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23

Yes, if an object is stationary, then starts moving in the positive direction, all derivatives of the position can become positive when the object starts moving. But so what? It's not a paradox. It's just a mathematical statement. All the derivatives are positive. No big deal. It's hard to tell why you find this confusing. You might be thinking that there ...


22

The correct thing to say would be that "if v=0 and dv/dx is finite then a=0". A simple example, to help illustrate what's going on, is the well known case of constant acceleration "-g" near the earth's surface. In this example, we consider "x" to be the height above the ground, and assume the initial x is zero. In this case $$ x=-\frac{gt^2}{2}+v_0t $$ $$ ...


7

No, it doesn't imply that $a = 0$. If, at some value $t = t_0$, the acceleration is non-zero while the velocity is zero, the position function is either a minimum or maximum. That is, $x(t)$ is stationary there: $$x(t_0 + dt) = x(t_0)$$ which means that at $t = t_0$ $$\frac{dx}{d\dot x} = \frac{dx}{dv} = 0$$ thus $\frac{dv}{dx}$ is undefined at $t = ...


5

You can apply chain rule if $v$ is differentiable wrt $x$ and $x$ is differentiable wrt $t$. I think there are no other conditions,as this post on MathSE seems to say, http://math.stackexchange.com/questions/688152/necessary-conditions-for-the-chain-rule-of-differentiation-to-be-valid#= and this condition is not always available. When $v=0$,make sure ...


3

I quite sure it's not theoretically possible. Without doing any actual calculations, I recall that accelerating a massive object to light speed would require an infinite amount of energy. Some energy certainly can be gained with the "slingshot" method, but definitely not an infinite amount. Specifically, it's the Lorentz factor that prevents objects from ...


2

Have a look here: https://en.wikipedia.org/wiki/Non-analytic_smooth_function There are functions which are identically zero for negative arguments, non-zero for positive arguments and smooth everywhere.


2

Who says all derivatives have to change continuosly? With a suitable setup, acceleration can jump from zero to something. Location and velocity can not change discontinuously, but acceleration can. No, there are no "infinitely many derivatives of velocity". In your question, you write "acceleration has to increase from zero and therefore the third ...


2

You can achieve this in two distinct ways: 1) use a reference (I'll give an example below) 2) measure other quantities that can be converted into an acceleration. Here are some possibilities: 1a) Gravity. At rest, the accelerometer should measure the local g. 2a) Put it on a rotating platform. Knowing the rotational speed and distance to the pivot, you ...


2

The coordinate velocity does indeed change discontinuously, but only if the acceleration changes discontinuously i.e. the jerk is infinite. Since for any physical system none of the time derivatives of position can be infinite, in a physical system the coordinate velocity can't change discontinuously. But let's ignore this for now and examine why we get a ...


1

Einstein's equivalence principal states that an accelerated reference point is indecipherable from a reference frame in a gravitational field, so an accelerated reference frame will act in the same way that a gravitational field with the same acceleration would act. As for if all reference points are equally valid, the answer is generally "yes" with some ...


1

I want to take another tack than that of the other answers. This will be one big handwave rather than a rigorous mathematical argument, but I hope it gets the idea across intuitively. First off, as I noted in a comment, and as hft notes, you are using "v" to mean both "velocity as a function of time" and "velocity as a function of position". That's ...


1

Assuming the total force is $10~\text{N}$, the $7~\text{kg}$ mass is accelerated by $1.0~\text{m/s^2}$, so it is experiencing $7~\text{N}$ force in x-direction. Therefore, $F_x = .7F$. The angle is roughly $45^\circ$.


1

The answer is, you'll have to be more specific. When a fist hits a bag, the force varies a lot in terms of time. If you imagine trying to punch a bag as slowly as possible, there is a slow onset as your fingers just graze the leather, followed by something more "meaty" as your bones start to press against your skin which is pressing against the actual ...


1

force = acceleration * mass, hence acceleration will be $a=50N/22kg \approx 2.27m/s^2$ Distance it moves might be found by integration: $\int_0^{1.2}v(t)dt=\int_0^{1.2}atdt$, since speed $v(t)=at$ Answer to (B) then is 1/2*1.2*(2.27*1.2)=1.63, which seems pretty close to what you have got


1

Your calculations are correct. They differ from your model (which uses ABS braking) however, because they don't take into account the duty cycle of the braking. If this is added to your calculations, then the two results should be similar.


1

You need to solve system like following: $\begin{cases} d_0=\int_{0}^{\tau}v_0 \exp(-kt)dt=\frac{v_0}{k} (1-\exp(-k\tau)) \\ v_f=v_0\exp(-k\tau)\end{cases}$ You need to solve for $\tau$ which is time till stop, and $k$ which is essentially rate of acceleration.


1

First, you can use your velocity expression to determine the time that it takes to reach a certain final velocity, $v_f$. $$v_f=v_0e^{-k\Delta t}$$ Given that $\Delta t$ is the amount of time spent decelerating. If you solve this for $\Delta t$ you'll find: $$\Delta t=-{1 \over k}\ln\left({{v_f \over v_0}}\right)$$ Next, you can determine an expression ...



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