# Tag Info

0

Speed has no direction while velocity does. For example, if I say that I'm running at 10 mph, I have given you my speed. If I say that I'm running 10 mph north, then I have given you my velocity. Acceleration is change in velocity. Imagine this: I am in my car and you look at me before I even press the gas pedal. You close your eyes then open them (imagine ...

1

The effects of inertial acceleration is best understood from the physics of Newtonian mechanics. A good site to support your understanding of this physics: http://hyperphysics.phy-astr.gsu.edu/hbase/traj.html I recommend studying the trajectory calculator as it applies to your problem.

1

I do not know where you read that, but "net acceleration of the body is = acc.due to gravity - acc. of the body" does not much sense to me. Usually, you calculate the net acceleration of a body once you know what forces are acting on it. For a body under free fall, the acceleration will be "g", the acceleration of gravity because it is the only force that ...

0

The acceleration of the plane will cause an offset in the motion of the mass/spring system; if it was centered on x=0 when the plane was stationary, its motion will now be centered around $x = -\frac{m\cdot a}{k}$ where $k$ is the spring constant, and $F = m\cdot a$ is the force needed for the mass to accelerate with the plane. If the plane is accelerating ...

0

There are various ways of calculating acceleration and jerk from your data set. THe value of $dt$ you want is the time step generally, in your case $100$ ms. The method that you suggest $$a=(v2-v1)/dt$$ will give you an answer and is reasonnable, but it will not give you the best value I expect. It might be better to use something like Savitsky-Golay ...

3

The total force acting on a raindrop equals $g$ minus air resistance which increases with velocity. In other words, as the raindrop speeds up, air resistance increases which decreases the acceleration (until eventually the acceleration equals zero and terminal velocity has been reached).

4

Who says all derivatives have to change continuosly? With a suitable setup, acceleration can jump from zero to something. Location and velocity can not change discontinuously, but acceleration can. No, there are no "infinitely many derivatives of velocity". In your question, you write "acceleration has to increase from zero and therefore the third ...

3

Have a look here: https://en.wikipedia.org/wiki/Non-analytic_smooth_function There are functions which are identically zero for negative arguments, non-zero for positive arguments and smooth everywhere.

25

Yes, if an object is stationary, then starts moving in the positive direction, all derivatives of the position can become positive when the object starts moving. But so what? It's not a paradox. It's just a mathematical statement. All the derivatives are positive. No big deal. It's hard to tell why you find this confusing. You might be thinking that there ...

0

Use Newton's 2nd law. F=ma over here a= 5+10 (external acceleration provided and gravitational acceleration) Check out related videos in youtube.You will understand the concept behind this problem.(I have given a link of one video below) https://www.youtube.com/watch?v=hPEx3gxtPK4

2

You can achieve this in two distinct ways: 1) use a reference (I'll give an example below) 2) measure other quantities that can be converted into an acceleration. Here are some possibilities: 1a) Gravity. At rest, the accelerometer should measure the local g. 2a) Put it on a rotating platform. Knowing the rotational speed and distance to the pivot, you ...

1

force = acceleration * mass, hence acceleration will be $a=50N/22kg \approx 2.27m/s^2$ Distance it moves might be found by integration: $\int_0^{1.2}v(t)dt=\int_0^{1.2}atdt$, since speed $v(t)=at$ Answer to (B) then is 1/2*1.2*(2.27*1.2)=1.63, which seems pretty close to what you have got

0

Yes you can! For example: when you are sitted, doing nothing. From the reference system of the earth you are firm, with no velocity BUT there's the gravity acceleration. We can say that we have everytime a potential to have a velocity if we have a force. For the reason that says that for every force there's an other force with same value but with a contrary ...

0

I think that the fact that the clocks in the two ships appear synchronized to the initial observer in S distracts us and makes us think of the situation in non-relativistic terms. In general, most SR paradoxes can be eased by remembering the relativity of simultaneity. Let's say that the 2 spaceships have clocks at their centers, and the observer in S finds ...

0

Light speed is something of a cosmic speed limit. Nothing can exceed the speed of light, and only massless particles can travel at light speed. Any particle with mass would require an infinite amount of energy to accelerate to the speed of light.

3

I quite sure it's not theoretically possible. Without doing any actual calculations, I recall that accelerating a massive object to light speed would require an infinite amount of energy. Some energy certainly can be gained with the "slingshot" method, but definitely not an infinite amount. Specifically, it's the Lorentz factor that prevents objects from ...

1

The answer is, you'll have to be more specific. When a fist hits a bag, the force varies a lot in terms of time. If you imagine trying to punch a bag as slowly as possible, there is a slow onset as your fingers just graze the leather, followed by something more "meaty" as your bones start to press against your skin which is pressing against the actual ...

0

Imagine the rigid body was rotated around the accelerometer. It would be hard to detect to motion (though probably not impossible) making one need two.

1

Assuming the total force is $10~\text{N}$, the $7~\text{kg}$ mass is accelerated by $1.0~\text{m/s^2}$, so it is experiencing $7~\text{N}$ force in x-direction. Therefore, $F_x = .7F$. The angle is roughly $45^\circ$.

0

Your $2.0 \frac{m}{s^2}$ is the acceleration in the direction of the force, at $60°$ angle with the x-axis. You need to find the component of the acceleration in the direction of the x-axis. If your acceleration is completely in the y-direction, your object will have $0$ acceleration in x-direction. If it is completely in the x-direction, it will have those ...

1

I want to take another tack than that of the other answers. This will be one big handwave rather than a rigorous mathematical argument, but I hope it gets the idea across intuitively. First off, as I noted in a comment, and as hft notes, you are using "v" to mean both "velocity as a function of time" and "velocity as a function of position". That's ...

1

Einstein's equivalence principal states that an accelerated reference point is indecipherable from a reference frame in a gravitational field, so an accelerated reference frame will act in the same way that a gravitational field with the same acceleration would act. As for if all reference points are equally valid, the answer is generally "yes" with some ...

2

The coordinate velocity does indeed change discontinuously, but only if the acceleration changes discontinuously i.e. the jerk is infinite. Since for any physical system none of the time derivatives of position can be infinite, in a physical system the coordinate velocity can't change discontinuously. But let's ignore this for now and examine why we get a ...

-1

Initially a small force needs to be applied to get the desired velocity but this small force is neglected and we start our observation after the body acquires the velocity and the force is removed .

0

Note that when you apply chain rule , you assume dx not to be zero . This will clear it up for you .

5

You can apply chain rule if $v$ is differentiable wrt $x$ and $x$ is differentiable wrt $t$. I think there are no other conditions,as this post on MathSE seems to say, http://math.stackexchange.com/questions/688152/necessary-conditions-for-the-chain-rule-of-differentiation-to-be-valid#= and this condition is not always available. When $v=0$,make sure ...

7

No, it doesn't imply that $a = 0$. If, at some value $t = t_0$, the acceleration is non-zero while the velocity is zero, the position function is either a minimum or maximum. That is, $x(t)$ is stationary there: $$x(t_0 + dt) = x(t_0)$$ which means that at $t = t_0$ $$\frac{dx}{d\dot x} = \frac{dx}{dv} = 0$$ thus $\frac{dv}{dx}$ is undefined at $t = ... 22 The correct thing to say would be that "if v=0 and dv/dx is finite then a=0". A simple example, to help illustrate what's going on, is the well known case of constant acceleration "-g" near the earth's surface. In this example, we consider "x" to be the height above the ground, and assume the initial x is zero. In this case $$x=-\frac{gt^2}{2}+v_0t$$$\$ ...

0

This exceeds the speed of light, which seems incorrect. During acceleration, the speed of light may seemingly be exceedet from the viewpoint of the accelerated observer. This is why one talks about "uniform relative velocity" when talking about inertial frames, in which the speed of light may not be exceedet. When you accelerate, you change your ...

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