, this is essentially the Bohr model, ...



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