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13

Your error is simply that you are assuming that $v(x)$ is differentiable with respect to $x$ at $v=0$. The chain rule needs that all derivatives involved exist before you can apply it. In the case of just letting go of something, the function $v(x) = \sqrt{2gx}$ is not differentiable at $x=0$, which is where $v=0$, so you are not allowed to apply the chain ...


9

Yes, it is possible to rise theoretically w.r.t a ground frame. But keep in mind that the rope-man system's center of mass must keep moving downwards because of the only external force acting on the system(gravity). The lighter M gets, the harder it will be for the man to rise, and it will become impossible in the limit the rope becomes massless. This is ...


4

I find no trouble in thinking of velocity and acceleration vectors as arrows. First some definitions: Velocity is a vector. Speed is it's magnitude. Acceleration is a vector. It's magnitude has no new name. We agree that acceleration is present if there is a change in velocity: $$\vec a= d \vec v /dt$$ That is, any change. So if magnitude (speed) and/...


3

When an object starts at rest, the change in velocity when it has made an infinitesimal displacement is infinite - in other words, $\frac{dv}{dx}$ is undefined. You can see this most easily by plotting the curve of $v$ as a function of $x$ for an object starting at rest: $$x = \frac12 a t^2\\ v = at\\ x = \frac12 a \left(\frac{v}{a}\right)^2\\ v = \sqrt{...


2

The drone flies by its propellers exercising force against the air inside the airplane. It flies with respect to the air inside the airplane. Since the air is being carried by the airplane, the drone will fly with respect to the airplane. It'll fly forward or in whichever direction you point it, with respect to the X. Assuming the airplane is airtight.


2

Let's consider the frame in which initially both the masses are at rest to be the frame $O$. In frame $O,$ momentum conservation is trivially followed because of the symmetry of the problem. For the energy conservation, we require that $M = m \sqrt{1-v^2}$, where $m$ is the initial rest mass of each of the particles and $M$ is the final rest mass of each of ...


2

I want a more mathematical way to see it. Acceleration is defined as the rate of change of velocity with time. Velocity, being vector, can change just by changing direction keeping the magnitude constant. In a circular motion if the angular frequency $\omega$ is constant, then the magnitude of the velocity i.e., speed is constant but the velocity changes ...


1

Here is how to answer this question (rather than the specific answer, which I don't think you're after). Firstly they tell you that the two bits of the path are the same length ('the first half of the path') and are straight. Call the length $l$. They also give you the angles, $\alpha_1$ and $\alpha_2$ of the two bits of the path to some reference line. ...


1

Consider what you stated: $$a=v\frac{dv}{dx}$$ Now rewrite it: $$\frac{dv}{dx}=\frac{a}{v}$$ If $v$ is tiny, then you know that $\frac{dv}{dx}$ must be enormous in order to produce the acceleration that you darn well know exists. Accleration at the top of that trajectory is surely $9.8$ $m/s^2$. As $v$ gets smaller and smaller as it gets to the apex, $\...


1

The kinematic equations are derived from differential equations. This means that you can start with the acceleration, then move to velocity, and then position given that the necessary information is available (like initial velocity or position). I find that understanding this helps me visualize the connection between the three. You can not rely on the ...


1

Why your computed average acceleration is wrong? the average acceleration is defined as: $\overline a=(v_2-v_1)/(t_2-t_1)$ where the $v$'s are instantaneous speeds. If you start with zero initial speed you can simplify it to $\overline a= v /t $ $v$ is still the instantaneous speed at $t$. For a constant acceleration $v=at$ so in this case you ...


1

The glitch in your logic is that you supposed acceleration to be distance/time squared while using your formula: $$ \Delta x = v_0t+\frac{1}{2}at^2 $$ The above formula gives $$ a= \frac{2\Delta x}{t^2}$$ supposing that vo is equal to zero. This makes the acceleration equal to the average acceleration in your case since the car is moving in one direction ...


1

Gravity acceleration is... acceleration, measured in $\mathrm{m/s^2}$. It is the rate of change of the velocity.. Velocity is measured in $\rm m/s$. It is the rate of change of the position. The vertical velocity and the acceleration due to gravity of a body are collinear but they can have different magnitudes as well different orientations. Think about a ...


1

You already have a relation between $\theta$ and $\alpha$. If this is correct, all you need to do is to rearrange it. If it is not possible to rearrange it to get the given relation, then either your equation is wrong, or the given equation is wrong, or both. If the particle grazes the upper plane at maximum height, where the velocity vector is horizontal,...


1

If velocity at bottom of slope is all you want then energy method in Gert's and Dr Xorille's answers is an elegant and easy way to get at it. However you seem to be tangled up with resolving vectors. In working with vectors you should set up a coordinate system and stick to it until the end. If you have to set up more than one, and worse, have to switch ...


1

Because you can separate the vector space (assumption) of R x R into two separate vector spaces, each spanning the real line R with independent basis. When you are in the vector space R then you can't reach all elements in R x R. You can do that if you make a superposition of each of these R spaces into R x R and then you can reach all elements in this space ...



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