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0

I'm going to just use $32^∘$ below; it doesn't make a difference. Your equation isn't correct. You should have $F_x−f_k = 0$ or $Fcos32∘−f_k = 0$. The x-component of F is $F_x = F.cos32^∘$. Writing $F_x.cos32^∘$ doesn't make logical sense. Why? Shouldn't I be able to use $∑F=0$ in this problem to find the answer? Yes, you can.


1

Here the average acceleration can be understood as follows: The particle going from A to B along a half circle with speed 1m/s can be here viewed as the particle going from A to C and again to A with an initial velocity 1m/s as shown in the figure: The particle in the half circle, will move under a centripetal acceleration which is always directed towards ...


1

Is there a difference between "average acceleration" and centripetal acceleration? Yes, in fact they're almost completely unrelated. The average acceleration is defined as $$\vec a_\text{avg} = \frac{\Delta\vec v}{\Delta t}$$ It is one quantity that partially describes the motion of a particle over an extended time. In other words, average ...


0

The problem with centripetal acceleration is that it is not a vector, and cannot possible have a negative sign. It should remain "constant" in this case, but its direction is changing. But actually, it's not remaining constant, just the magnitude is remaining constant while the direction is changing. Centripetal acceleration will just give you the magnitude ...


0

From a mathematical standpoint, any collision in which no mass is lost is described by two equations: Conservation of energy: $ m_1v_1^2 + m_2v_2^2 = m_1v_1'^2 + m_2v_2'^2 + E $ Conservation of momentum: $ m_1v_1 + m_2v_2 = m_1v_1' + m_2v_2' $ You know the masses and initial velocities, so this reduces to a system of two equations in three unknowns ($ ...


1

To parameterize the degree of inelasticity you use the "coefficient of restitution" which is 1 for elastic processes and 0 for inelastic processes. This is described by $$ \text{coef. of restitution} = c_R = \frac{\text{final relative speed}}{\text{initial relative speed}} = \frac{v_2 - v_1}{u_1 - u_2} \,. \tag{*} $$ This also tells you how to compute the ...


1

Perfectly elastic and perfectly inelastic collisions are just limiting cases on a scale of how much kinetic energy is retained. As noted in @Nathan's answer, if you work in the center-of-mass frame, a perfectly inelastic collision results in 0% of the kinetic energy retained, while perfectly elastic collisions have 100% of kinetic energy retained. So, you ...


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 ...


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 ...


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.


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. ...


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.


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 ...


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. ...


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 ...


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 ...


0

This question has a nice answer. It was found by John Hobby in his paper Smooth, Easy to Compute Interpolating Splines and it is the system used by Metapost to create pleasing trajectories. To summarize the results of the paper, the goal is to find an aesthetically pleasing path $z(t)$ (in 2d) subject to the constraints $z(0) = z_0$, $z(1) = z_1$, $z'(0) = ...


-1

If you draw the figure, the vectors $\vec{PA}$ and $\vec {QA}$ are orthogonal. The triangle is $30-60-90$ so you can find the length of $PA$ and $PQ$. The observed $\dot \theta$s give you the projection of the airplane velocity on these two axes. Add these two projections to get the total 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 ...


0

[Preamble not really answering the question: Sounds almost like a classical philosophy question. To turn it around, you might ask whether its worth travelling at the speed limit of 5mph slower. And 5 mph slower than that. Or should you just walk? "] The accepted answer computes the time saving and the fuel cost. In the airline industry, this is formalised ...


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 ...


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 ...


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).


0

For hanging block $$Mg-T=Ma$$ For block on platform, consider the two$(M^-,m)$ as a system $$T=(M^-+m)g\\\implies a=\frac{Mg}{M^-+M+m}$$ Now consider the block $M^-$ at rest(change your reference frame), So now a pseudo force $ma$, should be acting on $m$ leftwards horizontally. Taking components parralel to plane: $$ma\cos\alpha=mg\sin\alpha\implies ...


4

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


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.


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 ...


0

Short answer: No, Long answer: let's say your plane is moving to the right. If the plane were purely horizontal (like a cube) and friction less, the block placed on top won't even move (there is no force on it except gravity and the normal force from the block underneath it). Now lets look at an inclined plane. The block placed on top again has only two ...


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.


0

In the drawing you give, it appears that the 'down' direction is the direction of positive displacement, i.e., the $x$ unit vector points towards the bottom of the page. Thus, a positive acceleration $a = \frac{d^2x}{dt^2}$ is towards the bottom of the page in the direction of increasing $x$. If you calculate a negative value for $a$, you know the ...


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 ...


0

I don't know if it helps, but in the fifth chapter of the first volume of the book Um Curso de Física Básica (It is in Portuguese, the title in English would be "A Course in Basic Physics") by H. M. Nussenzveig there is a good presentation of a very similar situation to this, ending with a discussion of the deflection of an electron beam in a cathode ray ...


0

You know how much it will take for the drop to fall from the end of the plates to the gutter. And since you should be able to calculate the distance "x" then you have a distance to be covered in a certain time thus obtaining the velocity it has (by requirement).


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 ...


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 ...


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 = ...


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 ...


0

The simplest way to look at this: you are accelerating with an acceleration. $g$ for a distance $h$, the decelerating over a distance $s$ (however far the rider can move his center of gravity to "absorb the shock"). Since distance goes as acceleration squared, the effective deceleration is $$ a_{decel}=\sqrt{\frac{h}{s}}$$ This is on top of the ...


0

Paired with: There is something very efficient about the economy of Soviet authors, who use very few words to say a whole lot. I read from these books and learnt well :)


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: ...


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 ...


1

I think the more direct way to show this is to appeal to conservation of linear momentum: I can't move or "jump" the mass without some commensurate opposite momentum. Feynman effectively mentions this later with the rocket. Linear momentum conservation really comes directly from the invariance of all known laws of physics under spatial translation (see ...


0

For a collision to be inelastic, all energy stored in the two (or more) objects are contained in those objects before and after the collision. This is in reality impossible as some of the energy will always be converted into some other form. In a car accident for example, some (allot) of the energy contained in the two cars will go into mainly the ...


0

The elastic and inelastic collision are models used in point mechanics of what really happens in the continuum mechanics world. In reality, the objects both deform when they collide. As contact is made, the object become coupled by non-interpenetration forces in the direction normal to their contact area, and friction forces in the direction tangential to ...


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 ...



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