When I practise problems, I come across ideal situations like constant velocities, constant accelerations, etc. But in real situations, objects usually don't magically gain momentum or acquire acceleration. Correct me if I'm wrong, but velocities often fluctuate due to a rate of change, i.e., acceleration, and accelerations must also be fluctuating as forces are not built up instantaneously. Therefore, there is a rate of change for each of these parameters. So, how many times do I have to repeat this process, i.e., differentiate x(t), for real objects(not particles) to reach a constant rate of change?
59395
This might be a very vague and unclear question, but let me explain. When an object at rest moves, or moves from point $A$ to point $B$, we know the object must have had some velocity (1st derivative of position) during that trip. It's also true that the object had to have accelerated to gain that velocity (2nd derivative)... if we extend this argument, is it logical to say that the object must have experienced jerk (3rd deriv.) to gain that acceleration? And so on and so forth... 3rd, 4th, 5th position derivatives? Or is my logic flawed somewhere?
99658
I remember the first time I saw this formula, $F = ma$, and it seemed perfectly correct and natural to me (in the sense that it is very straight-forward and logical, not that it is easy by any means). But then, I wondered, what if an object is moving with a constant speed? Maybe a car. If it travels with constant speed and hits me, it still hurts. So it's force must not be $0$.  One more thing related to this problem. If the same car is pulling another car behind it, if the car goes with constant speed, it implies that the first car doesn't exert any force to the second car at that moment, since it is not technically pulling it anymore (espacially if we don't take into consideration the friction).  So my question would be the following: How do these two phenomenons describe the same law of motion?   PS: If gravity wasn't behaving as some kind of acceleration pulling me constantly to the Earth's core, then if I jumped upwards, would I start slowly getting away from Earth?
107539
I recently came across a question asking the following:     If a motor is switched on, it quickly reaches a top speed. Why does it not just go faster and faster and faster?   I thought it might be related to the fact that the electrical power is constant, so I tried approaching with $P=\tau \cdot \omega$. However I still do not have any clue to reaching the conclusion.  Is it somehow related to the energy loss (due to friction)? Like the terminal speed of a parachute or something? But I was thinking that friction should be constant ($f=\mu \cdot N$) while air drag is related to the speed, so this may not be the answer.  Thank you.
87130
So in order for two things $A$ and $B$ to move apart, for example, relative to each other, $B$ can be set into motion away from $A$. This means that we have to increase $B$'s velocity and therefore the acceleration has to be positive. If the acceleration has to be positive and $A$ and $B$ were stationary before, that means that $B$'s acceleration has to increase from zero and therefore the third derivative of motion with respect to time has to be positive. The third derivative has to increase from zero and etc. etc. Is this just another way to state Zeno's paradox (and thus a dumb question) or does motion really involve an increase of the magnitude of infinitely many derivatives of velocity?
64188
A while back in my Dynamics &amp; Relativity lectures my lecturer mentioned that an object need not be accelerating relative to anything - he said it makes sense for an object to just be accelerating. Now, to me (and to my supervisor for this course), this sounds a little weird. An object's velocity is relative to the frame you're observing it in/from (right?), so where does this 'relativeness' go when we differentiate?   I am pretty sure that I'm just confused here or that I've somehow misheard/misunderstood the lecturer, so can someone please explain this to me.
61042
Last year, I took physics and over the summer I have started to wonder about why many phenomena work the way they do (such as why is $\mathrm{KE} =\frac{1}{2}mv^2$, etc.). I have found answers to all of my questions, except for this one:  Why are we allowed to use trigonometry to get vector components? I understand that if we draw a right triangle, then we can express the adjacent side with $\text{length of hypotenus} \times \cos(\theta)$. But why does this work with velocity and with forces? I have tried to reason through this many times, but I am not able to figure out why we are able to do this. Any ideas?   BTW: My attempts for reasoning through velocity were as follows: If we have a ball moving in two dimensions, and we are given that the ball has $\sqrt{3}$ the speed in the x-direction than the y-direction, then we can conclude that the ball will move in a 30-degree angle. I am not sure how to apply this to forces or acceleration though. For example, why is it that in the case of a ball spinning horizontally while attached to a string, that the y component of the strings tension gets smaller as the centripetal force increases? I can see how we can prove this by using trigonometry, but why does this trig work in the first place?
54571
Why does the object not go inward, into the circle if the acceleration is inward? I think its because the velocity to outward? So they sort of cancel each other out? But if the speed is kept constant and the acceleration is inward, won't the object eventially move inward?   Also, what keeps the direction of acceleration always point inward?
112402
Zeno's arrow paradox:     Zeno states that for motion to occur, an object must change the position which it occupies. He gives an example of an arrow in flight. He states that in any one (duration-less) instant of time, the arrow is neither moving to where it is, nor to where it is not. It cannot move to where it is not, because no time elapses for it to move there; it cannot move to where it is, because it is already there. In other words, at every instant of time there is no motion occurring. If everything is motionless at every instant, and time is entirely composed of instants, then motion is impossible.   The paradox says that "in any one (duration-less) instant of time, the arrow is neither moving to where it is, nor to where it is not". But it must be correct if passed time's length is 0 because, for example, if we take a bullet with constant velocity and take an instant time (dt~0) in its movement, obviously we will see that the ratio between the displacement with passed time will be equal to the velocity of the bullet. But if the passed time's length is 0, we no longer say we divide time into small part because it doesn't matter how much times you sum these instant times, you never reach the first time length that you divided.  But if my explanation was correct, it wouldn't be a paradox, so what is the problem with my explanation?
3247
From what I understand, the  states that there is a fundamental natural limit to how accurately we can measure velocity and momentum at the same time. It's not a limit on equipment but just a natural phenomenon.  However, isn't this just an observational limit? There is a definite velocity and momentum, we just don't know it. As in, we can only know so much about the universe, but the universe still has definite characteristics.  Considering this, how do a wide range of quantum mechanical phenomena work? For example, quantum tunneling - its based on the fact that the position of the object is indefinite. But the position is definite, we just don't know it definitely. Or the famous light slot experiment? The creation of more light slots due to uncertainty of the photon's positions?  What I am basically asking is why is a limit on the observer, affecting the phenomenon he is observing? Isn't that equivalent to saying  because we haven't seen Star X, it doesn't exist? It's limiting the definition of the universe to the limits of our observation!
30152
I was given a problem for homework where we needed to calculate the time for a falling object to reach a certain velocity when accounting for drag force. I did it by setting up acceleration as a function of velocity and integrating (it was a differential equation).   However, this is an introductory physics course, with no knowledge of calculus required going in. We haven't even done derivatives yet, strictly speaking. I was fortunate enough to have taken calculus before, so I was able to recognize and solve the differential equation.   When I asked my classmates how they did it, they said they messed around with numbers until they got something that worked (it was online with no points deducted for wrong answers). For most of them, they just divided the terminal velocity by acceleration due to gravity, which makes no sense, since we weren't even asked for time taken to reach terminal velocity, but 63% of it. That method just happened to round to the same number as the correct one.   My question is, is there some way to find this value using elementary physics, or did my professor give us an unfair problem? The TAs weren't any help and I have class during her office hours.   The question itself is as follows:     The terminal velocity of a 4×10$^{-5}$ kg raindrop is about 9 m/s . Assuming a drag force $F_D=−bv$, determine the time required for such a drop, starting from rest, to reach 63% of terminal velocity.
2678
Note: For the purposes of my question, when I refer to free fall assume it takes place in a vacuum.   From my (admittedly weak) understanding of the equivalence principle, falling in a gravitational field is physically indistinguishable from floating in interstellar space. This would make sense to me if gravity simply caused an object to move at a constant velocity. Moving at a constant speed, or floating in space, are just two different ways of describing an inertial frame, and are fundamentally no different. But free falling in a gravitational field means accelerating continuously, and doesn't an accelerating body experience a force? Then isn't free falling fundamentally different from floating in space?
62225
On , it is written that      So, we see things falling with an acceleration which we call the acceleration of gravity,and thinking that we live in a straight line , uniformly moving or stationary inertial frame, we attribute that acceleration to a force, the force of gravity. Whereas in reality,objects falling towards the Earth are moving along geodesic paths with no acceleration and according to modified version of law of inertia, have no force acting on them. They fall simply because the curved space-time near the Earth ...   Now, why do the objects falling towards the Earth move along the geodesic paths with no acceleration? That means the objects don't have any force acting on them, but why? A body in a free-fall moves with acceleration $g$, so, why is it written like that? Why does the author use Law of inertia on freely falling body?? Law of Inertia can only be applied when no external force acts on the body. So, is the freely-falling body accelerates under force of gravity or moves uniformly while moving through geodesic paths as quoted by the author?
82475
Motivating the non-relativistic momentum $$\mathbf{p} = m\mathbf{v}$$ is quite easy: it is meant to represent the quantity of motion of the particle, and since the mass is one measure of quantity of matter it should be proportional to mass (how much thing is moving) and should be proportional to velocity (how fast and to where it is moving). Now, in Special Relativity the momentum changes. The new quantity of motion becomes $$\mathbf{p} = \dfrac{m\mathbf{v}}{\sqrt{1-\dfrac{v^2}{c^2}}}$$ Or, using $\gamma$ the Lorentz factor $$\mathbf{p} = \gamma(v) m\mathbf{v}$$ where I write $\gamma(v)$ to indicate that the velocity is that of the particle relative to the frame in which the movement is being observed. The need for this new momentum is because the old one fails to be conserved and because using the old one in Newton's second law leads to a law which is not invariant under Lorentz transformations. So the need for a new momentum is perfectly well motivated. What I would like to know is how can one motivate that the correct choice for $\mathbf{p}$ is the $\gamma(v)m\mathbf{v}$. There are some arguments using the mass: considering a colision, requiring momentum to be conserved, transform the velocity and then find how mass should transform. Although this work, it doesn't seem natural, and it is derived in one particular example. On my book there's even something that Einstein wote saying that he didn't think it was a good idea to try transforming the mass from $m$ to $M = \gamma(v)m$, that it was better to simply keep $\gamma$ on the new momentum without trying to combine it with the mass. So I would like to know: without resorting to arguments based on transformation of the mass, how can one motivate the new form of momentum that works for special relativity?
41547
I'm curious as to whether or not the velocity of the center of mass of a system comprised of two spheres can change after the two spheres collide. Looking at the equation for the velocity of the center of mass for a system of particles:  $$V_\text{CM} = \frac{m_1v_1 + m_2v_2 + \cdots + m_nv_n}{m_1 + m_2 + \cdots + m_n}$$  It looks like, if after the collision, one of the particles changes direction, and the negative terms outweigh the positive terms in the numerator, the velocity of the center of mass would change direction (and possibly magnitude). However, I can't think of any examples where this would happen in elastic or inelastic collisions. I'm not even sure if its speed can change. It would make sense that it would if kinetic energy (velocity) was lost in an inelastic collision, however, I can't come up with any conditions to make this happen.  I could really use some insight.
92255
Referring to unidimensional motion, it is obvious that it doesn't always make sense to write the speed as a function of position. Seems to me that this is a necessary condition to derive formulas like:  $$v^2=v_0 ^2 +2\int_{x_0}^{x}a\cdot dx$$  In fact, in the first step of the demonstration (the one I saw, but I think that this step is crucial) it's required to write $a=dv/dt=(dv/dx)(dx/dt)$, that doesn't make sense if $v$ isn't a function of $x$.   When can one rigorously write $v=v(x)$?
53105
The force that propels a bullet should at least be similar in intensity to the force experienced as recoil and the force exerted on the target. But these forces seem really low (the equivalent of a shove).  If all it takes to make a small object achieve bullet speed is the equivalent of a shove, why can't humans throw objects at these speeds?
10214
I'm having trouble understanding the motivation between wanting to transition between the two descriptions.  I figure that switching from  is useful if we want to recover a particle's trajectory maybe? Is there any other reason to do so? and then what are we measuring by switching from Lagrangian to Eulerian?