# Tag Info

## New answers tagged acceleration

0

Let's define an $x$ axis for our problem, as below: At $t=0$ the smaller mass $m$ is at distance $x_0$ from $M$ and at rest ($v=0$). The larger mass $M$ which we assume stationary exerts at force $F$ on $m$: $F=\frac{\mu m}{x^2}$ with $\mu=MG$. As a result of this force, $m$ experiences an acceleration $a$ and respecting the vector directions and ...

0

$m_1g - T = ma$ $3T -m_2g = ma$ Solving these, and taking $m_1=m_2$, We get, $a=g/2$ Which is same as Daniel's answer.

0

The mass m2 is hanging from what is known as a Luff tackle, which has a 3:1 ratio. This means that, from the string's viewpoint, the system acts exactly like a mass of size $m2\over 3$; the inertia is one third of m2's, and the force generated by gravity is equal to $gm2\over 3$. Since m1 and m2 are equal, this means that the system is equivalent to a ...

0

The no math answer to this one is to realize that acceleration is the rate of change of velocity. At the top most point, the velocity is indeed zero. However, it is changing momentarily after that. If the acceleration was zero, the ball would have had no change in velocity and would have stayed up in the air forever.

0

You are correct, in that the velocity is zero, so its direction doesn't mean anything, but just because the velocity is zero doesn't mean the acceleration is zero. And it's not zero, it's -9.8m/s/s, as you acknowledge, so the direction of acceleration is meaningful.

2

You throw the ball upwards with velocity $v$ and it returns to your hand with velocity $-v$. Let's draw a graph showing the velocity as a function of time: Acceleration is defined as: $$a = \frac{dv}{dt}$$ so it is the gradient of the line in this graph. The velocity-time line is straight so the gradient is constant which means the acceleration is ...

0

I think you are subconsciously mixing up velocity with acceleration. Let me give you an example. Imagine these are the measured speeds of a particle thrown vertically into the air at different times: time, speed 0s, 50m/s 1s, 40m/s 2s, 30m/s 3s, 20m/s 4s, 10m/s 5s, 0m/s 6s, -10m/s 7s, -20m/s 8s, -30m/s 9s, -40m/s 10s, -50m/s <-- it hits ...

1

When you shoot the ball upwardly, gravity acts on it with a force $mg$ where $m$ is the mass of the ball and $g=9.81 ms^{-2}$ the Earth's gravitational acceleration. If the initial upward velocity was $v_0$ then the instantaneous velocity $v$ is given by: $v=v_0-gt$, so after some time $t=\frac{v_0}{g}$ the balls's velocity becomes $v=0$. However, we know ...

0

At the topmost point, the velocity vector is a null vector whereas the acceleration vector has constant magnitude $-9.8\,\mathrm{m/s^2}$ and constant direction downwards i.e. towards the centre of earth.

0

Consider this. The twin paradox, but with a twist. The twin that accelerates is the one that is younger on return. Statement one It doesnt matter which direction the accelerating twin goes. Ie if she leaves from the equator and goes due north from the earth and returns, it doesnt make any difference if she had gone due south and returned. Statement two ...

3

The universe is described by a scale factor, normally indicated by the symbol $a(t)$, that is a function of time. We take the scale factor to be one right now, so in the past $a$ was less than one and in the future $a$ will be greater than one. Roughly speaking, if $a$ has the value $\tfrac{1}{2}$ it means everything was half as far apart as it is now, and ...

1

Your equation 1.1 can be used with constant velocity. Here you have to use the $2^{\text{nd}}$ equation. ie $a = 2d/(t^2)$. So, the answer is $118.4 \, \text{cm}/s^2$.

1

An accelerating object has a changing velocity. Obviously so since the object starts with zero velocity and the velocity increases with time according to the SUVAT equation: $$v = u + at$$ So your equation 1.1 is no use here. It calculates the average velocity. This could actually be used to calculate the acceleration, but the working is a bit involved ...

1

You're confusing cause and effect. The better equation here is $A={F \over M}$ When a force is applied to an object, the object is accelerated. The object gains kinetic energy from the force, but it doesn't have a property of acceleration. The gun accelerates the bullets with some acceleration for some time period. One gun could accelerate the bullet ...

2

In method two,you are writing $F_{net}=ma$ separately for each block.So it should be $F_t=m_1a$ and $m_2g-F_t=m_2a$,that is,you must use the mass of just the one block you are considering,not both the blocks together.This agrees with method 1,which is also correct.

0

Method #1 is correct. Not exactly sure what's going on with the attempted solution in Method #2, but one thing to try in situations like this is to try out some special cases to see if the answer makes sense. For example, let's take the special case where the mass m1 is zero. It should be apparent that the acceleration in this special case should simply be ...

0

A couple of other things that can be considered: - As the car's speed approaches zero, static friction will take over from dynamic friction in the brakes, which will suddenly increase the deceleration if constant pressure is maintained on the brake pedal. This effect can be avoided by reducing the pressure at the last moment. - The effect of the springs ...

0

Hint: Begin by showing that the general solution of the linear first-order differential equation $$a=\frac{dv}{dt}=-0.5v$$ with constant coefficients is $$v(t)=v(0)e^{-0.5t}.$$

4

It isn't clear from your question exactly what you are integrating and how, but this is the way to tackle problems like this. You know that: $$\frac{dv}{dt} = -kv$$ The way to solve equations like this one is to rearrange it by dividing both sides by $v$ and multiplying both sides by $dt$ to get: $$\frac{1}{v}dv = -k\,dt$$ Now we can integrate both ...

0

You've confused velocity and speed. Speed is independent of direction, and is therefor always positive. For a 2-dimensional analysis, you have to consider the two different components of the velocity, Vx and Vy, but the speed S is $$S=\sqrt{V_x^2 + V_y^2}$$ Let's take a version of your example. An object travels along the x-axis at 5 m/s in the positive ...

0

The kid pulls himself, so there are two tensions 2T to support his body: 2T-mg=(1/5)mg so T=(3/5)mg The force the kid applied equals the normal force n exerted by the seat, therefore T+n-mg=ma n=mg-T+ma=mg-(3/5)mg+(1/5)mg=(3/5)mg

0

First a comment that you don't need GR for problems like this: SR is fine for any kind of accelerated motion in flat spacetime. Imagine the motion as a sequence of boosts. Think of it at first discretely. At each step, the object undergoes the same boost: let's make it a small one of relative velocity $\Delta v$, with a boost matrix $\Lambda(\Delta v)$. So ...

2

Let me try to add a missing intuitive piece in the explanation to the discussion to @Gert's answer of why there must always be radial acceleration for any circular motion. Remember what acceleration is: Change in velocity. $\vec a=d\vec v/dt$. More mathematically it is change in the velocity vector. If velocity is changed - either magnetude or direction - ...

1

Given a particle moving uniformly on a circular path $\vec r(t)$, without loss of generality, we can parametrize the motion as: $$r(t) = R \big( \vec e_x \cos(\omega t) + \vec e_y \sin(\omega t) \big),$$ where $\omega$ is the angular velocity of the motion. By using Newton's axiom $\vec F = m \vec a$ we can calculate the force necessary for the particle ...

4

Particle A is subject to the centripetal force $F=m\omega r^2$ just the same as particle B. In both cases the wall provides the reaction force to keep both particles in uniform circular motion. There can be no circular motion without a centripetal force.

-1

You can differentiate certain types of functions an infinite number of times and they never will become zero. See this related question: http://math.stackexchange.com/questions/1210819/infinite-number-of-derivatives. After reading your edit of the question, I'm thinking this may be more a philosophy than physics question. What comes to mind is Zeno's ...

0

However it is known that a force can't be applied instantly. There is no fundamental factor limiting the rate of change of a force. There are only practical limitations. How short EM pulse can you create? That fast can current-carriers in surrounding bodies experience pulse in force.

1

The rate of acceleration is called jerk. The reason you don't see a name for derivatives of acceleration in classical mechanics text is that they are not particularly important concepts for the study of Newtonian mechanics. It is not important if we can or can't apply acceleration instantaneously. However allowed jerk rates are analyzed and studied in ...

2

How does dark energy allow the universes expansion to accelerate? I hope that it is clear to the questioner and the readers that the horse pulling dark energy is the experimental observation that the expansion of the universe is accelerating. Dark energy is proposed as the reason why the expansion is accelerating. It is called "dark" because it is not ...

1

Read the article on this link https://www.scribd.com/doc/279174920/Decreasing-Mass-Cosmology-and-the-Accelerating-Expansion-of-the-Universe and you'll see that 'dark energy'isn't neccessary for the accelerating expansion of the universe

2

The equation you're using is a kinematic equation. Kinematics, you might recall, is how we mathematically describe motion; it doesn't deal with the causes of that motion. If an object is accelerating at $70 \sqrt{2} \text{ m/s}$ and decelerating at $-24.5 \text{ m/s}^2$ due to a force or a combination of forces, then it will come to rest after traveling ...

0

Suppose you apply an upwards force $F$ on the falling body. Gravity produces a downwards force $-mg$ so the net force is: $$F_{net} = F - mg$$ The net acceleration is then: $$a_{net} = \frac{F_{net}}{m} = \frac{F}{m} - g$$ The acceleration you have calculated is the net acceleration $a_{net}$ given by the equation above.

2

I think you're mentally confusing "acceleration" and "force. I think your thought process is "If I'm going to apply an acceleration to an object, that acceleration will be fighting against the acceleration of gravity, so the two accelerations will partially cancel each other out". The problem is that acceleration isn't something you can "apply" to an ...

1

So, it turns out that I am out of coffee. Anyway, since we already know that: $w = m \times g$ (Newton's Second Law) We know that $w = \frac{Gm_Em}{R_E^2}$ Dividing the entire equation by $m$, we get: $\frac{w}{m} = \frac{mg}{m}$ Which is equal to: $\frac{Gm_Em}{R_E^2m} = g$ (Substituting the values) And finally: $\frac{Gm_E}{R_E^2} = g$

6

I suspect this may not make much sense to non-GR heads, but the Einstein equation relates the curvature of spacetime to an object called the stress-energy tensor. The stress-energy tensor describes the properties of the matter/energy that is causing the curvature. In most cases we're only interested in the amount of matter/energy present i.e. its density, ...

-3

Nobody has any certain answers to this, but IMHO there's an obvious issue with the cosmological constant, which is "the value of the energy density of the vacuum of space". If it's really constant, we've got energy being continually created as the universe expands. That goes against the grain of conservation of energy. I'm not happy with that because I ...

1

You've almost got it! The constant thrust comes from a mass rate $\mu$ of fuel being expelled at a velocity $v_f$ as opposed to the speed of the rocket itself $v$. Therefore the equation is instead: $$(m_0 - \mu t) \frac{dv}{dt} = \mu v_f - \alpha \frac {(m_0 - \mu t)}{r^2},$$ where $\alpha = G M.$ Hence the gravitational term you wrote as \$G m_e (W_0 + ...

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