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18

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 $$ x=-\frac{gt^2}{2}+v_0t $$ $$ ...


6

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 ...


4

When you are lifting an object, you are exerting a force that balances the force of gravity on the object. By $$ F = m g$$ where g is the acceleration due to gravity, you see that a greater mass causes a greater gravitational force that has to be balanced by the force you apply to the object by holding it or lifting it at a constant velocity. Using the more ...


4

Let's consider a circular orbit in Schwarzschild coordinates, taken to be in the equatorial plane for simplicity. The test particle's position has components $x^\mu = (t, r, \pi/2, \phi)$, where $t$ and $\phi$ vary linearly with time/proper time and $r$ is constant. Then the $4$-velocity is $u^\mu = (\dot{t}, 0, 0, \dot{\phi})$, where dots denote derivatives ...


2

If you start with the bag stationary at 300m then drop it the bag is going to fall straight down, and its maximum height would indeed just the 300m point it started from. However you're not starting with the bag stationary. You're starting with the bag moving upwards at 13 m/s. So the bag is going to start at 300m then move up, come to a halt, then start ...


2

You can certainly create a pressure gradient. Depending on the acceleration, that gradient could be as large as you like and could lead to a very low pressure at the front, which might approach a vacuum. The equation is very simple: $$ \Delta P = - \rho g \Delta h $$ So for a $1m$ tube, filled with ambient air ($\rho = 1.2754\ kg/m^3$) and a $1g$ ...


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

The equation governing the motion of body at right is, $m_1g-T=m_1a$ [since $F_N=ma$] ($Equation_1$) The equation governing the motion of body at left is, $T-m_2g=m_2a$ ($Equation_2$) Adding both equations we get, $a=\frac{m_1-m_2}{m_1+m_2}\times{g}$ The net acceleration produced in string at right is due to net force acting on the body at right. This ...


1

I'm not exactly sure what you are asking. If you're wondering about how we know that bodies of different masses fall at the same rate if we ignore other factors like air resistance, then you might want to take a look at experiments like these. If you are interested in how we arrive at the conclusion that the acceleration is equal to gravity, we can ...


1

If you know the mass of the car, then you can use $$F=ma$$ to calculate the force given the acceleration or the acceleration given the force. It doesn't make much sense to use Newtons to measure acceleration, because $F$ Newtons is related to $a$ $\text{ms}^{-2}$ by $m$ kilograms. Your equation is only right if $m=1$ kg is understood.


1

I suspect your teacher glanced at it very quickly, and didn't realize you were using timesteps. When you work with timesteps you are doing a discrete approximation of the differential equations which describe the relationships, namely: $\frac{dv}{dt} = a$ and $\frac{dx}{dt} = v$ which in the discrete form, and arranged to match your code, become: $\Delta ...


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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