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1

In principle you could get the displacement from accelerometer measurements, if you also had an estimate of the orientation of the phone at all times. You would need to use the phone orientation to convert each instantaneous acceleration measurement into the same coordinate frame, and then subtract off a constant component representing gravity, then ...


4

A thought experiment: after N steps, each of which create a change in angle $\Delta \theta$, we should end up with a normal distribution of angles with a standard deviation of $\frac{\sigma}{\sqrt{N}}$. When you change the step length, you therefore need to scale the standard deviation by the square root of that change, so that after moving the same distance ...


3

In order to solve this question, we need to determine the distance from the eyes of the observer to the horizon, as this will be the relevant distance the car travels within those 2 seconds. Let's define a few variables: $d =$ distance between observer and horizon $v =$ velocity of the car $t =$ time for car to travel from horizon to observer $h =$ distance ...


1

Distance is: $$ \begin{aligned} x & = \int v(t)\,{\rm d} t \\ x & = \iint a(t)\,{\rm d}t\,{\rm d}t \\ t & = \int \frac{1}{v(x)}\,{\rm d}x \; \Rightarrow \mbox{solve for $x$}\\ x & = \int \frac{v}{a(v)}\,{\rm d}v\\ \end{aligned}$$ The first is used when speed is prescribed as a function of time. The second when acceleration is prescribed as ...


0

I'm going to assume you only want to find the distance whenever the acceleration is a constant value. If not, the suvat equations no longer apply. (Furthermore, it is important to emphasise what is defined by 'distance'. The value that you are actually calculating is the displacement, i.e. the distance away from the starting point of the motion. This is not ...


0

What I think you're getting stuck on here is that you're still thinking about things in terms of forces and you seem to be trying to relate energy back to force via work. I know most classes start by introducing everything as force and acceleration, but you can solve this problem entirely by working with energy values. The idea here with conservation of ...


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 ...


0

I'm going to assume you have some familiarity with linear algebra, as the math becomes much less tedious than trying to do three-dimensional trigonometry with the $x$, $y$, $z$-coordinates directly. You are looking for a function describing the line of the horizon. Since it is a circle and thus a one-dimensional object, I'm going to call it $\vec{h}(t)$, ...


1

Consider the following results: From the definition of scalar product of four vectors, $$ \tag{1}(p_1 p_2)^2 \equiv (p_{1\mu}p_2^\mu )^2 = (E_1E_2 - \textbf{p}_1 \cdot \textbf{p}_2 )^2.$$ The usual dispersion relations: $$ \tag{2} E_i = \sqrt{ | \textbf{p}_i |^2 + m_i^2}.$$ The velocity $\textbf{v}_i$ in terms of momentum and energy: $$ \tag{3} ...


1

Molecular motion is a manifestation of heat. Even at absolute zero, molecules in their ground state, have a zero-point energy ($\ne 0$), in the form of molecular vibrations. As temperature rises, rotational and other vibrational modes get excited. Translational motion arises as the solids melt into liquids or sublime into gases. (For materials, like Helium, ...


1

Translational motion just means linear 'velocity'. That is, motion 'across-ways' (ie: in x-, y- and z- directions). So the question can be rephrased as 'why do molecules have velocities'? Well, if they didn't, everything would be realtively motionless, there would be nothing happening in the universe and you wouldn't be able to ask the question and we ...


5

If the car starts out going in a straight line, it will drop a little bit in the time it takes to cross the gap. If the drop is larger than the height of the chassis above the ground, the car will crash into the opposite wall. When the drop is less than that (small gap, high speed) and the wheels are able to absorb the shock, it is possible that the car will ...


8

The reason that you get slip at even the smallest forces results not from the fact that the tire is slipping against the ground, but that the tire is elastic. There is no way to completely eliminate slip with an elastic tire. Let's see why this is. To measure the slip, lets put twenty little green splotches of die evenly spaced on the circumference of the ...


2

The first one is correct, $L_{new}=2L$. In your second expression $L=\frac{v^2}{2a}$, when changing $a$ you must also find a new end speed $v$, $$L_{new}=\frac{v_{new}^2}{2a_{new}}=\frac{1}{2}\frac{v_{new}^2}{2a}\neq\frac{1}{2}\frac{v^2}{2a}=\frac{1}{2}L$$ A note from reading the comments to the question: I am assuming the objects fall during the same ...


0

Both graphs have 0% force at 0% slip. That implies that no force is being applied to the tire. As applied force is increased until a certain point the increase in slip remains nearly linear because the coefficient of friction remains nearly constant. Thus effective force also increases nearly linearly. After that point the coefficient of friction gets ...


1

Acceleration is simply a rate of change of velocity. So the magnitude tells you, how quickly velocity changes.


-2

In physics, magnitude is the size of a phusical object, a property by which the object can be compared as larger or smaller than other objects of the same kind. More formally, an object's magnitude is an ordering (or ranking) of the class of objects to which it belongs.


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 ...


0

the velocity is tangential to the curve and given by $$\vec{v} =\frac{d\vec{r}}{dt}$$ and the position vector $\vec{r}$ is given as $$\vec{r}=x~\hat{i}+y~\hat{j}$$ Given $y=f(x)$, $$\vec{r}=x~\hat{i}+f(x)~\hat{j}$$ So, $$\frac{d\vec{r}}{dt}=\frac{dx}{dt}~\hat{i}+\frac{d}{dt}f(x)~\hat{j}$$ Simplifying, ...


1

Looks like you wish to express matrix addition with matrix multiplication. This is done with adding extra (service) dimension, which components are always 1 for vectors. $\begin{bmatrix} a_1& a_2& a_3 & 1\\ \end{bmatrix} \times \begin{bmatrix} 1 & 0 & 0 & 0\\ 0 & 1 & 0 & 0\\ 0 & 0 & 1 & 0 \\ b_{41} & ...


1

The rotation matrices you used only apply to rotations around cartesian axes (x,y,x). You cannot generalize it to spherical coordinates in the way you did. See page 4 here for the rotation matrix in spherical coordinates.


0

According to your example s = d|r|/dt = 1, and r(t = 0) = 0, s.t. (1) |r| = t Now, since x^2 = y, r^2 = y + y^2 , s.t. you get a simple equation of 2nd degree in y, y^2 + y - r^2 = 0 . Solve this equation, (2) y = [-1 +- sqrt(1 + 4r^2)/2. Substituting here (1), (3) y = [-1 +- sqrt(1 + 4t^2)]/2, and therefore, (4) x = sqrt(y) = sqrt{[-1 +- sqrt(1 ...


0

You can derivate each component separately: $r'(t)=<x'(t),y'(t)>$ However your example is different, you do not give x(t) and y(t), but only the shape of curve and the constraint of constant speed. This contraint is $r(t)^2=x(t)^2+y(t)^2=s^2$ using that you should be able to get the solution after some algebraic work


3

The answer is correct is you do not take into account the sign. Suppose that the acceleration were constant then $x=1/2 a t^2$, and to reach $x=1$ at $t=1$ then $a=2$. Of course, in this example the mass did not stop, If it stops, it needs to descelerate. In the symmetrical case of constant accekeration and constant desceleration it is easy to see that the ...


1

Just use the standard equation: $y=y_0-v_0t-1/2 a t^2$, you have all the data to solve for $a$ on each of the two cases ($v_0=0$, I assume, because it was dropped from rest). The two accfelerations should be the same and equal to g, however I am assuming this is not the case due to either air friction, or buoyancy.


2

The planet will follow an elliptical path and as an ellipse is a 2D figure it can only be traced to a plane and from two points infinite planes can pass, so you need to decide first which plane you want which is the missing information, because of which you are not able to do so. If finally you form a plane $\vec r.\vec N=p$ then the direction of the ...


0

There are only exact solutions when only using two bodies (namely Kepler orbits), however when you use any more bodies there will be no general solution. These systems of more than two bodies can be approximated numerically, like you tried, by using discreet time steps. But now you enter the realm of numerical integration for ordinary differential equations. ...


0

A simulation does not have a prescribed position $x(t)$. The forces are known, and the position and velocity is found by considering a small time step $\Delta t$ and finding the acceleration $a=\frac{\sum F}{m}$ to act on this step. For each step then $$ \begin{aligned} t & \rightarrow t + \Delta t \\ x & \rightarrow x + v \Delta t \\ v & ...


0

First, I take it from your question that you are working to make a simulation of bodies in 2D with gravitational forces between them, like say earth and moon - or sun and earth - is that correct? If this is the case then you are going to have a problem using $v=v_0+at$ and similar formulae because the acceleration will not be constant. $v=v_0+at$ and ...



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