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2

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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Perhaps this will elucidate your first point: $\frac{dy}{dx}=\tan \alpha$ is in fact the gradient of $y=f(x)$ at $(x,y)$. Of course this also means: $dy= \tan \alpha dx= \frac{dy}{dx}dx$ Divide both sides by $dt$: $\frac{dy}{dt}=\frac{dy}{dx}\frac{dx}{dt}$ $v_y=\frac{dy}{dx}v_x$ Also: $v^2=v_x^2+v_y^2$


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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 - ...


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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 ...


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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 ...


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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 ...


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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.


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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 $\theta$.


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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 ...


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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) &= ...


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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 ...


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Think of a position axis $x$ 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 $x=+7$ ...


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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 ...


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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 ...


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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, ...


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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: ...


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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 ...


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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 ...


0

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 ...


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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} = ...


0

A collision is an event in which two or more bodies exert forces on each other for a relatively short time. Collision is short-duration interaction between two bodies or more than two bodies simultaneously causing change in motion of bodies involved due to internal forces acted between them during this. Collisions involve forces (there is a change in ...


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$$adx=vdv$$ $$-\int_{R}^{x}G\frac{M}{R^3}xdx=\int_{0}^{v}vdv$$ $$v=\sqrt{2 (-\frac{GMx^2}{2R^3}+\frac{GM}{2R})}$$ $$v=\frac{dx}{dt}$$ $$t-0=2\int_{0}^{R}\frac{1}{\sqrt{2 (-\frac{GMx^2}{2R^3}+\frac{GM}{2R})}}dx$$ $$t=1267.5s$$


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This is not so bad - we simply parametrize the bullets' paths in 2D space as a function of time. Then, instead of having a line that is constant throughout time, the problem is more similar to the question you are asking - at any given time, you have a point (the bullet), and the path that the bullet takes is given by a line. But you want to find out if one ...


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There are often lots of ways to solve problems and finding the simplest way is something that comes with experience. The reason your teacher keeps setting you so many problems is to give you that experience. One of the common tricks is choosing a suitable reference frame to do the calculation. As watched by a stationary observer it's a somewhat complicated ...


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Your approach is correct; your ability to read data from a graph is suspect (the divisions are 2 m/s each). The initial velocity is -12 m/s, and at time t=9 s it is up to 18 m/s That should change your answer...


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Well the body will accelerate as long as you are applying a force on it. So if u apply a force for forever it will accelerate forever but there is an interesting thing that happens. Force applied for ever You would notice that if u apply a force for an infinite amount of time then by F=ma there should be acceleration for an infinite amount of time leading ...


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Hints: 1) Maximum speed occurs when the first derivative of velocity with respect to time is zero. 2)For this you will have to differentiate your initial function as the first derivative of velocity is acceleration. 3) For the remaining parts, I suggest that you find an explicit relationship between velocity and time by solving the Differential ...


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7 TeV is not that much kinetic energy, that has been covered by your question and previous answers. However, in the context of a proton, with a rest mass of $1.672×10^{−27}~\mathrm {kg}$ (very, very little mass), when a single proton has 7 TeV then it is travelling at a specific speed: $$E= mc^2$$ \begin{align}E& = E_0 + \text{KE}\\ \text{KE}&=E- ...



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