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You can find the position (as a function of time) for the balloon with simple kinematics: $y_B=200-\dfrac{1}{2}gt_B^2$ (with $g=+9.8\:m/s^2$). After five seconds, the archer shoots the arrow upward, the position of which can also be represented with kinematics: $y_A=40t-\dfrac{1}{2}gt_A^2$ Now, they're on separate time scales, where $t_i=0$ corresponds ...

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There are some very useful elementary equations that describe basic motion with constant acceleration, and these are: $$v=u+at,$$ $$v^2=u^2+2as,$$ $$v=ut+\frac{1}{2}at^2,$$ where $u$ is initial speed, $v$ is final speed, $s$ is displacement (how far the object has moved) and $a$ is acceleration. You must now think about your problem to determine which ...

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How long does it take to stop? $v/a$ or roughly 2 seconds. How far does it go up in 2 seconds? ${1/2}at^2$ or 20m, roughly. You figure it out exactly.

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At the instant you release the arrow, it begins to lose velocity due to gravitational acceleration, which can be represented by a vector pointing opposite to the arrow's direction of flight. The arrow loses about 9.8 m/sec of its velocity every second (this is the magnitude of gravitational acceleration at the Earth's surface). Solve for the time it would ...

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You need to read the problem again. This is a typical problem that beginning physics students have: they don't read the problem enough. You should read a problem at least 3 times before you do any work. Read slowly. You will see that you're supposed to find the angle of the velocity vector, not the angle of a line between the starting and ending points.

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This sounds like trying to hit a fixed target at a different elevation when you know the horizontal and vertical distance to the target (2D projectile motion). Basically you have 2 equations you have to solve for simultaneously, the horizontal and the vertical distances. You have 3 unknowns – the time, the initial velocity of the projectile and the angle of ...

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In GR, the speed of light is not constant, it varies with the curvature of space--time. So the constancy of this universal speed depends on space--time's having constant curvature. Which it doesn't, but this is locally a useful approximation, and in order to address the OP's intention, we will from now on assume that the Universe is a space of constant ...

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I must object by saying, I could go about questioning everything, but one must rather understand how? and why? things are the way they are. One must understand the Maxwell Equations, the problems and issues that come with it, special relativity, how it solves the problems, how to derive the Lorentz factor and how it breaks down at the speed of light. There ...

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Solve for $F_b$ from the horizontal braking distance. Assume $F_b$ is constant, then during braking kinetic energy has been converted to friction work: $$F_b \Delta x = \frac12 mv^2$$ where $\Delta x=123\:\mathrm{ft}$ is the braking distance and $v=60.0\:\mathrm{miles/hour}$. I've not checked the rest of your work. You don't need to invoke friction ...

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I am giving the solutions of original task (to get the speed at point B). I am not sure if the questions are necessary to perform the task. If you are sure the path taken does not matter (and I will assume that per your statement). So, let us consider a straight line path. Vertical component of F overcomes gravity and causes vertical move. Only horizontal ...

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Using the first $2$ statements or conditions, apply the principle of conservation of energy and you will be able to calculate the resistive force offered by the target material. Next use the second condition and again apply the principle of conservation of energy to get the required answers, mainly the velocity of the bullet after penetrating the target.

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some context is missing here but it could be the difference between Eulerian and Lagragian quantities, i.e. the spatial derivatives vs the particles derivative. It is mostly used for continuous materials (e.g. fluids), but can extend to other cases (e.g. field or stream of objects). The spatial (i.e. Eulerian) velocity in a field is the one at a given ...

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I think that this is a very interesting problem which is conceptually difficult. You do not need to worry about the FBD for the truck. The box should be your main focus. Diagram 1 is the FBD as long as the box does not slide relative to the truck. With the aid of diagram 1 work out the maximum acceleration $a$ the box can have as a result of the static ...

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Just recall that kinetic friction always opposes the motion; if this were not true you could use friction to generate free energy. Static friction opposes the applied forces.

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If the shore is really far away, you can't tell the water is moving. Then it is immediately obvious that the boat must have spent the same length of time moving away from the crate as moving towards the crate - 1 hour each way. If we look at just the crate, we see that it moved 3 km in 2 hours (it was found "5 km downstream from the turnaround point" which ...

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Your question implicitly assumes that a any object will "boomerang" (you mean return to thrower I guess) only depending on the rotation speed. This is not the case. In very simplistic terms, the boomerang motion depends on its shape and material, besides on its speed. Actually even the speed is not as simple, because it needs a proper combination of ...

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The equation for your curve is given by: $$\frac{dv}{dt} = \frac{F(v)}{m}$$ where $F(v)$ is the net force on the car, which is a function of the velocity. we solve the equation by integrating to get: $$\int \frac{dv}{F(v)} = \frac{t}{m}$$ The trouble is that the net force $F(v)$ is a complicated function that doesn't generally have a simple analytic ...

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The magnitude of centripetal acceleration is $\frac{v^2}{r}$ instantaneously. It applies no matter the speed on your circular path. (Technically it's true for any curve, but $r$ would be changing on non-circular curves, making calculations more difficult.) The tangential acceleration is constant, so you can write a function for $v$. Then you have two ...

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This is not a proper derivation. At a fundamental level, there are at least three important points that are not taken into account by this approach: as you consider a second mass point, it is somewhat difficult to adjust (in a non-arbitrary way) the derivation to obtain the correct energy term related to the angular momentum and/or rigid body rotation ...

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If you take your final expression $$x(t) = \underbrace{\left(x_0 + \frac{b - a}{2}\tau^2\right)}_{x_0^*} + \underbrace{\left(v_0 + \tau(a - b)\right)}_{v_0^*} t + \frac{b}{2} t^2, \quad \text{with}\ t>\tau,$$ then $x_0^*$ and $v_0^*$ would be the position and velocity at $t=0$, however this is only meaningful if $\tau<0$ (however in that case $x_0$ ...

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Firstly it's worth noting that such a discontinuity can never be 100 % real. To go from acceleration $a$ to $b$ instantaneously ($\Delta t = 0$) would require an instantaneous change in the net force responsible for the accelerations and that isn't possible in the material world. Secondly, I think you are over-thinking your problem. Just write the ...

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This equation holds whenever there is constant acceleration. Here are 2 ways of deriving that equation, which I hope help you understand it. Energy conservation The change in kinetic energy must be equal to the work done on the particle. $$\frac{1}{2}m v_A^2 - \frac{1}{2}mv_B^2 = \int F\cdot dx$$ For a constant force and mass $\int F\cdot dx = F (x_A - ... 0 Draw a graph with time along the horizontal and velocity up the vertical. Let's start with an object in motion at constant velocity. Its motion on the graph will be represented by a horizontal line at some distance from the y=0 axis. After some period of time, it will have covered a distance equal to velocity x time. That distance will be represented on ... 0 Start with acceleration, which we assume to be a constant$g$. Also assuming 'up' is the positive spacial direction i.e.$g$is$(-)ve: $$a(t) = -g = -9.81 \,ms^{-2}$$ Integrate once to get velocity: $$\dot{a} =v(t) = \int_0^t -g dt = -gt +v_0$$ Integrate again to get the distance: $$\ddot a = x(t) = \int_0^t(-gt + v_0)\, dt = -\frac{1}{2}gt^2 ... 1 You don't need calculus to understand this and I think you are right to be trying to gain a deeper understanding than just memorizing some formulas. During that first second the body accelerates - it starts with 0 velocity and gains linearly giving 9.8 at the end of the first second, so at that point, it hasn't been moving at 9.8m/s for a second, it has ... 0 Intuitively, the body spends some time at every velocity between 0 and 9.8 m/s during the first second. From the formula distance = speed * time, if we call that time interval dt and add up all the contributions (using integral calculus) the answer is 4.9 m. 1 The concept you're after is the dot product between 2 vectors (your displacements). More specifically you want to use$$ \vec{a} \cdot \vec{b} = |a| |b| \cos(\theta) $$to find \theta. 1 To compliment John's answer I'll give you an example: the kinetic energy of a harmonic oscillator. First we need to determine the velocity v=\frac{dx}{dt}=\frac{d(Asen(\omega t +\phi_0))}{dt}=A\omega cos(\omega t+\phi_0). Because kinetic energy is \frac{1mv^2}{2} we substitute v KE=\frac{1}{2}mA^2\omega^2cos^2(\omega t+\phi_0) Because ... 3 If the velocity of a mass m at some moment of time is v, then the kinetic energy and momentum are:$$\begin{align} E &= \tfrac{1}{2}mv^2 \\ p &= mv \end{align}$$If the velocity is changing with time, i.e. it is a function of time v(t), then the kinetic energy and momentum will also be functions of time:$$\begin{align} E(t) &= ... 0 Kinetic energy's quadratic makes perfect sense if our reality is not actually first order in space, and is instead simply a measurement of the relative rate that an object is passing through time. The space of our existence then becomes the space of simultaneous time, at any given point in time, as it progresses. In this scenario, changing the kinetic ... 0 Think of a position axisx$starting at zero, positive to the right and negative to the left. It is like a number line. A positive velocity means that the value of$x$is increasing eg going from$x=+3$to$x=+5$or$x=-7$to$x=-4$or$x = -3$to$x=+4$. A negative velocity means that the value of$x$is decreasing eg going from$x=-3$to$x=-5$or$x=+7$... 0 well you have given answer in your own question! Velocity and acceleration are both vector quantities, meaning they have magnitude and 'direction'. The sign (+/-) will depend on the direction. To simplify, let me give you an easy example.. Case 1: An object is moving down from the top of a mountain. The acceleration (in this case 'g') will act in the ... 0 When you say "minimizing the danger of it breaking on the ground" I am assuming you mean you want to reduce the kinetic energy of the object when it hits the ground (Also I have assumed there is no air resistance in the problem). In order to do that the object must not have any horizontal or vertical velocity component at the moment of release wrt the ... 2 As you figured out, the horizontal and the vertical part of the motion are independent. If you throw the bottle upwards, it will go upwards for some time, then turn and fall back. When it reaches the height of your hand, it will have the same velocity as when you threw it, just the opposite direction (downwards) - so this doesnt help as well. Clearly, ... 0 The energy required to accelerate an object by a given velocity increment is linear in the initial velocity in the non-relativistic limit (where$E_k=\frac{1}{2}mv^2$applies). It is even more energy intensive for the relativistic case when the velocity of light (c) is approached. That is because the relativistic expression for kinetic energy is: ... 1 Your train is travelling east at a speed of 20 m/s, since you walk to the back of your train you are only travellint east at a speed of 18.6 m/s with respect to the ground. Your friend is travelling west at a speed of 28 m/s, or a speed of -28 m/s to the east. From your friends perspective ground is travelling east at a speed of 28 m/s and you have an ... 1 Given the acceleration$a = \sin \left( \frac{\pi t}{T}\right) $, by integration you get $$v(t) = \int_0^t a\,{\rm d}t = \frac{T}{2} \left(1-\cos\left(\frac{\pi t}{T}\right) \right)$$ $$s(t) = \int_0^t v\,{\rm d}t = \frac{T}{\pi^2} \left(\pi t - T \sin\left(\frac{\pi t}{T}\right) \right)$$ Since the last one cannot be inverted for$t(s)$, we can ... 1 If the masses were not accelerating, it would be the case that B exerts an upward force on A equal to $$F_{AB} = (0.4)\cdot(9.81) \mathrm N = 3.92 \mathrm N$$ in order to cancel the downward force of gravity. Since A is accelerating upward, it must be that B exerts a greater force equal to $$F_{BA} = (0.4) \cdot (9.81 + 0.5) = 4.12 \mathrm N$$ So ... 1 Newton's third law says the the force on A due to B is equal and opposite to the force on B due to A. This in turn means that the changes of momentum of A and B are the same in magnitude but opposite in direction. This is how the momentum becomes rearranged. B loses some momentum and A gained an equal amount. So when two atoms collide you can think of ... 0 The equation used to find the displacement in motion with constant acceleration in 1 dimension is -$ s = ut + \frac{a{t}^{2}}{2} $where u is the initial velocity , s is displacement and a is acceleration (constant). You can re-write the equation as -$ t^2 + \frac{2u}{a}t - \frac{2s}{a} = 0 $or$ (t + \frac{u}{a})^2 - \frac{2s}{a} - \frac{u^2}{a^2} ...

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You can rearrange using the quadratic formula to get: $$t = \frac{-u±\sqrt(u^2+2as)}{a}$$ Thanks AccidentalFourierTransform for the answer.

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From when I worked among missile engineers, accelerometers were used, along with gyroscopes (mechanical or laser). I don't know of 6th order differential equations. I do know of 3rd order, namely in the steering by swiveling the engine nozzle. Specifically, the engine nozzle angle is off-center by a certain amount, causing an angular acceleration (2nd ...

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The difference between an office chair's 5 wheels/supports and a regular chair's 4 legs is that the latter has all of its load going straight down. The legs only need to be strong enough not to shatter. In fact, a chair could easily get away with 3 legs but for the stability. In contrast the office chairs legs support load perpendicular to their ...

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It is because the ball was traveling with you when you threw it. Imagine the following question: I am in a car traveling at 5m/s holding a ball. Where will the ball be relative to me in 10 seconds? Answer in my hand. To be obtuse: ball has velocity 5m/s. After 10s it will have moved forward 10 x 5 meters = 50. I have velocity 5m/s. After 10s it will have ...

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Consolidating some of the points made in the answers to the question you linked, and comments: When constructing a chair, 4 legs is easy when you use traditional (wooden) construction - 90 degree angles, and easy to make stackable. A little bit harder than three legs because you have to make sure they are all the same length (or the chair will wobble). ...

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This is how I understand it. There is a series of definitions used in physics, and one used in engineering mostly. I'll describe the one used in physics first: In mechanics, we describe the motion of bodies, and the causes that effect them. This includes the special case where the "motion" is no motion, i.e. bodies are stationary. The description of the ...

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The first postulate of Special Relativity - which is also the Principle of Relativity is "Laws of Physics are invariant in all the inertial frames." Now consider the following scenario: Suppose there are two events which are happening simultaneously in frame $O$ and are spatially separated by distance $l$. From your Equation $1$, the spatial interval ...

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No need to force $\tilde\alpha_1 = \alpha_1$. After taking advantage of the fact that (1), (2) and (5), (6) must be inverses of each other, your transformation reads, in matrix form, $$\left(\begin{array}{c} x'\\ t' \end{array}\right) = \alpha_1(v)\left(\begin{array}{cc} 1 & -v\\-\frac{v}{c^2} & 1 \end{array}\right)\left(\begin{array}{c} x\\ t ... 4 You do not need the kinetic energy. Working with the total energy \gamma m c^2 produces the same result. Assuming both the total initial energy \bar E_0 = \gamma_0 m c^2 and the additional energy E_i are known, write \gamma_1 mc^2 = \frac{mc^2}{\sqrt{1-\beta_1^2}} = \bar E_0 +E_i for \beta_1 = \frac{v_1}{c}, then$$ \sqrt{1-\beta_1^2} = ...

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