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I hope I have found how to evaluate the relative acceleration. If we place the origin in the centre of Earth according to a geocentric model, the curve described by Venus is $$\mathbf{r}_{VE}(t)=R_E\begin{pmatrix}\cos(\omega_{SE}t+\varphi_1)\\\sin(\omega_{SE}t+\varphi_1)\end{pmatrix} + ...


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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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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.


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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.


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When you're using the equation F=ma, the F is ALWAYS the total/resultant/net/unbalanced force, NOT one of the individual forces. It describes the effect (the acceleration) that happens due to the cause (the total force on an object). Here you happen to be right because (at least horizontally) there is only one force, the frictional force, so you should get a ...


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So I'm assuming you're saying that the work done on the car in distance $d$ has to be equal to its kinetic energy $\frac{1}{2} m v^2$. Then, using $W = F d$: $$ F d = \frac{1}{2} m v^2 \\ m a d = \frac{1}{2} m v^2 \\ d = \frac{v^2}{2 a} $$ So, yes, this equation is correct. Your relation between the two forces is also correct. Since mass drops out, the ...


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Let us denote $\ P$ the point corresponding to the end of the rod with no force applied, then we apply a force $\ \overrightarrow{F} $ on the other end of the rod, such that is parallel to the plane and perpendicular to the rod. Now, with respect to the center of mass, the only force whose torque is different from zero is $\ \overrightarrow{F}$: ...


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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 ...


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in your answer you ignored the initial velocity the bag will rise a little until its velocity is zero then it will fall again , it is like he found the total distance


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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 ...


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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 ...


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I can think of a light beam as a pulsating stream of water from a hose traveling at the speed of light. If there's a hole in the side of my space ship and the hose of streaming water is pointed directly, perpendicular to my ship's direction of travel, at the hole, then only a portion of the pulsated water will enter the hole. Now concerning the portion of ...


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Concepts Assume the projected area, $A$, of the book is about equal to the projected area of the paper as they both fall towards the earth. The fundamental principles are (1) Different gravitational force, $F_m=mg$, that acts on each object's mass, m (2) Similar opposing drag forces, $F_d=-(1/2)C_d A \rho v^2 $ which act on each object in air (3) ...


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The acceleration of any object due to gravity is $g = 9.8m/s$ and this constant does not depend on the mass of the object (or the speed of the object or anything else for that matter, as long as you're on/near earth). Pushing an object away from the earth is another story. While the acceleration of objects towards earth does not depend on their masses, their ...


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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 ...


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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 ...


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From what I understand, the question is pretty straight forward. For acceleration: $$L=\frac{a_1 \cdot t_1^2}{2} \implies t_1 = \sqrt\frac{2L}{a_1} \implies v=\sqrt{2a_1\cdot L}$$ For deceleration: $$a_2\cdot t_2+ v=0 \implies t_2=-\frac{\sqrt{2a_1\cdot L}}{a_2}$$ The minimum time is thus $t_{min}=t_1+t_2$.


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Yes, this is called the Equivalence Principle, and is foundational to the study of General Relativity. Mathematically, we have Newton's law of universal gravitation, which states that the gravitational force between two masses $M$ and $m$ is given by $$\mathbf{F}=-\frac{GMm}{r^2}\hat r.$$ where $G$ is Newton's constant. But we also have Newton's second law. ...


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(Classical Physics only) Any massive body has a property known as inertia, thus even a body floating in outer space would require some kind of force to be accelerated. Using Newtons second law, you would find $$\tag{NII} \sum \vec{F} = \frac{\mathrm{d}}{\mathrm{d}t}\vec{p},$$ which for constant mass and one-dimensional motion simplifies to $$\tag{NII'} ...


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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 $1g$ ...


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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 ...


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Your solution is correct at the level of your course. We don't see what else is in your code, so there might be another problem. I can only guess what he means by "acceleration / 2". Perhaps he didn't read your program carefully enough; he might have been expecting a solution involving $1/2 a t^2$, and when he didn't see it, moved on. Your solution ...


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$s = v \cdot t$ is only true if $v$ stays constant. If $v$ changes with time, the proper relation is $$ s(t) = \int_{t_0}^{t} v(t') \mathrm dt'$$ For $v = a \cdot t$ with a constant acceleration $a$ this becomes $$ s = \frac{1}{2} a t^2 $$ So instead of calculating the position from the velocity you could calculate it from acceleration directly.


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There is no limit in acceleration, but since acceleration is the rate of change of velocity per time, and the time dilatation of the falling particle relative to an outer observer goes to infinity at the event horizon (where the escape velocity would be the speed of light), the falling object never makes it through that border from the viewpoint of an outer ...


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yes you will be pushed up to the ceiling of elevator because you are accelerated inside the elevator due to gravity and the elevator itself is accelerated due to external faster acceleration, that makes you stuck on the elevator ceiling then they ceiling will push on you with normal force making you move with the elevator downward


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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.


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If the wheel is undergoing some sort of angular acceleration, then the magnitude of its velocity - its speed - would be expressed as a function of time, $v\equiv v(t)$. So what does this mean for the period of rotation? That means the period also becomes a function of time. However, since the period still represents the time required for the wheel to rotate ...


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You asked: Why do both vehicles experience the same magnitude of force? The larger principle at work is conservation of momentum. (Noether's Theorem, symmetry, and all that jazz.) During the small time frame of the collision we generally assume that there is no transfer of momentum into or out of the system of the car and truck. Changes of momentum ...


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Let's be clear on the terminology here, because taken literally, your question is a bit like asking whether angular momentum is green or non-green. Polar coordinates exist in space. Curves of constant $r$ are circular and curves of constant $\theta$ are radial, in the usual way. Taking their tangent vectors and normalizing them, we get ...


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Having pondered this for a while, I think the simplest answer must be that the formula $$\boldsymbol{a} = (\ddot{r} - r \dot{\theta}^2) \textbf{e}_r + (r \ddot{\theta} + 2\dot{r}\dot{\theta})\textbf{e}_\theta$$ can only be produced when we differentiate with respect to the inertial observer. It is only due to our differentiation of the terms in the inertial ...


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Yes speed will increase $9.8$ m/s every second provided that there is no other force acting as resistance to the fall. In reality air resistance depends on speed and the faster someone falls the more the resistance of the air around them. Eventually any object falling through the atmosphere will reach 'terminal velocity' the speed at which the force ...


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If I understand what you asked, yes that is the situation for which acceleration due to gravity is applied. However, there is a terminal velocity due to air resistance, and the Acceleration and resistance will reach equilibrium. The wikipedia page discusses skydiving first, and that is what the activity of purposefully jumping out of a plane is called.



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