# Tag Info

20

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 $$x=-\frac{gt^2}{2}+v_0t$$ $$... 10 No they aren't. Suppose we have some velocity v(t). The differential with respect to time is just the acceleration:$$ \frac{d}{dt}v(t) = a(t) $$Now differentiate it with respect to distance s:$$ \frac{d}{ds} v(t) = \frac{dt}{ds}\frac{d}{dt} v(t) = \frac{dt}{ds} a(t) = \frac{a(t)}{v(t)} $$7 No, it doesn't imply that a = 0. If, at some value t = t_0, the acceleration is non-zero while the velocity is zero, the position function is either a minimum or maximum. That is, x(t) is stationary there:$$x(t_0 + dt) = x(t_0)$$which means that at t = t_0$$\frac{dx}{d\dot x} = \frac{dx}{dv} = 0$$thus \frac{dv}{dx} is undefined at t = ... 5 You can apply chain rule if v is differentiable wrt x and x is differentiable wrt t. I think there are no other conditions,as this post on MathSE seems to say, http://math.stackexchange.com/questions/688152/necessary-conditions-for-the-chain-rule-of-differentiation-to-be-valid#= and this condition is not always available. When v=0,make sure ... 4 If you bring an object from infinity to a distance R then the potential energy change is:$$ \Delta U = -\frac{GMm}{R} $$Assuming your object starts at rest, the potential energy change is equal to the change in kinetic energy, so we have:$$ \frac{GMm}{R} = \tfrac{1}{2}mv^2 $$so:$$ v^2 = \frac{2GM}{R} $$You want v \ge c, so:$$ \frac{2GM}{R} ...

3

The initial velocity and acceleration here are in opposite directions. The magnitude of velocity (represented by $S=|\vec v|$) decreases upto a certain instant. (i.e. where $\vec v=0$). Edit: Also, consider these graphs. ($t^.$ being the time where $v=0$) Note how the velocity increases but the magnitude of it (in the $S$, $t$ graph) decreases till $t^.$. ...

3

She asked us if the body was accelerating or slowing down Acceleration is defined as the time rate of change of velocity and, in this example, the acceleration is constant and positive. So, the full answer is: the velocity of the body is always increasing while the speed is decreasing for $t<1$ and increasing for $t>1$. In this plot, the ...

2

I think the answer is that we simply don't know. The Milky Way is one of the hardest galaxies to observe because we're embedded in it, and so are lots of opaque dust clouds. The good news is that assuming the Gaia mission delivers we'll soon know a lot more.

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

Let's say car and bike be at rest at $1pm$ so, $v_c=0$ and $v_b=0$. Calculations for motion of car: Since car is moving with constant acceleration, At 1:00:00pm, $v_c=0m/s$, $S_c=0m$ At 1:00:01pm, $v_c=4m/s$, $S_c=4m$ At 1:00:02pm, $v_c=8m/s$, $S_c=12m$ At 1:00:03pm, $v_c=12m/s$, $S_c=24m$ At 1:00:04pm, $v_c=16m/s$, $S_c=40m$ Calculations for motion ...

1

The simplest way to look at this is to consider separately the horizontal and vertical velocity/position. For a projectile launched at angle $\theta$ and velocity $v$, the components are: Horizontal velocity $$v_h = v\cos\theta$$ Vertical velocity $$v_v = v\sin\theta$$ The position at time $t$ is then given by $$(x, y) = (v_h\cdot t, v_v \cdot t - ... 1 The symbol s_{NN} is in OP's context of RHIC the Mandelstam s-variable in a Nucleus+Nucleus collision. The s-variable is also known as the square of the center-of-mass energy. 1 You don't have to use the standard equations of motion, you can just deduce them from Newton's laws: You have a force \mathbf{F}=m\times -9.8 \: \mathbf{\hat{e}_y} acting on the arrow. Newton's second law is \mathbf{F}=m\frac{d\mathbf{v}}{dt}. Integrating for the x and y coordinates yield:$$\frac{dv_y}{dt} = -9.8\int_{v_{yi}}^{v_{y}}dv_y = ...

1

When you take the average of two quantities, you need to consider the "weighting". In this case, the time spent at each of the velocities matters, and becomes this "weight". In general, when you have a weighted average you multiply each value by its weight, and divide by the sum of the weights. When all the weights are $1$, that reduces to the familiar ...

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: $\Delta ... 1 The answer is negative. The gravitational acceleration is the same for all the atoms. Thus, the internal structure of the body is not affected. Using the terminology in your question, the average kinetic energy of atoms inside this object doesn't change. This is the reason for which given a body left free in the gravitational field, we usually consider the ... 1 Difficult to give quantitative answers without a more definite hypothesis of the signature you are interested in. There has been lots of work done that studies the dynamics of stars with respect to Galactic disk. Possibly the most celebrated is the work of Kuijken & Gilmore (1989) who studies large samples of K-stars towards the south Galactic pole. The ... 1 Velocity has units of m/s and acceleration m/s² so the root of acceleration would be √m/s. This unit does not make any sense. If you just want the numerical value, then v=a·t → √a=a·t then √a/a=t. 1 Your calculations are correct. They differ from your model (which uses ABS braking) however, because they don't take into account the duty cycle of the braking. If this is added to your calculations, then the two results should be similar. 1 You need to solve system like following:$\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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