Acceleration Definition and 1000 Threads

My guess was simply that as acceleration changes from the north to east direction, the total magnitude change of v is ##v \sqrt 2##. Acceleration is ##\mu g##, so time would be ##\frac {v \sqrt 2} {\mu g}##. This agrees with the textbook solution. What I do not understand is the trajectory...

I understand based on the equation F = ma that if there is no acceleration, the forces on the object all balance out to 0 in all directions. What I don't get is for example, slowly lowering a heavy stone slab at a constant velocity v, and raising it way above my head as high as I can at a...

This is the problem set. I am stuck from this point... If anyone could give me a hand I would really appreciate it. I know this is probably really simple, but I don't know any of this and have been trying my best with youtube, and other peoples posts. PS this is for high school

If the engine is constant, then the wheels of the car exerts a constant force on the floor. And F = ma, So the car should be accelerating rather than maintaining the same speed. What is going on here?

My initial approach to this question was breaking the components of acceleration in the x and y axes and applying the three equations of motion to find the final velocity as well as the final position. As we were expected to find the net final velocity of the particle, I found the resultant of...

The answer should be 10 m/s^2 but I don't know how to solve it

Doesn't friction always oppose the motion? From the clockwise rotation here, shouldn't the cylinder be moving to the right? so why are the acceleration and friction in the same direction to the right, and in the same direction as the motion? (attached image for reference)

Why is the center of mass acceleration zero? Did I do anything wrong?

i solved it like this... s = ut + 1/2 at^2 t= 1.08 (from part a) u= 65 sin4.30 a= 9.81? or -9.81 the answer said -9.81 why? wouldn't acceleration change from -9.81 to +9.81 because it moves up then down??? its soo confusing...

So I thought the stone would initially experience acceleration in the backward (leftward) direction then continually accelerate in the inward direction of the tire (i.e. upward then rightward then downward then leftward, etc.) as the tire moves forward. But the answer is immediately upward...

This is the question. To this point everything is clear. I have problem with following part: The authors claim that each part of the remaining rope is under constant acceleration. So it is in free fall and only gravitional force acts on it. If we release a rope like above, before it hits the...

Basically, I tried to find the solution by calculating P=Fs/t, where F= 2250 and s is the distance traveled in the 12th second and that result differs from the result I get when I calculate the power using P=Fv. ##F=ma=1500*1.5=2250N## ##s_{12}-s_{11}=...

So for Q1, I answered down (towards Earth) but the solution says there is no acceleration there. For Q2, I answered mgh, but the solution says it's mgh/t, which is power, right? I just want to make sure I'm not super confused. Thank you.

This is from an old exam. The velocity of a particle moving along a straight line is v = 4 + 0.5 t. What is the instantaneous acceleration at t=2? The solution is supposedly 2 because a = dv/dt = t. But I thought dv/dt here would be 0.5. What am I missing? Thanks.

Is the reason behind coriolis acceleration is that as you move far from the centre of a rotating frame the tangential velocity increases?

Pretty straight forward, ...reason of posting is to check why i am having a negative value for ##a##. From my study, i know that ##R(||)## to plane ##F - 40 \cos \dfrac{π}{3} = 4a## ##a = -5 m/s^2## or can i as well have the equation ( friction and tension are at equilibrium) as, ##40...

I'm taking college physics without calculus this semester and it's been quite the challenge to say the least. We recently covered free body diagrams and while I understand the different vectors in the FBD, making calculations is killing me. Specifically Newton's 2nd law. The problems range...

I tried to use the energy route but the acceleration was not a plausible one. It was just way too big.

Does it exist an invariant way to define acceleration in Newton physics like the proper acceleration in GR ? In Newton physics if an accelerometer attached to an object reads 0 it does not mean it is actually not accelerating (since gravity is a force). To define inertial motion the concept of...

First of all, I wish everyone a Happy New Year. I am interested in your expertise on a special constellation, which I will first briefly describe. If you observe an object that is approaching the event horizon of a black hole, it is said that at some point the distant observer will have the...

##\displaystyle R=\frac{mv}{qB}\implies v=\frac{RqB}{m}## where ##v## is the speed of the proton ##\displaystyle\frac{dv}{dt}=\frac{Rq}{m}\frac{dB}{dt}## On substituting the values, I get ##\displaystyle\frac{dv}{dt}=9.58\times 10^4\ m/s^2## This answer, however, is incorrect. Where have I...

[FONT=times new roman]1. [FONT=times new roman]The first equation between velocity ##v## and time ##t## can be derived using the graph I have drawn for the purpose as shown on the right. Since acceleration ##a_0## is a constant, the graph of ##v-t## is a straight line. The slope of the line is...

Easier case: Elevator is at rest. We need to prevent box from free fall so friction should be bigger than "mg".(And they can be equal) When we push with force F we know that the maximum static friction is ##u_sF##. "mg" should be smaller than ##u_sF## or should be equal to it so the minimum...

Hi, I am having problems with task b I then defined the velocity vector and the acceleration vector as follows ##dot{\textbf{r}}'(t) = \frac{1}{||\dot{\textbf{r}}(t)||} \left(\begin{array}{c} \dot{r_1}(t) \\ \dot{r_2}(t) \end{array}\right)## and ##ddot{\textbf{r}}'(t) =...

A. Correct answer is radius = 1770m, acceleration = 2.73*10^-3m/s. B. I don't know how to approach this problem. I don't know if I should start with forces, energy, or basic kinematics.

How would we define a value for acceleration if only the direction is changing and not the speed?

If an elevator is moving upward, what does a downward acceleration mean? When applying the free body diagram, will the positive direction be upwards since the elevator is moving up?

I have a reference system A with three clocks of the same type. Two clocks are at rest in the origin of A and could be synchronized without any problems. The third clock rests at a distance in the x-direction. Is it possible to synchronize this third clock by accelerating the second clock at...

Here is only my solution: ##A_1 \frac{\mathrm d h}{\mathrm d t}=-A_2\sqrt{2hg}##, so by integrating we get ##h(t)=\left(\sqrt{h_0}-\frac{A_2}{2A_1}\sqrt{2g} t\right)^2.## Setting ##h(T)=0## we get ##T=\frac{A_1}{A_2}\sqrt{\frac{2h_0}{g}}.## By doing the first time derivative of ##h## we...

I don't get how is the 4th case different from the 1st case? In both cases the weights are hanging and are not accelerating, but somehow in 4th case the force meter shows 0N while in 1st shows 10N. All other meters show 10N but the last one. Now, I don't know hot to solve last one. I tried...

Hi, I was looking over one of the sample examples in Halliday and Resnick, the one about the scale in the elevator. There is something that bugs me about it, and I'd like to know if you agree. The example has to do with finding the reading of a scale that is measuring someone's weight in a...

TL;DR Summary: Find initial vertically upward speed of the ball Find horizontal speed of the speed Find angle How to: Find initial vertically upward speed of the ball Find horizontal speed of the speed Find anglei try to solve it but it didn't work

In my approach i have distance as ##(x)## and velocity as ##(x^{'})##, then, ##(x^{'}) = kx^2## where ##k## is a constant, then acceleration is given by, ##(x^{''}) = 2k(x) (x^{'})## ##(x^{''}) = 2k(x)(kx^2) ## ##(x^{''}) = 2k^2x^3##. Correct?

I am currently studying Newton's laws and mechanics. I have this question: Why is distance=half a*t^2? Where did the 1/2 come from? Can someone explain this without using calculus?

1) a=0 and v=0 2) 0.7 m/s

This is a follow-up question emerging from another thread in the Sci-Fi Writing and World Building forum. Specifically, @DaveC426913 had criticised another book in which the plot is set in motion by a plan to turn an interstellar colony ship around and return back home. In my setting, a similar...

The correct answer is obtained by rearranging Δ x/ Δt = v. However, I assumed there would be some acceleration in the y direction so I tried to use the kinematic equations. To find the time I simply rearranged Δ x/ Δt = v, assigning v=5.2 m/s and Δ x = 650. I assumed there is no acceleration in...

The acceleration near the earth, due to the force of gravity is g. Now every particle when moving in a curve trajectory had a centripetal acceleration towards the center (say the sun) a=(v^2)/R. If this is true why we measure weight only with the account of g? I guess when R is big it might be...

Hey guys, Can someone help me understand how to understand this problem intuitively please? How I understand is that I need to look the acceleration relative to the lift as if it were f.e. on another planet with a different acceleration. this gives me a = g - 5. But then again if I didn't look...

I’m an absolute beginner and I need someone to show me where I’m wrong. Knowing the formula of acceleration ∆v (change in velocity) / ∆t (change in time) where ∆v = ∆x (distance) / ∆t, a common way of relating acceleration to distance is to say a (acceleration) = (distance/time)/time =...

If the truck accelerates at 1.47 fps until it reaches its top speed of 8.8 fps, then it takes approximately six seconds to reach the top speed (1.47 x 6 = 8.82). In its first six seconds, the truck covers 30.87 feet (the truck reaches 8.8 after traveling 23.2 feet). To cover the remaining 57.13...

I know to break it down into its x and y components and then use Pythagorean: Acceleration in the x direction is Fx/m ---> (7.50 x 10^6*cos55) / (4.50 x 10^5 kg) = 9.56 m/s^2 Acceleration in the y direction is: (Fy - mg)/m ---> ((7.50 x 10^6*sin55) - (4.5 x 10^5* 9.8 m/s^2)) / (4.5 x 10^5 kg)...

When using magnetism to accelerate the Maglev, and neglecting the usual frictions, and also relativistic effects, is there any limit how fast it can accelerate to? Or is there any sort of increasing "drag" of any sort native to magnetism, which would get in the way of acceleration, as it goes...

We have 2 objects, m1 and m[SUPlB]2[/SUB] Friction is present between the two objects but not between m1 and the floor. A force is exerted on the bottom object which causes it to accelerate parallel to the floor. The thing I'm wondering for while now is, how do I prove that the acceleration of...

What is the initial acceleration of mass 5M .The pulleys are ideal and the string inextensible. My attempt- 2Mg-T=2Ma (for 2M) T=Ma (for M) Solving we get T=2Mg/3 T-N=5MA (for 5M) N=2MA (for 2M) Solving we get A=2g/21 but the given ans. is 2g/23

im not sure where to start

Here are two similar, quite simple thought experiments, followed by assumptions on final clock readings. In the end, my most important question to them. Exp1: - we have two space ships, ss1 and ss2, both have clocks on board, named cl1 and cl2 - we have a third clock, cl3, somewhere located in...

Question picture: My solution: Where: S is the lineforce Ff is the force as a result of friction a is the resulting acceleration F is the acting force The answear is supposed to be a=(F-2mg(mu))/(m+M) Any idea what i could have missed? Thanks for your help on beforehand!