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You have to be careful to use the same sign conventions for $x$ and for $v$. If you're careful enough you can talk about $v$ as a vector with only one component, and take the sign seriously. Let's choose the positive-$x$ direction to be "towards the right". If the bicycle's speed is to the right, $v > 0$, then when the friend catches the bus she's to ...

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You must assume circular orbits, as stated in another answer. The ground speed will also be variable due to inclination, unless it's equatorial. An extreme example is an east-west orbit versus a west-east orbit. In one case, the ground velocity is added, in the other case, it's subtracted. But with the query limited to equatorial orbits, we can continue. ...

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Geostationary satellites travel in circular orbits about the equator. If you express the velocity as angular velocity, it will equal the angular velocity of Earth's rotation, about $7.3 \times 10^{-5} radians/second$. If you had another satellite in circular orbit about the equator, but not geostationary, you could subtract the angular velocity of the ...

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What I was doing wrong was putting manufacturer-provided motor torque directly into the formulas; actually the overall gear ratio must be taken into account, and it results from data taken around on internet that for electric vehicles it is =~8 (dimensionless). Hence the proper expression to use for $v_f tanh(\frac{F}{mv_f} t)$ is not $v_f ... 1 So, you are correct, that it is not quite as simple as that. Have you studied Special Relativity (SR) yet? One of the most fundamental ideas is the limiting speed of light. Nothing moves faster than c, and this means that velocities must add differently. Think of the reverse, if you were on the rocket-ship traveling at$0.5c$and shot a laser-beam at$c$, ... 0 There are a number of wrong assertions in your statement. You say, "surely ... continue accelerating to infinity." Since$ a = \frac {F}{m} $, as long as you apply net force F, you only get constant acceleration a. The acceleration value not only does not go to infinity, but actually does not change, unless the net force changes. When there is no net ... -1 The relation is:$\sum F=0$comes from$\sum F=ma$If$\sum F=0$, so that either$m$or$a$must be$0$. Currently, there is no substance of mass$0$. So,$a=0$. Means that no force is applied to the object. If it is moving, it will continuously move with constant velocity. If the$v=0$, the object will stay stationary. Hope you understand. 2 There is no law that says the sum of forces on a given object must be$0$, that is simply the condition for mechanical equilibrium. If an object has constant$0$velocity (or, more generally, any constant velocity), then its acceleration ($\frac{dv}{dt}$) is$0$and, by Newton's second law as you have it, the net force acting on it is$0$. However, if all ... 0 I am adding another answer in order to show the steps needed to solve such problems in the general sense. An acceleration run is split into parts of differing acceleration domains. Given a starting condition for time, distance, speed, acceleration of$t_0, x_0, v_0, a_0$here is what you can do Constant Torque with Air Drag$t=t_0 \rightarrow t_1$,$v=v_0 ...

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I'll try to give a back of the envelope calculation. Suppose: Mass of the bullet: $m = 10 \mathrm{g}$ Speed: $v=300 \mathrm {m/s}$ Length: $l =5 \mathrm{cm}$ This gives a momentum: $p=mv=3 kg \frac{m}{s}$. You should be able to stop it in: $t =\frac{l}{v}=0.1 \mathrm{ms}$. That means, you would have to apply a force of: $F=\frac{\Delta p }{\Delta ... 1 The answer you are looking for is no. The point of bullets is to kill people. If you tried, you would find the bullet goes by too fast to see, much less reach out and catch. If the bullet hit your fingers, most likely they would be shot off. The point of movies is to entertain. An impossibly heroic hero is more entertaining than reality. -1 If you can attain the speed as that of bullet or greater than that, either by any means, you can catch the bullet! Speed of bullet is approximately around$896 m/s$. If you can sit in North American X-15 vehicle (see the picture), you can attain the speed of$2016.11 m/s$, you can catch the bullet provided conditions are better:) -1 If you are in a high speed train travelling at almost the same velocity as the bullet, then you can grab it without hurt you. Now you can let the train slow down gradually, the bullet is in your hand and you are still alive. 1 It was said that a French pilot did this during World War I, although this is possible in principle, I can't find any reliable references about it (perhaps in French?). The initial speed of a bullet is at least hundreds of meters per second, and catching it using fingers is incredible. But when the bullet has been flying for a period, its speed may decline ... 1 They are not the same thing. Suppose you are standing still and your friend is moving at 5 meters per second on a train. Let's say your frame is the$K'$frame and your friend's frame is the$K$frame. Now you throw a ball in the direction of the trains motion at eight meters per second. Then the speed you see is$\mathbf{v}'=8\mathrm{m/s} \hat{x}$, where ... 3 There is one formula relating the speeds of any two "platforms" (say$P$and$Q$) between each other: $$V_{P}[ Q ] = V_{Q}[ P ].$$ And there's of course the well known symbol for "speed of light (in vacuum)", as determined of light signals exchanged by members of any one platform between each other:$c$. The speed of any one platform ($Q$) as determined ... 1 There are two formulas for adding velocities. The first is typically called Galilean relativity and the second special relativity. The first is simple, if you stack your tank on top of a train then the speed of the shell is the sum of the velocities,$v_1+v_2$. These things can be added as much as you like. You have an aircraft carrier moving at$v_1$... 0 There are two things that limit the maximum traction (F) of a car. One is given by the friction formula, F = μR (Traction = friction coefficient x weight of car), above which the wheels start to spin. The other is the power equation. P = Fv or F = P / v (Traction = power / velocity), which the engine isn't powerful enough to exceed. Note that the max ... 1 What you're asking is essentially whether anything can rotate faster than the speed of light. Just like how it would take infinite energy to accelerate an object to the speed of light in a straight line, it would also take an infinite amount of energy to rotationally accelerate an object to the speed of light. In any practical sense, this tower would be ... 2 You have to split the time domains into the gears needed to reach 60mph. For each gear, there have to be assumptions on the power delivery of the car. Typically 1st gear is traction limited, so you can assume constant acceleration up to the speed where peak power occurs. The relationship between power speed and acceleration is$P(v) = m \,v\, a(v)$. So run ... 1 Let us first simplify a bit, and get rid of the vertical movement of$m$. It is much simpler to assume that$m$is standing with zero velocity in the path of$M$, which will collide it at$t=0$. This will be the same problem along the horizontal axis, and avoids to superimpose it with another problem in the vertical direction. The conserved quantities are ... 0 The relation$v=\sqrt{Fd/m}$suggests that$v$depends on not only mass but also the external force acting on a body multiplied by distance travelled by the body divided by its mass. 3 Your friends are correct. If there is no force in the left-right direction, then linear momentum will be conserved in that direction. Because the new composite object has more mass than the original object, it will have a lower speed to the right. What about energy? Kinetic energy is not conserved in this case, because the collision is inelastic. The ... 1 Your equations of motion are: $$\Delta y = v_{0}\sin{\theta} \Delta t - \frac{1}{2}g\Delta t ^2 \\ \Delta x = v_{0}\cos{\theta}\Delta t$$ You can rearrange to eliminate$\Delta t$: $$\Delta t = \frac{\Delta x}{v_0 \cos{\theta}}$$ Substituting this into the$\Delta y$equation yields: $$\Delta y = \Delta x \tan{\theta} - \frac{g(\Delta ... 0 Perhaps your question is whether the speed of approach of the two particles is 2c, is this so? Yes, it is 2c and this does not violate the principles of relativity, because such speed is not the speed of a particle, but it is just a derived value. On the other hand, the speed of a photon is c regardless of the inertial frame, and is calculated by the ... 0 I'll qualify this by saying it should be a comment to the question so please don't shoot me based on that... taking the extended commentary of the question into account regarding using doppler shift, you would be measuring relative to the object or emmissions that produced the doppler shifted particles, which is a longer explanation of the accepted ... 1 Well that's a result using differentiation and derivation. Have you studied calculus? If not, there is a simple way to look at it.$$\frac{\Delta p}{\Delta t} = \frac{\Delta (mv)}{\Delta t}$$(Yes there should be a \Delta t in the denominator, too.) Now, what does \Delta(mv) mean? It represents the change in the quantity mv. For current situations, ... 1 You are right, there is a \Delta missing in front of the t. \Delta v = v_2 - v_1. If the mass is not changing, then \Delta (mv) = mv_2 - mv_1 = m(v_2 - v_1) = m\Delta v. Hope that helps. The equation that includes \frac{\Delta m}{\Delta t} is not Newton's second law. The second law is valid only for systems of constant mass. An equation like ... 3 1) Yes indeed, the absence of a \Delta in the second expression is just a typo. 2) The last expression is derived assuming that mass is a constant. If it helps, just set the mass equal to 4, or something. If we want to know how the quantity 4v changes, we really only need to know how the quantity v changes. Suppose v changes from v_1 to v_2. ... 1 Yes. It should be:$$\frac{dp}{dt}=\frac{d(mv)}{dt}$$I'm using d instead of \Delta because I am thinking about the limit where the changes in p and t are very small. Then these are called infinitessimal changes, and denoted by a d. Usually, when one considers simple problems in Newtonian mechanics, what one does is study a given object with a ... 8 There's a simple way to look at this that doesn't involve any maths. Suppose the two cars are parked and are stationary, and you accelerate past them in your car. If you are accelerating forwards then from your perspective it looks as if the two cars are accelerating backwards (at the same rate). But the cars are at rest, so the distance between them can't ... 0 No, B will perceive the same value of v for A. According to special relativity, it is impossible for A and B to determine which one of them is moving from their reference frames. Both will have the same experience from their perspectives. A will see B contracting along the direction of motion and its clock moving slower, and vice versa. If B ... 0 by definition, any object is stationary relative to itself, regardless of any other frame. from the perspective of that object, no dilation exists, and time is experienced at it's full normal rate rather than a fractional amount as relative to another objects perspective. the rate of time doesn't increase to infinity, but rather can decrease by a fractional ... 0 Oke in order to understand this more intuitively lets calculate the following example. We have 3 rockets of 100 kg side by side floating in space. Rocket 1 and rocket 2 have the same speed and rocket 3 has a speed v_{int} different from rocket 1 and 2. whe call rocket 1 our observer and where gone try to get rocket 2 to they same speed as rocket 3 In ... 0 You're messing up with simple time dilation. Time intervals are relative quantities. Two observers may not be agree with measured time intervals of an event. You see other moving observer's time dilated. Also, you see other observer's time dilated if she is deep in Gravity well than you are. Meaning, you find other observer's measured time interval more than ... 1 There is no absolute stationary object, an object may only be stationary with regard to an observer. If e.g. the relative velocity of an object is zero in our reference frame, we observe an object which is not moving with regard to our own reference frame. In this case Lorentz factor is 1, that means that there is no time dilation at all. -2 Lorentz factor, the speed-dependent dilation function, is$$\lim_{v \rightarrow 0} \sqrt{1-v^2/c^2} = \sqrt{1-0^2/c^2} = \sqrt{1} = 1.$\$ We see a considerable time dilation at v=0.

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