New answers tagged

-1

if you know x and t, then you have vo*cos(alpha)=x/t you can choose alpha or v0 for different height put it in y and you get both.


0

With the spring attached to a wall (and not the ceiling) one might assume that the mass is sliding on a friction-less horizontal surface. Then the starting kinetic energy = the final potential energy: (½) m vo^2 = (½) k x^2 where k is the spring constant and x is the stretch. If the spring is hanging vertically, the rest position occurs where its force ...


0

It depends what sign convection you choose, suppose you have chosen upward displacement as negative, then velocity and acceleration direction are different, hence you can take a=(+) , for basic problem remember if direction of velocity is opposite the force like gravity then by assigning upward displacement as positive, you can take a=(-) ,.


0

Initially the mass has kinetic energy $\frac 1 2 mv^2$ and the spring has potential energy $E_{pot}=0$. So $E_{total} = \frac 1 2 mv^2$. When the spring is fully extended, the velocity of the mass is $0$, so $E_{kin}=0$, and the potential energy is a function $f$ of the spring's maximum extension $x_{max}$. So $E_{total}=f(x_{max})$. Since there is no ...


0

The equation you wrote is correct. All you need to do is take the partial derivative. I think this is where you got confused. You can check the difference between partial and total derivative. So here is the answer: $\dfrac{\partial KE}{\partial t} =\dfrac{\partial ( \dfrac{1}{2}\cdot m \cdot \dot{\vec{r}} \cdot \dot{\vec{r}})}{\partial t} = 0 $ $\dfrac{\...


-1

1st one is correct.The equation describes the acting of the external force,as theres no external horizontal force acting on m1,its not in the equation,however it is accelerated by the tension of the string. horizontally on cart M, which pushes on block m2. They therefore accelerate at a = F / (M + m2). In the absence of friction, F cannot impart any ...


-1

Gravity pulls down on M and m1, but the ground holds them up. These vertical forces are irrelevant in the absense of friction to convert them to horizontal forces. Gravity also pulls down on m2. For it not to move, the tension in the string must be just enough to hold it up. T = m2 g Force F pushes horizontally on cart M, which pushes on block m2. They ...


2

I think you are querying the necessity of the statement Infinite accelerations are not allowed To get zero displacement one has to have a positive displacement and a negative displacement and hence at some stage the velocity must be zero as shown on the left hand graph which is continuous and well behaved. Now what about going from a velocity $...


1

In part. This is the equation of position as a function of time for uniformly accelerated motion with $t$ shifted by $t_0=2s$, i.e. in terms of $(t-t_0)$ rather than just $t$. Note that the velocity at $t=t_0=2s$ is $u$ but that’s not the initial velocity. Since the acceleration is constant with $t$, you can read it off at any $t$, including a $t=t_0=2s$, ...


1

So the answer should be given by taking a particular frame of reference....imagine if you are in the elevator, you will feel that the acceleration of the stone is g+a downwards but now imagine that you're on the ground looking at the stone falling, you will feel that the acceleration is simply g downwards hence I think that the answer was given taking ground ...


1

Your teacher is right. You are confused about acceleration and velocity, which can be tricky. Acceleration is the change of velocity with respect to time and nothing can be accelerated without a force acting on it. Think about it. Your car won't drive unless the engine is powering it. Your bike won't go unless you pedal. So as soon as a force starts/...


2

You're confusing velocity with acceleration. The acceleration "a" acts on the stone only until you let go. "After the release" means "after you let go", hence you no longer impart any acceleration to the stone, and "a", the acceleration of the lift, no longer affects the stone. In your balloon experiment, the stone has velocity "u", but in the lift the ...


0

There isn't a single point of contact, when walking there is multitude of points or a whole surface that makes contact with the ground. But otherwise yes, those points of the boot, if it does not slip, have zero velocity with respect to the ground.


-1

In the last step $v^2=2v^2$ means $ v=0$. So you cannot divide both sides by $v^2$ because that is division by 0. That is why you get nonsense.


7

You're using $v$ in two different ways. In the first few equations, $v$ means the final velocity after a period of uniform acceleration, assuming one starts from rest. In the equation $v = s/t$, $v$ means the average velocity instead. That's half as much, which is why you're off by $2$. This is a general warning for learning physics. In other high school ...


0

You always hear that an increase in fluid velocity causes a decrease in pressure, but actually it is the other way around: the lower pressure causes the fluid to accelerate into the lower pressure zone. (This applies to steady state flow at subsonic speeds.) In common physics problems, where a fluid speeds up and then slows down again back to its original ...


2

I took another stab at it and I was able to solve it using the following equations: $$x_f=x_i+v_xt$$ $$y_f=y_i+v_y{_i}t +\frac 12at^2$$ Applying and rearranging these to my problem For x-direction I get: $$d=0+v_i\cos\theta\cdot t$$ $$v_i=\frac{d}{\cos\theta\cdot t}$$ And for y-direction: $$0=h+v_i\sin\theta\cdot t -\frac 12gt^2$$ Apply previous equation and ...


2

Consider a streamline along which the fluid velocity is increasing. This means that the fluid is accelerating, so there must be a force. Forces in an ideal fluid are related to differences in pressure, so the pressure must be decreasing.


2

You had the correct tangential, $c$, and radial, $\dfrac {c^2t^2}{b}$, accelerations If the radius is constant then in polar coordinates $\vec v = b\,\dot \theta \hat \theta = c\,t\, \hat \theta \Rightarrow \dot \theta = \dfrac{ct}{b} \Rightarrow \ddot \theta = \dfrac cb$ and $\vec a = -b\,\dot \theta^2\, \hat r + b\,\ddot\theta \,\hat \theta =-\dfrac{c^...


0

Both are correct as it is just a matter of wording although I would favour change in position and displacement most of the time. Consider one dimensional motion along the x-axis with the unit vector $\hat x$ defining the positive x-direction. A body starts at position $+3 \,\hat x$ with velocity $+6\,\hat x$ and after undergoing constant acceleration ...


0

You correctly wrote the expression for the four-velocity and $\gamma$ but it'd be much easier to use straight forward the time coordinate (Rindler) transform for a uniformly accelerated frame (without integral calculus and 4-velocities) as $$t=\frac{c}{a_0}\mathrm{asinh }\left(\frac{a_0}{c}\tau\right)$$ where $\tau$ is proper time in (in P's frame) and $t$ ...


39

$$v_\text{average}=\frac{\Delta s}{\Delta t}$$ $$v_\text{instantaneous}=\lim_{\Delta t\to0}\frac{\Delta s}{\Delta t}$$ If the time interval gets infinitesimally small $\Delta t\to 0$, then you are dividing with something very, very tiny - so the number should become very big: $$\frac{\cdots}{\Delta t}\to \infty \quad\text{ when } \quad\Delta t\to0$$ If the ...


40

Suppose you are travelling at a uniform velocity and you cover 1 meter in 1 second. Your average velocity is $$\frac{1\ {\rm m}}{1\ {\rm s}} = 1 \frac{\rm m}{\rm s}.$$ If you consider a 1 millisecond interval within that 1 second, you cover 1 millimeter. Your average velocity in that milisecond is $$\frac{1\ {\rm mm}}{1\ {\rm ms}} = 1 \frac{\rm m}{\rm s}...


1

You’re right. You’ll get the change in position $\Delta x$, which is the displacement. Maybe your teacher is confused about “displacement” ($\Delta x$) vs. “position” ($x$).


0

I believe not only this question is too linguistical or philosophical for this SE, but also that the concept of motion is understood as a postulate within Physics. You don't define motion. You calculate and predict it.


1

The trajectories of the thrown object and its consecutive bounces will be parabolic in their shape. In this case the ball will lose some of its kinetic energy each time it make contact with the Ground. However, the angle it makes with the ground each time will be equal to the angle it was original throw at. This means that all of the trajectories will be ...


0

In almost every case, we choose to model an object as a point because we don't care about the object's internal structure for the calculation we're trying to make. This can have several causes; for example: The effect of the object's internal structure may simply be too small to be detectable in the results; The desired precision of the calculation may be ...


-1

When we are dealing with their mass. For example, Newton's law of gravity deals with two objects as the distance between their masses. The "point" is the center of mass of the object which is just a theoretical mathematical point. Remember that there is a difference between total mass and mass density. When you observe an object in space you are looking ...


1

Each derivation rests on the assumptions used. The standard kinematics equations you mention first, depend on the assumption of constant acceleration. My problem is that your friend hasn't stated what assumptions he used to get to $x(t) = x_0 + v(t)\,t$. So let us differentiate both sides to see what kind of acceleration is needed (using the product rule) ...


5

To give a purely qualitative answer consider the meaning of your friends third line $$ x(t) = x_0 + v(t) \cdot t \;, \tag{1} $$ (where I've made the multiplication explicit). This claims that you find the position at moment $t$ by taking the initial position ($x_0$) and adding to that the elapsed time times the velocity the particle has at moment $t$. So, ...


2

I think I understand it now, mathematically speaking, but is there a more conceptual answer? OP evidently seeks a conceptual answer to why $x(t) \ne x_0 + v(t)\cdot t$ when $v(t) = v_0 + at$ and $a$ is a constant. Consider the simple case that the initial position and initial velocity are zero. Stipulate that $v(t) = at$ where $a$ is a constant and it ...


17

Given velocity $v(t)$, the distance moved after a certain time $t$ is not $v(t)t$ - this formula works at constant velocity, but when the velocity is changing, the correct expression is $\int^{t_f}_{t_0} v(t) dt$. Therefore your friend's third line is incorrect.


7

The error is just that $v(t)t$ is not the anti-derivative of $at$. This is easily checked by just taking the derivative. $$\frac{\text d}{\text dt}\left(v(t)\cdot t\right)=v(t)\cdot\frac{\text d}{\text dt}(t)+t\cdot\frac{\text d}{\text dt}(v(t))=v(t)+at\neq at$$ It's a simple calculus mistake.


4

The displacement is only the velocity multiplied by the elapsed time if velocity is constant as you suggested. To derive the equation for varying velocity you must consider the infinitesimal case where the elapsed time is so small that you can consider velocity constant. In this case, a small displacement $dx$ is given by the product of velocity v(t) by ...


0

Whether or not the person is "saved" would depend on the duration of the free fall which would determine the persons velocity upon impact. For example, If a person jumps from some height above the ground the person can lesson the impact force by bending the knees immediately on contact. This reduces the average force on the person by the work energy ...


0

R.H=h/4 [t2÷t1] ² t1=Time of free fall t2=Time of flight h=Free falling height We can calculate the first Rebouce by this equation.


0

One way to understand this is to realize that $\vec{v_1} -\vec{v_2}$ is basically adding $\vec{v_1}$ with $\vec{-v_2}$ i.e. $\vec{v_1} -\vec{v_2} = \vec{v_1} + (\vec{-v_2})$. So, as the first answer explained, if you have two cars $A$ and $B$ travelling east at velocities $\vec{v_1}$ and $\vec{v_2}$ respectively, the person sitting in car $B$ will see car $A$...


2

Let's say you are observing from your point of view two objects, traveling in their own directions and at their own speeds. So you have two velocity vectors $\vec{v}_1$ and $\vec{v_2}$. By subtracting one from the other, you get the relative velocity between those objects: $\vec{v}_2 - \vec{v}_1$ will be the velocity of object 2 as observed by object 1 (...


0

You have set up a false comparison. You are computing two different energies. In the relativistic case, when you you write $$E^2=m^2+p^2, $$ for a single particle, the energy term is the sum of the kinetic energy and the mass energy: $$E=K+m.$$ In the non-relativistic situation you have defined the energy simply as the kinetic energy and ignored the ...


0

That's because force is a vector quantity. Unlike a scalar quantity like for instance temperature, a force always has magnitude and direction. Therefore when adding two forces, we have to add them like vectors. If you're interested in why we use vector addition for forces, this answer, which draws from Newton's Principia Mathematica, might be for you. A ...


2

Your source should've worded that more carefully. This is not the acceleration of the particle in the rest frame; instead, it's the acceleration in a momentarily comoving inertial frame, that is, an inertial frame in which the particle is momentarily at rest. In the actual rest frame, which is a non-inertial frame, the velocity and acceleration of the ...


0

If you are able to specify acceleration as a function of position or velocity, then it is still true that $$a=\frac{\text dv}{\text dt}=\frac{\text d^2x}{\text dt^2}$$ because this is just the definition of acceleration. You encounter the position case in things like simple harmonic motion (like a mass on a spring). Then you can think of acceleration as a ...


1

Acceleration is the derivative of velocity with respect to time, by definition. It doesn't matter what factors affect it, that's still what it is. You could have acceleration as a function of the amount a spring is stretched, acceleration as a function of how much you press the gas pedal, acceleration of a sail boat as a function of how fast the wind is ...


0

First, some information to motivate my answer. In circular motion, it is useful to break forces into components: the radial component $F_r$ (points towards or away from the center of the circle around which the particle is moving) and the tangential component $F_\theta$ (points tangent to the circle around which the particle is moving). For planar motion, ...


0

Average speed has its main meaning in situations when the actual speed is almost constant in the observed time period, i.e. it doesn't vary too much in the substantial part of the observed period. (As you can see, it is very subjective — what are “too much” and “substantial part”? Statisticians would probably use something as e.g. 5%.) when total time is ...


4

Why do we use velocity instead of speed for different physics problems? I recognize how they are different but why use one over the other? My cottage is 100 km due north of my house in Toronto. If I drive 100 km/h, will I arrive at my cottage in an hour? What if I drive east? What if I start in Montreal? Thus velocity.


4

On the motorway, your passenger asks: "How fast are you going?". You look at the speedometer and answer: "109 km/hr". Had he asked 1 minute later, then you would have said: "111 km/hr". Had he asked 1 minute earlier, then you would have said: "110 km/hr". You are telling him the instantaneous speed. But what if he asks you afterwards: "How fast did we ...


0

This average speed / velocity doesn't give accurate information about motion of an object then why it is taught? Primarily because that's what you'll actually use in the real world. For example, if you drive for an hour and cover 60 miles, I suspect you'll describe that you were "averaging 60 mph". I am sure that on such a trip there were times where ...


1

You have a few misconceptions. The friction which you draw in the AB direction actually acts in the opposite direction, toward the center of the curve. This friction component is the force component which has magnitude of precisely $mv^2/r$. That $mv^2/r$ value simply tells the magnitude of force necessary radially in order to travel a certain curved path at ...


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