Acceleration Definition and 1000 Threads

$\tiny{299}$ For $t \ge 0$ the position of a particle moving along the x-axis is given by $v(t)=\sin t—\cos t$ What is the acceleration of the particle at the point where the velocity is first equal to 0? $a. \sqrt{2}$ $b. \, —1$ $c. \, 0$ $d. \, 1$ $e. —\sqrt{2}$Ok well originally it was...

I have searched the forum a bit but could not find a discussion of this Astrophysics letter: 'Evidence for anisotropy of cosmic acceleration', Jacques Colin et. al, 18 Oct 2019. Can someone please direct me to any forum discussion?

In this experiment, I still can't figure out why the graph between time period and distance from point of oscillation is like that. Why does it first decrease and increase so steeply? I got the second part because it goes near the centre of gravity and time period becomes almost infinite there...

I have solved part a using the conservation of energy, getting a (correct) answer of 47.9 km/h, but I am unable to make headway with part b. Based on the flywheel rotating at 237rev/s when the car is moving at 86.5 km/h, I obtained omega = (237*2pi)v/24=62v. Differentiating both sides should...

Now, if I recall correctly, lighter objects with smaller inertia do go faster in terms of acceleration and farther than objects with heavier mass or inertia when the same force is exerted on them. But what about same accelerations? If a light object and a heavy object were to undergo the same...

for example take a look on this graph

I solved this problem using second Newton law for translational motion and the same law for rotational motion, and got $$a= \frac {F} {m+ \frac {I} {R^{2}}} (cosϕ−rR)$$ where m is spool mass. Now, we have three cases: (a) ##cos\phi>\frac{r}{R}##, when spool is accelerating to the right, (b)...

This is what I've done so far:

Here's my calculations for (a) and (b)

Let me preface this by saying I have no background in physics or any of the above other than hours and hours of reading. Could someone explain why (if a method was developed) we couldn't use EFA as a thrust system for a rocket? My reasoning is instead of using fuel to fight gravity and push...

Well, using (1) is easy to see that, at a given time in C ##t## both curves are described with the same value of ##\tau##, i.e. ##\tau_A=\tau_B=\tau##. So the corresponding positions at a given time ##t## are $$x_{AB}=x_{0,AB}+\frac{\cosh{(a\tau)}-1}{a}$$ and therefore $$\Delta x \equiv x_B -...

But we know it not true becouse then it was almost impossible to run on a train

Let´ s call ##N_x## the magnitud of the force between the rod and the box and ## N_y## the magnitud of the force between the rod and the surface. ##N_x = ma_c## ##N_x= ma_r## ##mg-N_y=ma_y## The following I think is to find a relation between ##a_r## and ##a_y## and that can be found by...

Hello, so I am working on a projectile motion lab but I'm not sure what to do right now. Essentially, the lab consisted of my classmates and I using an air table to show that the vertical and horizontal components of projectile motion are independent. During one of our trials, we placed a puck...

I attempted the question with d=vi x t + (at^2)/2 gravitation acceleration= -9.8 and I got the solution of 22.724. Should I use the value of -9.8? or should I just use 9.8? should I use the equation above? I feel like what I am calculating is not displacement but distance... thank you

I have computed that the acceleration in my problem is a(t) = -gj - k/m(|r(t)| - L_0) * r(t)/|r(t)| Where a(t) is the acceleration vector, g is the gravitational acceleration, j is the unit vector in y-direction, k is the spring constant, m is the mass, r(t) is the position vector, |r(t)| is...

So, this may be a really stupid question, and I strongly feel as though I'm missing something here. How can it be that objects of different masses have the exact same acceleration when mass is in fact resistance to acceleration? And then, if in (a vaccum) I throw upwards M and m ( a bigger and a...

Please refer to the image

I tried this but I don't know if it makes sense: Average velocity from A to B = 22/2 = 11m/s Average velocity from B to C = 104/4 = 26m/s (26-11)/6 = 3.75m/s

Summary:: This is a question about finding the acceleration of a point in a mechanism Hi, I have a question about the mechanism shown in the attached picture: Question: We are told that \omega = 6 rad/s and the first part is asking me to find the acceleration of point P on the piston when...

Suppose we have an accelerometer carrying a charge. The charge density everywhere in the instrument is uniform, or at least what I mean to say is, the charge on any component is proportional to that component's mass. Now, in an inertial reference frame, we place the accelerometer in an electric...

I got the correct answer which is √g. I solved it as if it were a circular motion with radius of 1. But got no idea why my solution did work.

When the box travels a ## X## distance, the wedge travels ## \frac{X}{2}##. So ##a = 2A## Using the wedge as a non inertial frame: I didn't use (4). Using (2) on (3) and then on (1) I got: ##2mA=mgsin\alpha +mAcos\alpha + \frac{-mgcos\alpha sin\alpha +mAsin^2\alpha +MA}{2cos\beta -...

Hi everyone I had this argument with someone told. About angular acceleration. His opinion: Since both pulleys are connected by a string then both of them must have the same angular acceleration. My opinion: Since a2 "acceleration of the second mass" is half a1. Then angular acceleration of the...

Just started learning about uniform circular motion. I really don't understand how we get aΔt2/2 on the side. I also searched on the internet for a similar derivation, but there are none so simple. Thanks for your help! P.S There is a mistake in calculation in second line (textbook error).

I'm working on a project where we have a mass (50 kg) sitting on a spring (350 N/mm) and are subjecting it to a sudden impulse (20g) along the spring axis to simulate a shock. We have the profile of the acceleration defined as: ##a(t) = x''(t) = P\cdot \sin^2 (\pi \cdot t / T)## Where P (peak...

In the case where ground clearence isn't the limiting factor; Since motorycles are steered using countersteering, a motorcycle cornering at constant max g can achieve any decrease of lean angle (bringing the bike up) by turning harder towards the inside of the turn momentarily, which is not...

Therefore, if someone were to ask what the magnitude of centripetal acceleration is at the top of the wheel at a given instant (relative to the ground): ##v_{cm} = v_{translational, center-of-mass/wheel}## ##ω = ω_{point-of-contact}## ##v_{top} = 2(v_{cm}) = 2(rω)## ##a_{c(top)} =...

Since the acceleration ##\vec a## is given by ##\vec a = \frac{d^2 \vec x}{dt^2}##, it is a function of ##t## only. Of course, the derivative implies that ##t = t(\vec x)## so we can also in principle express ##\vec a## in terms of ##\vec x##. But how can we express an acceleration dependent on...

Because the friction is the same in both parts, the calculated acceleration from (b) should be the same for (c) I knew I could find Vf, and thought I could do it with an energy equation Ei=Ef mgh=1/2mv^2 gh=1/2v^2 (2)9.81(1.5)=1/2v^2(2) (square root)29.43=(square root)v^2 v= 5.424 Then...

We are given that ##v' = \frac{1}{10}v^2 - g##. I tried using implicit differentiation so that ##v'' = \frac{1}{5}vv' = \frac{1}{5}v(\frac{1}{10}v^2-g)## and set this equal to 0. Hence we have 3 critical points, at ##v= 0##, and ##v = \pm \sqrt{10g}##. Calculating ##v''(0)=-120##, we know the...

I was just reading a set of thermodynamics lecture notes and came across the following In most thermodynamics problems I have done, it is indeed assumed that the piston does not accelerate so we can simply equate forces on the piston. However, I don't fully understand the line of reasoning...

Initially we are given the statement Vav = (x-x0)/t, so far so good. But, we encounter the following paragraph... "We can also get a second expression for Vav that is valid only when the acceleration is constant, so that the v-t graph is a straight line (as in Fig 2-14 - [I've omitted the graph...

Hello, I hope you are all having a great day ! I've got a physics test in a couple of days and I have some questions:1. In a calculation, if the acceleration is in m/s², I presume the speed also has to be in m/s and not in km/h ? 2. So with this graph (v with t), I have to find the total...

Well, I just had this thought earlier, and I want to share it. Here it is. So, we all know about inertia, right? The resistance to acceleration, or change in motion. Well, there is also a concept about derivatives of acceleration, mainly jerk and yank. If you don't know, jerk is said to be the...

So I figured out the equation, but it is probably wrong because the answer doesn't tally. Since the string is inextensible, I can assume that tension is the same for both sides, and acceleration for both masses is the same too So: I can say that the acceleration of 2kg block =acceleration of 7kg...

a) Eg = Gme/r^2 r = √Gme/Eg r = √[(6.67x10^-11 N*m^2*kg^2)(5.98x10^24 kg)]/(4.5 N/kg) r = 9.41x10^6 m h = r2 - r1 h = 9.41x10^6 m - 6.38x10^6 m h = 3.03x10^6 m that's over 3000 km. Did I not use for right equation? Is Eg not 4.5 N/kg? Also for b), isn't the force of gravity the centripetal...

I've been attempting to solve this problem for three days now. I have thrown away my old attempts (like, scrumpled up into the bin), but my old attempts involved: Trying to set up simultaeneous equations relating the journeys between EH and FG to find the deceleration, but the reason why this...

So far what we know about the circular motion is that an object moving in a circle experiences a force towards the center of the circle and as a result accelerates towards this center. But we also know that an object always moves in the direction of resultant force - if two tractors moving at...

ω(10)=(1.3)∗(1.0−e^(−10/22) )= 0.475 rad/s 0.475 rad/s=0 +α(10second) α=0.0475 rad/s^2 ∫ω(t)=Θ =1.3t + 28.6e^(-t/22) | (t=10s, t=0) total angle by which the wheel rotates over this period of t=10 seconds = 2.55 rad Θ= 2(pi)(8m)= 1.3t + 28.6e^(-t/22) 0=1.3t + 28.6e^(-t/22) - 2(pi)(8m) t=34...

33.33 RPM (1min/60second)(6.28 radian/1 revolution) = 3.45 rad/s linear velocity=3.45 rad/s * 0.1524m=0.526 m/s linear acceleration= (0.526 m/s)^2 /(0.1524m)=1.814 m/s^2 1.814 m/s^2=(0.1524m)(rotational acceleration) rotational acceleration=11.9 rad/s^2 ω1= (3.45 rad/s)+(11.9rad/s^2)(10s)...

Hello, I hope you are all very well. I am in second year of High School and I have a practical work in physics. The experiment was to release a long tape with a mass of 40g at the end from a certain height. An instrument would hit the tape 50 times/s and put a mark each time. From that we have...

Well, first a wrote the equation for acceleration in non inertial systems. ##a_I=a_o+\dot \omega \times r+\omega \times (\omega \times r)+2(\omega \times v_{rel}) +a_{rel}##. Then, ##a_o=0## (because the system doesn't move), ##a_i=0## (because it is measured from the non inertial system)...

Summary: what is the equation for a pumkin's acceleration when the air pressure in not constant. My Daughter and I are going to Pumpkin chunkin for the first time. I would like to get two orange shirts and scribble: what part of this don't you understand: (DifEq?) acceleration of a pumpkin...

1. For the car to apply brakes, we have ##v^2=2ar⇒a=\frac{v^2}{2r}=μg\;\;[ma=μmg]⇒v=\sqrt{2μgr} ## 2. For the car to go in a circle ##\frac{mv^2}{r}=μmg\Rightarrow v=\sqrt{\mu gr}##. We find from above that the maximum velocity ##v## possible to avoid a collision is ##\sqrt{2}## times as much...

All I know is that it involves Newton's 2nd law F=ma. Im wondering if I use average acceleration (vf-vi)/time I'm not sure where else to go from here.

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I tried this problem 3 times. I only have two attempts left. First time: Centripetal acceleration: 7560 m/s^2 Centripetal Force: 4.7 Newtons Second time: Centripetal acceleration: 25.032 Centripetal Force: 4.7 Third Time: Centripetal...

I calculated the average velocity in a previous problem and got 1.146788991m/s over a time of 8.72s. I know I can’t use a_ave=(Vf-Vi)/(Tf-Ti) because I don’t know the final velocity and have no way to find it. Do I multiply average velocity by time?