29
The problem of perception as to "What is a new Equation of Motion?" seems to originate with the dogmatic teaching of the Three Equations Of Motion as a Set of Results to be Learned.
They are in fact three results derived from the distillation of Newton's Laws:
$$\mathbf f = \dfrac {\mathrm d} {\mathrm d t} (m \mathbf v)$$
which differential ...
28
There's nothing wrong with your calculation. But plug in $v=u+at$ and you find:
\begin{equation}
x=(u+at)t-\frac{at^2}{2} = ut + \frac{at^2}{2}
\end{equation}
which is already one of your three equations of motion. So your teacher is probably just saying that it is redundant to call this a "fourth" equation, because it is essentially the same as ...
16
Formula $W=Fs$ is the formula for work, that is the energy transferred to the body by the force. It is the difference of kinetic energies between the final and initial state. If you're starting from $v=0$ and assume $E_{kin}(0)=0$, you have
$$
E_{kin}(v)
= E_{kin}(0) + ma \cdot \frac12at^2
= 0 + \frac12 mv^2
$$
If you start from the velocity $u$ and change ...
7
Just to add that when you use these equations it's good to think in a physical way as you go along. So:
$$
v = u + a t
$$
think: "yes the final velocity is the initial velocity plus the change owing to acceleration. This is basically the definition of $a$."
Next
$$
x = u t + \frac{1}{2} a t^2
$$
think: "the distance is travelled is how far it ...
6
You say that "I conceptualized it as the description of the path from the reference point to the location of the object" and you found that led to a nonsensical answer. Therefore your concept is flawed.
The location is the endpoint of the path, not the path itself. Indeed, it is the endpoint of infinitely many alternative paths.
In quantum physics ...
5
To demonstrate why you can't do this, let's step up one derivative, and explore an anecdote about why it doesn't much sense to add distance and velocity directly.
I ask you, "How far away is your house from Jill's house?"
You answer, "It is is five kilometers plus seventeen meters per second away." ($5 \,\text{km} + 17\,\text{m/s}$.)
I ...
5
Displacement which you understand, and time which you understand, form the ratio velocity which is displacement divided by time.
Consider an object that travels $100 m$ in ten seconds. That means its moved from its starting point to another point $100 m$ away in $10 s$. We then say that it’s velocity is
$$v = \frac{100m}{10s} = 10 m/s$$
meaning it travels $...
4
The general solution to
$$\ddot x+\omega^2 x=0$$
is
$$x(t)=A\cos\omega t+B\sin\omega t.$$
(It’s a second-order differential equation, so there should be two integration constants.)
If you want $t=0$ to be when the pendulum is at top of its swing, choose the cosine. If you want $t=0$ to be when the pendulum is at the bottom, choose the sine. For other cases, ...
3
It might be described as pedesis or as random motion.
The picture shows the velocities of many particles (as in a gas the molecules all move around). Velocity is a vector: it has a speed and it has direction. In 3 dimensions this means that it has three components mutually at right angles: for present convenience, call them up/down, left/right, towards/away.
...
3
You need an initial velocity. Otherwise, if the object starts at $x=0$ then $a=x=0$. Since $v=0$, there will be no motion.
3
Keep in mind there is a difference between a vector and a component of a vector. The velocity is a vector, the x-component of the velocity (or the "$x$-velocity" in the language your book uses) is just a component of a vector, which is a scalar.
The velocity vector can be expanded in terms of unit vectors $\mathbf{e}_x, \mathbf{e}_y, \mathbf{e}_z$ (...
3
My interpretation of the question is: is velocity something that really exists in nature or is it just a mathematical construct that we use for calculations?
In classical physics the distinction between reality and math is indeed somewhat blurred, and gives a taste of philosophy. However, from the quantum mechanical standpoint the distinction is very "...
3
When we do intuitive physics we use the categories our brains come "prepackaged" with, things like size, duration, force, warmness or colour. As we refine our understanding we start measuring these categories (sometimes splitting them, like colour into intensities at different wavelength). From these more formal variables we can derive other ...
3
What is 'Real'? How do you define 'Reality'?
This is not what we deal with in Physics... We just try to get an intuition with the physical quantities while dealing with them. So the question is about whether you can get an intuition for velocity or not.
Conservation of an entity is purely incidental and it's just a quality of the universe we live in.
3
Notice that the relative velocity is 2.0m/s. You know the bear was 26m behind the tourist. From this you can find the time the bear will take to catch up with the tourist. This is also the maximum time the tourist has to arrive at his getaway vehicle. Knowing his speed, you can then solve for the maximum distance he was alway from his vehicle.
3
Maintaining a constant velocity requires no energy.
3
First of I think it is great that your taking apart equations and putting them back together in a different ways as if they were LEGO pieces. I think most of us (with an interest in physics) did that in high school and it can be very insightful.
So instead of fretting over the characterization of $x = v t - \tfrac{1}{2} a t^2$, debating if it is an equation ...
3
I may be missing something, but this seems entirely straightforward and you are overcomplicating things. If the object moves parallel to the $Ox$ axis then it is a one-dimensional movement, so you can ignore all but the $x$ component.
The scalar velocity is then also the velocity along the $x$-axis, $|v|=v_x =t/(t^2+1)$. The acceleration is just the time ...
3
What is the value of acceleration when the string breaks?
The string will break when the tension in string reaches the tensile limit say ($T_0$).The tension in string for given acceleration can be calculated easily from FBDs.
Why does not the ball stays fixed in its initial position forever, so that the tension becomes high in the string, and it eventually ...
3
Assume the object hangs vertically before the bus starts to accelerate. When the bus starts to accelerate there is instantaneous relative movement of the object relative to the bus seen by observers both on the bus and on the ground. To the observer on the bus, the bus is not moving and the object moves due to the fictitious force present in the accelerating ...
3
There are several presentations of Zeno's paradoxes. One version is as follows (taken from Wikipedia):
Suppose Atalanta wishes to walk to the end of a path. Before she can get there, she must get halfway there. Before she can get halfway there, she must get a quarter of the way there. Before traveling a quarter, she must travel one-eighth; before an eighth, ...
2
No, you can't.m/s tells us how fast something is moving/ rate of change of distance and m/s/s tells us the rate of change of speed. we cant subtract them because these two are very different from each other.
2
Algebraic formula aside, we can see that $\langle \frac{1}{t} \rangle$ has dimensions of time$^{-1}$, or per unit time, which is a rate. As @Semoi pointed out in the comments we can look at your algebraic formula as well for a uniform probability measure:
$$\left\langle\frac{1}{t}\right\rangle_{\operatorname{Unif}([a,b])} = \frac{1}{b-a}\log\left(\frac{b}{a}\...
2
Reference : My answer there Need help understanding an equation of motion for a pendulum.
$=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!$
A choice for the general solution is the sinusoidal
\begin{equation}
x_{\bf s}\left(t\right)\boldsymbol{=}...
2
The $s$ in equation $W=mas$ can be replaced from 3rd equation of motion.
$$W=\frac{ma(v^2-u^2)}{2a}$$
$$W=\frac{1}{2}m(v^2-u^2) \rightarrow K.E.=\frac{1}{2}m(v^2-u^2)$$
Or else, if you still want to substitute $s$ from 2nd equation of motion:
$$W=ma(ut+\frac{1}{2}at^2)$$
$$W=ma(ut)+\frac{1}{2}m(a^2t^2)$$
From 1st equation of motion; $t=\frac{(v-u)}{a},\,\, ...
2
You need to distinguish between work done and absolute work done. Work done by one body on another can be negative. If the work done by $A$ on $B$ is negative, say $-W$, then this just means that $B$ does a positive amount of work $W$ on $A$.
For example, a compressed spring $B$ expands and accelerates a block $A$. We can say that $B$ does positive work $W$ ...
2
There is no sensical way to add or subtract them together since they are terms that have completely different units and mean totally different things.
Velocity is the rate of change of displacement whilst acceleration is the rate of change of velocity or in other words the rate of change of the rate of change of displacement. If we think of something that is ...
2
It's about how you define average. The formula $$ \bar{v} = \frac{\sum_i v_it_i}{\sum_i t_i}$$
is called a weighted average or a weighted mean, where times $t_i$ are the weights, as opposed to the arithmetic average/mean that is given by the first formula. In some contexts you need to use one, in some contexts the other. You can average over time (then you ...
2
We put $y = -h$ and $x = R$ in equation of trajectory.
$$-h = R\tan\theta - \frac{gR^2}{2u^2}\sec^2\theta$$
$$-h = R\tan\theta - \frac{gR^2}{2u^2}(1 + \tan^2\theta)$$
Now we have a quadratic equation in $\tan\theta$. As $\theta$ is real the determinant of quadratic equation should be always $\geq 0$. The problem is quite easily solved when you do $b^2 -4ac &...
2
The linear motion of an object can be used to determine one contribution to the angular momentum of the object relative to a point which does not lie in line with the velocity vector. Also, with an extended object, it is sometimes useful to work with an instantaneous axis of rotation. For that, the position vector from the axis to the center of mass must ...
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