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If you have a uniform acceleration the average velocity $\bar{v}$ is just $$ \bar{v}=v_0+\frac{at}{2}. $$ because the final velocity is $v_{initial}=v_0$ and $v_{final}=v_0+at$, i.e. $$ \bar{v}=\frac{1}{2}(v_{initial}+v_{final})=v_0+\frac{at}{2}. $$ Then you get $$ d=\bar{v}t=v_0 t+\frac{at^2}{2}. $$


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But what about the case where the velocity only changes its direction keeping its magnitude constant? The first equation you give is a vector equation which, in this case, means there are three equations: $$\vec a = \frac{d\vec v}{dt}\Rightarrow a_x = \frac{dv_x}{dt}; a_y = \frac{dv_y}{dt}; a_z = \frac{dv_z}{dt}$$ If we have the additional ...


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There’s a slight issue with some of your wording. Measuring and calculating a quantity are two very distinct tasks - one is experimental in nature, and one lies upon mathematical formalism and theory. That is not to say that they are entirely separate, but it’s worth noting from a pedagogical standpoint. Fundamentally, the answer to your question relies on ...


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Generally speaking, when we graph experimental data it's nice for the graph to be a straight line. This is because it's easy to tell if a graph is a straight line just by putting a ruler on it. If a graph is a curve it's hard to tell whether it's a parabola, ellipse, sine wave or some other curve without diving into some maths. In this experiment you are ...


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Yes, this is certainly true. Mass is defined by $m^2=E^2-p^2$ (in units with $c=1$), where $(E,p)$ is the momentum four-vector built out of the mass-energy and momentum. (This defines what's known as invariant mass, as opposed to "relativistic mass.") Mass as defined in this way is not additive, and depends on the motion of the particles within a system. As ...


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The relation between Kinetic Energy and momentum is derived in the following way. I hope this answers your question.


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Since both velocity and acceleration are vectors, it is possible as in the case of circular motion for the magnitude of the velocity vector to remain constant but for the direction of the vector to change. Since acceleration is the rate of change of the velocity vector, acceleration would be non-zero even if the velocity vector magnitude is constant, but ...


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The video shows a supposed perpetuum mobile of the first kind. That can't exist, because it would violate law of conservation of energy. So there should be something wrong with it. I've taken a still from the video (the frames of the video are all the same except for rotating colours of the objects): According to the description, part of the machine is ...


2

The job of calculus is to handle quantities that vary over the domain of the problem at hand. Often, and particularly in introductory physics, we care about quantities that vary in time. We cannot put them into our equations as constants. We also often care about the interrelationship of these quantities. So for example now velocity is given by ...


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Like mentioned in the comments, you got the initial velocity wrong. You source says: Further to this, the selected separation strategy foresees a fixed separation velocity of approximately 0.187 m/s. This imposes limitations on the capability of directing Philae towards the comet, i.e. it restricts the domain of possible positions and velocities ...


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It's no different than any other case with a general time dependence of acceleration. You just need to know that acceleration is the rate of velocity change: $a=\dfrac{dv}{dt}$ and velocity is the rate of position change $v= \dfrac{dx}{dt}$. Integrate twice and you get $$v=v_0+\int_0^{t} a(\tau)d\tau$$ $$x=x_0+\int_0^t v(\tau)d\tau$$ For constant $a$, this ...


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The multi-dimensional analog of simple harmonic motion is an object subject only to a harmonic potential, $U = \frac 1 2 k ||\vec r - \vec r_0||^2$, where $k$ is a positive constant (oftentimes called the spring constant), $\vec r$ is the object's position, and $\vec r_0$ is the position of the center of the potential. By choosing the origin to be the center ...


1

It came from the wall. If a ball hits a wall at constant velocity, then it's not going to "slow down and eventually come to rest". It's going to bounce back, and in fact, if the collision is elastic, it won't even necessarily slow down. The force on the ball from the wall serves to change the direction of the velocity (and possibly decrease its magnitude ...


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How else can you recover displacement from a non-linear velocity without integrating? How else would you calculate acceleration of a non-linear velocity without differentiating? ADDED: If your question is how to solve $\displaystyle v(t) = \frac{dx}{dt} = b\sqrt{x}$, let us know.


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When you say displacement, do you then mean displacement of each particle? Or the displacement of the wave / wavetops? I guess the latter. If this is a rope being swung to create the wave motion, then remember to keep a sharp distinction between particle motion and wave motion. As the particles in the rope move up and down (transverse) the wavetops are ...


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Calculate the kinetic energy of the water coming out in unit time. That is the power you need. $$Power = \frac12 (\rho A v) v^2 = 31.4 kW$$


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I'm not going to give you the answer, but try eliminating the velocities by substituting information provided to you about the momentum relationship. The velocities should not be in the final solution.


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As mass B accelerates, the other two "stay behind". The string will no longer be at 90 degrees to the motion, and tension in the strings will pull back on B. So in order to maintain constant acceleration on B, the force needs to increase. The tension in the two strings is the same, by symmetry. As the force on B (and thus, the system) changes, the center of ...


1

Let's assume that the balls are identical, and the strings are identical, and are elastic (obey Hooke's Law) and massless. You have stated that you've somehow arranged for B to have a constant acceleration. This means that the net force on B is constant. There are three forces on B: one due to the string connected to A, one due to the string connected to ...


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That's a good start, but point 2 can be made a bit more precise. Let's say the asteroid starts at $\mathrm{B}_0$ at time $0$ and has constant velocity $\dot{\mathrm{B}}$. Then its position at time $t$ is $$ \mathbf{B}(t) = \mathbf{B}_0 + t \dot{\mathbf{B}}. \tag{1} $$ Your $\mathbf{C}$ is just some point on this line, which we would know if only we knew ...


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You have to turn this into a 2D problem. To find 3 unit vectors describing a local 3D coordinate system where the planar problem is along the local $\hat{i}$ and $\hat{j}$ axes, with the plane perpendicular $\hat{k}$. Subscript A denotes the asteroid, and B the bullet, with $\vec{r}$ positions, $\vec{e}$ directions and $v$ speeds. The notation is reverse ...


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The change in velocity can be calculated by vector subtraction. ($d\vec{v} = \vec{v_f} - \vec{v_i}$). Divide by the time between the two velocities to generate an acceleration. The direction of the acceleration will be the same as the direction of the difference vector. The magnitude of the acceleration will be the same as the magnitude of the difference ...


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This is a project, so I don't feel comfortable answering every bit of this. But one way to model when something leaves the ground is to consider what the value of the normal force $N$ must be at that instant. Then if you can find an expression for $N(t)$, you're set.


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Here is the intuitive explanation: When a particle is moving, it will "run into" things. Thus, the "random force" from impacting another particle is not completely random: it is in part correlated to the motion of the particle before the collision - the force of the impact is more likely to be in a direction opposite to the current motion than any other ...


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The sum of the torques is zero, $\sum \tau =0$. Thus, since the centre of mass is in the middle of the pole, that's where the weight $W$ will act. $F_y$, is the normal force exerted by the pole on the paper. We have: $$mg\frac{l}{2}\sin{\alpha}-F_y\ l\sin{\alpha}=0$$ Hence, the normal force $F_y=\frac{mg}{2}$. Friction is given by $f=\mu N$, and so the ...


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You can cast your problem in terms of what is the velocity of the car at any instant of time after it started. The answer is $v = a_{\rm eff} t$ from the first kinematical equation. In the same time, the car would've traveled a distance $s = (1/2)a_{\rm eff} t^2$ from the second kinematical equation. The other option is - if your inclined plane is of any ...


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If i understood your question correctly, you seem to not know the displacement of the car but you still need to increase the velocity of the car. Velocity of the car has to be expressed in terms of a quantity. I suggest you use velocity as a function of time, $$v_f = v_i + at$$ $a = g \sin \theta $ as you told correctly. So $$v_f = v_i + g \sin \theta ...


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Let's say for simplicity $F(x) = 3 x^2$, just so we have an example to talk about. As you wrote $$f(x) = \int F(x) \mathrm d x + C = x^3 +C\,.$$ And now the initial conditions come into play (the $x_0$, $v_0$ and so on) or in some cases boundary conditions. Those are always needed to find a specific solution to a differential equation. So if I also know, ...



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