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83

Alright, let's start with your direct question. Since $d = vt$ the time it takes to travel a certain distance is inversely proportional to your speed $ t \propto v^{-1} $, and so the fractional change in time is proportional to the negative fractional change in your speed. $$ \frac{dt}{t} = - \frac{dv}{v} $$ So, if we consider typical a typical highway ...


30

A pendulum is a day to day example of this. If you watch a pendulum swinging from left to right as it passes the mid point the velocity and acceleration are: The acceleration always point towards the mid point, so as the pendulum passes through the mid point the acceleration reverses direction but the velocity does not.


13

Let me first go through this without friction or air drag. You say $v_y$ along the $x$-axis and the train moves with $v_x$ along the $z$-axis. This is a little inconsistent. I will use the velocities, but not your description of the axes. So the train moves in the $x$-direction, the ball is thrown into the $y$-direction and it the $z$-direction is up-down. ...


11

Get on a car, turn it on and press the accelerator! (${{a}}>0$)... then press the brake (${{a}}<0$). Until the car is steady, the situation is as you described.


8

Here's my messy approach. It's not ideal, but it may work reasonably well: Approximate the position function $x(t)$ as a fourth degree polynomial: $$x(t) = at^4 + bt^3 + ct^2 + dt + e$$ This way, you have one extra degree of freedom to manipulate, which you can use to minimize unnatural motion. Let's assume the motion starts at $t = 0$ and ends at $t = ...


6

Let us try Ross Millikan's suggestion, cf. his comment above: Let us minimize the (higher-order) functional $$\tag{1} S~:=~ \frac{1}{2}\int_{t_i}^{t_f} \!dt~ a^2, \qquad a~\equiv~\dot{v},\qquad v~\equiv~\dot{x}, $$ for fixed endpoints $$\tag{2} x(t_i)~=~x_i , \qquad v(t_i)~=~v_i , \qquad x(t_f)~=~x_f , \qquad v(t_f)~=~v_f , $$ and fixed initial and ...


6

Of course. Acceleration is the rate of change of the velocity. For motion in a line, if the object is slowing down, the acceleration is opposite the velocity. If the object is speeding up, the acceleration is in the direction of the velocity. Imagine you're pedalling a bike, gaining more and more speed and then, suddenly, stop pedalling and apply the ...


5

Absolutely. Acceleration is the change in velocity, so when you say that the acceleration reverses in direction, it means that the object is transferring from either speeding up to slowing down, like a skateboard which just passed the bottom of a U-shaped ramp, or from slowing down to speeding up. Think of the skateboard as having just crested a hill. ...


5

If the speed limit is 60 mph, it would take 60 min to go 60 miles. To go the same 60 miles at 65 mph, it takes 55.4 minutes. A time savings of 4.6 minutes. Is it really worthwhile to speed if all you are going to save is a few minutes (even less for shorter distances).


5

This is a very heavily constrained problem - and it sounds a lot like a spline fitting problem. As you may know, a spline is a curve that has continuous third derivative - and as a side effect, it minimizes the curvature as it connects points. In this case, the problem would then be re-stated as a spline that goes through the two points given, with the slope ...


5

Well, the three equations that you mentioned are all derived from the definitions of velocity and of acceleration. I'm assuming you know calculus (if you don't, just ignore the next part and look at the conclusions, or even better- learn calculus!). $$v=\frac{dx}{dt}$$ $$a=\frac{dv}{dt}$$ Now, to find displacement we just look at the first equation: ...


4

It Depends on Your Model How closely do you want to model reality? The truth is, most collisions are some mixture of inelastic and elastic. (That is, momentum is transferred, but not always "cleanly," some of that energy gets transferred into deforming the objects.) You can see this sort of thing if you watch slow-motion videos of things striking other ...


4

If the total kinetic energy before the collision equals the total kinetic energy after the collision, the collision is elastic. Otherwise, it isn't elastic. given the mass, the velocity, and the 'angle' the two objects are going two be when they collide - how can I know if I need to compute an elastic or an inelastic collision? The mass, velocity ...


4

Aside from the effect on one driver you might consider the effect on traffic in general, that is what happens when everyone breaks the speed limit. As far as fluid dynamics is concerned, the side-effects of your speeding (if any) are felt by the people behind you. Reaction distance increases linearly with speed, but stopping distance must include a term ...


4

Fluid dynamics models might have a practical value in heavily congested areas, but then you can't speed, save for some random short bits after traffic signals, rendering it irrelevant. Guess they're more of a traffic distribution models, smaller roads will attract more traffic if the counterpressure (congestion) gets higher on the main pipe (road). It ...


4

I was thinking you could do a Bezier curve inspired approach (but with quadratic interpolation instead of linear interpolation). You know that at the beginning of time you want the path to look like $$ p_0 + v_0 t $$ and at the end of the trajectory you want it to look like $$ p_f + v_f (t-T) $$ So, let's try to smoothly interpolate between these two ...


4

If you draw the path, it looks like this: The person moved 10 m in 5 seconds.


3

The accelerometer measures the negative of gravity plus any upwards acceleration see NOTE#1 $$ acc = -(g+\ddot{x}) $$ and you want to integrate $\ddot{x}$ to get speed $v=\dot{x}$ and position $x$. So your expressions should be $$v(t)=-\int_0^t ( acc+g)\,{\rm d}t \\ x(t)=-\int_0^t \int_0^t ( acc+g)\,{\rm d}t\,{\rm d}t$$ You also know that the final ...


3

(Temporarily pretend you are on the second floor of a building.) Jump up and down in one place. After you leave the floor: your displacement is always positive (i.e., above the floor); your velocity is positive (rising), zero (at your highest displacement), then negative (falling back to the floor); and your acceleration is constant and negative -- it's ...


3

There is not such thing as a "partially elastic" collision. Classical collisions between particles can be separated into two categories: elastic and inelastic. Elastic collisions are defined as collisions in which no energy leaves the system (i.e. $E_i = E_f$). All other collisions are inelastic, as some energy is lost ($E_i > E_f$). A perfectly inelastic ...


2

$(4)$ is false. Note that by the same logic, $v = \frac{\mathrm{d}|\vec{x}|}{\mathrm{d}t} = \partial_tR = 0$. Please note that you should ask about a specific physics concept and not just throw "Here's a calculation, show me where I'm wrong" at us.


2

The acceleration always points in the same direction as the force. That's because Newton's second law tells us that: $$ \vec{F} = m\vec{a} $$ where the force $\vec{F}$ and the acceleration $\vec{a}$ are both vectors. So to work out which direction the acceleration is just ask which direction the force is. In the example you've given consider whether the ...


2

Like Ross said, you need to try to remove the overall constant part of your acceleration data. Most accelerators report gravity all of the time. So even if your phone was sitting still on a desk, it would report and acceleration of -9.8 m/s^2 "down", however down lined up with the axes of your phone. Looking at your data a bit, for the first 1/3 of your ...


2

If your accelerations are in $g$, you should multiply the second term in column C by $9.8$. The velocities will then be in m/sec and your integration to make D will be in meters. You may then want to apply any constraints you know of. If you know the velocity starts and ends at zero, you might want to add or subtract a constant acceleration to the data to ...


1

Let me make you precise about conventions, because your notations are unconventional. Suffix in a vector quantity is always represent the component along specific direction, component of velocity $v$ along x-axis is denoted by $ v_x.$ However, your question is precise irrespective of your conventions used. Insight into your problem You have made precise ...


1

Assuming that the train is an inertial frame of reference (non-accelerating) and if we neglect both air friction and the effects of gravity, then the ball will move in a straight line away from you at the same speed which you threw the ball at until some other force acts on the ball. This situation would look the same as one where you threw the ball while ...


1

It is a wrong assumption that the incline moves with acceleration $\frac{Mg}{M^-}$. I am guessing that you find this by saying Newtons 2nd law on the incline like this: $\sum F_x=M^- A^-=Mg$ and then isolating the acceleration (I call it $A^-$). But you did not sum all forces here - you only included the large block's pull $Mg$ but forgot the small block's ...


1

Imagine what happens when $\Delta \phi$ keeps increasing to make a full rotation of $360^ \circ$. Then the angle of $P_2$ increases by $360^ \circ$ so that $P_2$ comes back to $P_1$. Also we know that after the full rotation $\vec{v}_1$ must be equal to $\vec{v}_2$ again. Since $\vec{v}_2$ is going around in a circle at the same time $P_2$ does, its angle ...


1

The perpendicularity of the velocities and radii, plus the fact that both triangle are iscoceles, guarantees that the triangles are similar. In similar triangles, all corresponding angles are equal.


1

Feynman showed only that jump in the position of the particle would change mechanical angular momentum due to that particle. If the jump is due to other body or medium capable of carrying angular momentum, total angular momentum of the system could still be conserved. Continuous mass conservation is a basic idea in physics, inferred directly from experience ...



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