New answers tagged velocity
1
You have to distinguish between the distance the man swims, relative to the water around him, and the total distance the man travels, relative to an observer on the river bank. The total distance relative to an observer on the river bank is the distance the man swims measured relative to the water around him combined with the distance the water moves ...
-1
The mass of object changes when its speed approaches zero because according to Einstein postulates of theory of relativity all the laws are same in all inertial frames and speed of light remains constant in inertial frame in vacuum. All the concepts of relativity are based on these two postulates. As one can not add any speed in speed of light, the Lorentz ...
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1) Spline curves are designed for just this sort of thing. A spline is basically a set of cubic (typically) polynomials at each section of your data that go exactly through your points. But since they're just cubic polynomials, they're pretty smooth. Assuming the path is a reasonably small portion of the earth, and the points are close together on a ...
0
You will calculate the velocity for the x, y and z components separately.
So if the object has $v_x$ velocity initially, and it spontaneously gains acceleration in the negative y direction, this problem becomes very simple. Over time this object will gain velocity in the y ($v_y$) direction. It will also always have its' velocity in the x direction, and ...
2
I would say that $\left\langle v_i \right\rangle\left\langle v_j \right\rangle$ is the product of averages, whereas $\left\langle v_i v_j \right\rangle$ is the average of a product. Hope you get the difference.
Take an example of $v_i = v_j = v$, then $\left\langle v_i \right\rangle\left\langle v_j \right\rangle = \left\langle v \right\rangle^2$, while ...
1
Your premise violates Newton's first law of motion:
If there is no net force on an object, then its velocity is constant. The object is either at rest (if its velocity is equal to zero), or it moves with constant speed in a single direction.
For an object (a body) to be accelerating there must be an external force applied. One of the reasons for ...
6
think about this with an example: the sine and cosine functions. They both average individually to zero over an interval. You can multiply those averages and still obtain zero. But if you multiply sin by itself and then average, you get a very distinct non-zero result.
When the functions are arbitrary, the average of the product quantifies statistical ...
1
The right hand side of your first equation involving $x$ is wrong. I will use $v_y$ to mean your $x$. Then you would have $9+v_y^2$ for the squared speed of the rain in the first case as you say. When he starts walking faster the squared speed of the rain is $(6-v_y)^2 + v_y^2$. The reason for the $6-v_y$, is that the man sees the angle of the rain as ...
0
Let's see with help of an example.
Let the particle is at $(0,0)$ moving with speed of $2m/s$ at $t=0$ and is subjected to acceleration $-2\hat i\ \text{m/s}^{-2}$.Now see after $2 seconds$.
We see displacement is zero,and distance travelled =$2m$ .Also acceleration is constant but still $\langle speed\rangle\not=0$ whereas $\langle velocity \rangle=0$
...
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Look at it this way. An electron is a charged particle. A moving, electrically charged particle creates a magnetic field, and the particle itself already has an electric field. If the particle is accelerating, then you're going to have a ripple effect from the electric and magnetic field, or an electromagnetic ripple, ie an electromagnetic wave. So an ...
2
On a uniform circular orbit, even if the speed does not change in norm, it does change in direction so that the speed vector change over time and $\frac{d\vec{v}}{dt}\neq\vec{0}$. In fact, in polar coordinates, you have
$$\vec{a} = \frac{d\vec{v}}{dt} = -\frac{v^2}{R}\,\vec{e_r}$$
Imagine a car taking a turn at constant speed: if the turn is left, you feel ...
0
Displacement of elevator with Constant velocity:
$s=ut$
displacement of elevator with uniform acceleration:
$s=ut+0.5at^2$
0
I can't understand why you're asking "Is distance, velocity a function of time?". The question is quite ambiguous because, when we define velocity, acceleration, or jerk in classical mechanics, we're quite sure that we're taking the time derivative of the predecessor. For instance, if you require velocity, then you're taking the time derivative of distance.
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2
The answer to this question depends very much on what field you're studying. For instance, in many areas of physics, being time derivatives of position, most would take the velocity and acceleration equations and treat the whole system as a differential equation, then solve for distance as a function of time only. Similarly, they would then differentiate the ...
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