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Suppose the friction between the nail and the wood is $F(x)$ where $x$ is the length of nail driven into the wood. As a first approximation we will assume $F$ varies linearly with $x$, so $F=\lambda x$. Then the energy required to drive the nail a distance $d$ into the wood is $\displaystyle E = \int_0^d F(x) \space dx = \int_0^d \lambda x \space dx = \frac ... 0 Velocity is a speed and a direction, or speed is the magnitude (value) of the velocity irrespective of its direction. In physics, they are distinct concepts with precise mathematical definitions. Velocity belongs to a class of objects called vectors, which is commonly called something "with a magnitude and direction", while speed is a scalar: ... 3 Speed is the magnitude of the velocity. It's a scalar quantity, which basically means that it's a "normal" number. Velocity is a vector, which basically means that it has a direction. So if one car is going north at 30 m/s and another is going south at 30 m/s, they have the same speed, but they have different velocities. There are some physical ... 0 Velocity is a vector. It has both an magnitude and a direction. Speed is a scalar. It does not refer to a direction. Speed is the magnitude of velocity. For example, a car travelling 60 mph in the Northbound lane of a highway has the same speed as a car travelling 60 mph in the Southbound lane. However, if the velocity of the first car is$\vec{v_N}$, and ... 0 Displacement of a body is dependent on the frame Suppose a ball is moving on the ground. In the frame of ball the displacement of ball is zero and in the frame of ground there is a some finite displacement Distance between two points or the shortest distance between two points is not dependent on frame even if those points are moving. Distance between two ... 3 If the object starts falling from rest at height$h$, then its velocity at time$t$later is $$\vec v=-gt\hat z$$ and its vertical position is $$z=h-\frac12gt^2$$ where$g$is the magnitude of the constant and uniform downward gravitational acceleration. This neglects air drag, and assumes that the gravitational field is the same everywhere. These equations ... 1 The answer to your question on how we deal with them is that we don't deal with the, they are unphysical since they cannot arise in any actual physical scenario. As other answers have noted, however, they may be useful to apply in intervals in which the position is defined. It is also worth noting which force those position functions would they arise from ... 8 If$x(t)$isn't differentiable at some$t_0$, then$v(t_0)\equiv x'(t_0)$isn't defined. That's what it means for a function not to be differentiable. If you argue that the instantaneous velocity of a particle should always be well-defined, then you're saying that the position function should be everywhere differentiable. Similarly, if you argue that the ... 2 For the$x(t)=|t|$case, we, physicists, say that "there's no problem!" There is no problem differentiating$x(t)=|t|$. It's just$v(t)=\dot{x}(t)=-1+2\mathrm{H}(t)$, where$\mathrm{H}(t)$is the Heaviside step function! Even more, the acceleration is also well-defined:$a(t)=\ddot{x}(t)= 2\delta(t)$. No problem at all! (Sorry for mathematicians ... 0 Let's look at the system you are describing from a very rigorous perspective: we have a dimensionless particle of mass$m$moving along the real line with velocity$\dot x(t)$. All of a sudden, a force$\vec F(t)$acts on the particle for an interval$\delta t$, creating an angle$\alpha$with the real line. For pure mathematical convenience, we describe the ... 1 The acceleration can be broken down into two components, one parallel to the path and one orthogonal to it. The orthogonal component causes the path to bend, the parallel component causes the particle to accelerate (forwards or backwards according to the component's direction). The curvature of the bend depends on the speed of the particle; the faster it ... 2 The time taken to cross the river is$\displaystyle t = \frac d {v \cos(x)}$In this time, the downstream drift$D$will be$D = (u - v \sin(x)) t \\ \displaystyle \Rightarrow D = \frac d v \left( \frac u {\cos(x)} - v \tan(x) \right) \\ \displaystyle \Rightarrow \frac {dD}{dx} = \frac d v \left( \frac {u \sin(x)}{\cos^2(x)} - \frac v {\cos^2(x)} \right)$... 1 First break the$V$into two components The drift$Dcan be written as $$[u-vsin(x)]t=D$$ to find minimum D, we differentiate and then equate to 0 $$\frac{dD}{dt}=0$$ $$ut-vsin(x)t=D$$ differentiating both sides, $$[t\frac{d(u)}{dt}+u\frac{d(t)}{dt}]-[\frac{d(vsin(x))}{dt}+vsin(x)\frac{d(t)}{dt}]=0$$ [t\frac{d(u)}{dt}+u\frac{d(t)}{dt}]-[(v\frac{d(sin(x))}... 2 By looking at the graph, the displacement does not change. If this graph described a car, then you can imagine a car parked in your garage, and ten seconds later, it is still parked in your garage at that exact same position. The car did not move whatsoever. Its velocity would be zero. As for the definition for velocity, it is not \dfrac{d}{t}. Suppose \... 3 I'm not sure it's better, but you can write the time rate of change of the magnitude of the velocity squared as $$\frac{d}{dt} v^2 = \frac{d}{dt} \vec v \cdot \vec v = 2 \dot {\vec v} \cdot \vec v =2\vec a \cdot \vec v \,.$$ If \vec a is perpendicular to \vec v the change in v^{2} (and thus the change in |\vec{v}|=\sqrt{v^{... 0 Your sister defined velocity the wronged way. For starters, you can start by correctly defining the definition velocity. See here. 0 You could say R Represents displacement and the tip of the vector is the body whose motion we analyze as R does not change with time the tip of the vector stays at the same place as time progresses which means its velocity is zero. If she did not learn vectors yet you could still give the same explanation just without a coordinate system. 2 As shown in Figure-01 \begin{align} \mathbf r & =\mathbf r_{\rm o}+\boldsymbol v_{\rm o}\,t+\tfrac12\mathbf a\,t^2 \tag{01a}\label{01a}\\ \boldsymbol v & =\boldsymbol v_{\rm o}+\mathbf a\,t \tag{01b}\label{01b}\\ \vert\boldsymbol v\vert^2 & =\vert\boldsymbol v_{\rm o}\vert^2+2\mathbf a\boldsymbol \cdot\left(\mathbf r-\mathbf r_{\rm o}\right) \tag{... 2 Though the above generalisations are correct, what we do is that we resolute (or break) any given motion along mutually perpendicular axes (namely x, y and z axes) and then apply these formula separately along each of these axes as:v_x=v_{0x}+a_{0x}tx=v_{0x}t+\frac{1}{2}a_{0x}t^2v_x^2=v_{0x}^2+2a_{0x}x$$along x-axes. Similarly,$$v_y=v_{0y}+a_{... 2 The question is asking for the normal force at the "top" of the bridge, where the gravitational force and the normal force are both aligned in a straight line. Since the bridge is circular, there has to be a centripetal force that causes the car to travel in a circle, and that force is provided by the force of gravity. This means that the normal ... 0 Short answer: if the object is a point mass, then there will be no rotation. If the object is large, then it will rotate about its center of mass. Long answer: I will assume no friction to simplify the calculations. Consider the following setup, where the blue square of massM$slides down a circular path with center$O$: Suppose the moment of inertia of ... 0 What is wrong is assuming that dv/dx is finite when v=0. Try this for motion with uniform acceleration. For example, for object thrown upwards with some initial velocity. At the top turning point the expression fir dv/dx tends to infinity. 5 The answer is that in the real world very few things move with a set velocity. To model aspects of the world- which is what physics and engineering are mainly about- you need to account for the fact that velocities are always changing. There are countless circumstances in which velocities are not constant- objects falling under gravity, circular motion, ... 4 As you know acceleration is the rate of change of velocity. For a complete layman , who is not even concerned about Newton's laws, but just needs to describe some daily life events, one area where acceleration is helpful is in describing how fast an object would fall to the ground from a given height. Since, it is not falling at a constant velocity but at a ... 0 If we have a hypothetical ball and a hypothetical ground that are not compressed and no energy is wasted, this object acts like light hitting a mirror when it hits the ground. The acceleration changes at the moment of impact with the ground are such that in a moment the constant amount of acceleration becomes negative. 0 The path of a planet around a star is indeed an ellipse with the star on one of the focal points. For an ellipse with semi-major axis$a$, and semi-minor axis$b$the focus point is$f=\sqrt{a^2-b^2}$distance from the center of the ellipse. With polar coordinates$(r \cos \varphi, r \sin \varphi)$from the focal point the ellipse tracs the following curve $$... 0 Unfortunately the first sentence is incorrect, leading to problems using formulas from kinematics (which is the study of motion without its cause). The Newton is the force needed to accelerate 1 kilogram at 1 meter/second squared. To put it another way, when the mass is constant, Newton's second law is F=m a so with mass in kg and acceleration in m/s^2 the ... 1 the equation of ellipse where the axes at (0,0) is$$x=r(\varphi)\,\cos(\varphi)\\ y=r(\varphi)\,\sin(\varphi)$$where$$r(\varphi)=\frac{a\,\sqrt{1-e^2}}{\sqrt{1-e^2\,\cos^2(\varphi)}}$$and$$e=\sqrt{1-\frac{b^2}{a^2}}$$a,b are the half-axis of the ellipse The velocity$$\vec v=v_r\,\vec{e}_r+v_\varphi\,\vec{e}_\varphi$$with$$v_r=\frac{dr}{dt}=\frac{dr}... 3$\Delta t=\Delta x/v$. So in your question$\Delta t =\infty/\infty$which is a standard indeterminate form. It could be 0, it could be infinite, or it could be any finite value. You have to use some other information to find it, usually by taking a limit. 1 IF something could travel with infinite speed, then instead of travelling between A and B, it would be at every point in between. In fact, it would be at every possible location simultaneously. But whether or not it could "traverse infinity" is a malformed question. There is no infinite distance. 0 This would more suitably be a comment, but do you mean a nice equation that describes ellipses? Perhaps you can try polar coordinates - they might be easier to work with. 0 There are specific rules for when one can use quantum mechanics and one can use classical mechanics equations even for elementary particles. It all depends on whether in the dynamical variables of the problem the value of h, Planck's constant, which controls the Heisenberg uncertainty relations, can be assumed to be zero, or the calculation involves ... 1 Velocity is the first derivative of displacement with respect to time. So, on differentiating, you get$v$, which should be a linear curve for this case, as your velocity. If you integrate the area under this curve, you will get back to your displacement. To get the acceleration of the particle, you need to differentiate it again, for it is the second ... 2 Let the resp.$g$s be: $$g_m=\frac{g_e}{6}$$ $$v_{0,y}=v_0\sin\theta$$ where$v_0$is the initial (launch) velocity and$\theta$the angle with the horizontal. The vertical velocity component is: $$v_{t,y}=v_0\sin\theta-gt$$ which reaches$0$when: $$v_0\sin\theta-gt=0\Rightarrow t=\frac{v_0\sin\theta}{g}$$ The total time airborne is thus: $$2t=\frac{2v_0\... 0 To take an extreme example, keep increasing the speed of the block until it reaches orbital velocity. At that point the block is in orbit (even though it's right next to the surface, even touching the surface -- remember, you postulated no friction and no air resistance). For a body in orbit the normal force is 0. So the two extremes are normal force equal ... 2 It's not about what causes the gravitational force to be greater. It's that the normal force will be less (these two statements are the same mathematically, but colloquially they change the focus). Let's look at a simpler example. Take a box and put it on the floor. We can say the normal force acting on the box is equal to its weight. Now let's apply an ... 1 As pointed out in a comment in another answer to this question: 'oscillation' is not a rigorously defined term. But there is enough convention that the word is a very usable word. From a physics point of view: There are two conditions that are necessary in order to have oscillation at all. -There must be a form of restoring effect present, such that there is ... 2 The motion of pendulum described in the OP is often referred to as "damped oscillations". In this sense it is fair to say that "oscillatory" is used in broader sense than "periodic", which implies exact repetition. 1 In your notation you are mixing up the integration of v. For the state space representation one would write:$$\begin{pmatrix} \dot h \\ \dot v \end{pmatrix} = \begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix} \begin{pmatrix} h \\ v \end{pmatrix} + \begin{pmatrix} 0 \\ \frac{1}{m} \end{pmatrix} F_\text{ext}.$$This expresses the equations of motion of a ... 0 You have equations for motion with constant acceleration, specifically one which gives position given time, initial position, initial velocity, and acceleration. In this problem, you have two moving objects: the spaceship and the shock wave. These two objects will each have their own different equations of motion. The ship is doomed if there is any time when ... 0 No, using "rate" and "with respect to time" together is correct and encouraged. A "rate" is a change in one quantity with respect to another quantity. You can have quantities that change over time, like the concentration of products in a chemical reaction changing over time. But you can also have quantities that change over ... 1 It doesn't make sense that you have a value for the centripetal acceleration a while \omega tends to zero. Those two follow each other. Physically, if you have a centripetal acceleration a causing you to move around a curve, then you also must have a nonzero angular speed \omega around that curve. Otherwise you wouldn't be moving around the curve and ... 0 When any object falls through a fluid like air it experiences viscous force.This force produces a deceleration that can be expressed as :$$ \vec F = m \vec a\vec a = \frac{\vec F }{m}$$So magnitude of deceleration due to Viscous force is :$$|a| =\frac {|F_{viscous}|}{m}$$The viscous force depends only on mass and surface area so the viscous forces ... 1 The cotton would have greater air resistance so it would fall slower. However, neglecting air resistance, all masses in the same gravitational field fall at the same rate. Near Earth's surface objects fall at about 9.8 meters per second squared, neglecting air resistance. Here is an interesting clip of a feather and a hammer being dropped on the Moon where ... 0 It depends on air resistance. If you have two objects with different masses then the object with greater mass falls faster. But here , since the masses are equal , now the fall rate depends on- surface area and density of the fluid. If an object has more surface area, it has more volume which means it has less density. So more the surface area, more the drag ... 0 Notice that in equation 2, as the angular velocity goes to zero, the centripetal acceleration also goes to zero (as does the tangential velocity). Also in circular motion, the linear speed is the magnitude of the tangential velocity (unless the radius is also changing). The components, v_x and v_y, both change as a function of time (unless you are ... 1 I can't access freely to the PDF document you cite in your question, so I cannot give a complete answer to your question. However, here are some preliminary remarks: The relationships you mention in your question are limited to a uniform circular motion (on a straight line, you have a finite and w = 0). In theory, you don't need \omega , you can ... 0 Note that in both cases the force acts for the same time. Per the work energy theorem: the net work done on an object equals its change in kinetic energy, or for a constant force acting through a distance d$$W_{net}=F_{net}d=\frac{1}{2}mv_{f}^{2}-\frac{1}{2}mv_{i}^{2}$$So if you want to compare the work done on the two objects, you must base it on an ... 0 Yes, the work will be larger if the object is already moving in the direction of the force. The mechanical power (work per time) is given as:$$P = \frac{\Delta W}{\Delta t} = \frac{Fd}{\Delta t} = F v$$where$v = \frac{d}{\Delta t}\$ is the velocity. So, yes, it costs extra energy to apply the same force to an already moving object than if it was at rest! ...