All Questions
28 questions
-2
votes
0
answers
70
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Use of $dv/ds$ in defining acceleration [duplicate]
We can write acceleration as either
$dv/dt$ or $v dv/ds$.
And surprisingly the work-energy theorem arrives from the second definition.
I feel it would be fundamentally understanding towards work ...
0
votes
1
answer
89
views
In $a = dv/dt$, is $a$ the net acceleration? [closed]
While going through the calculus approach to accelerate, we have,
$$a = dv/dt, $$
I think, here, v and a should be in the same axis,
is my process correct?
in a planar motion in two dimensions, it ...
- 1
-2
votes
3
answers
96
views
Why is it wrong to find centripetal acceleration using change of velocity over change of time?
This question asks to find the centripetal acceleration by giving the initial and final velocity over the change of time.
As shown, my book combined two rules to find the acceleration. I utterly ...
- 377
-2
votes
2
answers
122
views
Why does $\vec{a}=\vec{\omega}\times \vec{r}$ as well as the velocity does?
Today I came in class and in one of the problems the teacher used $\vec{a}=\vec{\omega}\times \vec{r}$ which made me very confused because I don't know where it comes from, it seems pulled out of thin ...
- 69
0
votes
0
answers
45
views
Physical and Diagrammatic representation of $a$=undefined when $v$=0 according to $a$=$vdv$/$dx$
$a$=acceleration
$v$=velocity
$x$=position along x axis
$t$=time instant
My teacher derived the $a$=$v$$dv$/$dx$ formula as follows
Assume a particle at time $t$ is at $x$ position having $v$ velocity
...
- 31
0
votes
2
answers
414
views
Why does tangential acceleration become 0 when the velocity is max? [closed]
I know that tangential acceleration equal to zero when the circular motion is uniform, but why is it equal to zero, when the velocity is max or min? Because there is no relation between the value of ...
- 11
0
votes
1
answer
43
views
Are terms tangential acceleration and normal acceleration only used for instantaneous velocity?
Are terms tangential acceleration and normal acceleration only used
for instantaneous velocity?
- 23
-1
votes
2
answers
67
views
Instantanous and uniform velocity and acceleration [closed]
If the mathemical expression of instantanous velocity is $d/t$, what is the mathematical expression of uniform velocity.
If the mathematical expression of instantanous acceleration is $v/t$, what is ...
- 107
1
vote
7
answers
293
views
I'm having trouble understanding the intuition behind why $a(x) = v\frac{\mathrm{d}v}{\mathrm{d}x}$ [duplicate]
I was shown
\begin{align}
a(x) &= \frac{\mathrm{d}v}{\mathrm{d}t}\\
&= \frac{\mathrm{d}v}{\mathrm{d}x}\underbrace{\frac{\mathrm{d}x}{\mathrm{d}t}}_{v}\\
&= v\frac{\mathrm{d}v}{\mathrm{d}x}
...
- 339
0
votes
1
answer
87
views
How do I reconcile these two definitions of acceleration?
How do I reconcile these two definitions of acceleration?
$$a=\frac{d\bar{v}}{dt}=(\frac{dv^k}{dt}+v^i v^j \Gamma^k_{ij})\bar{e}_k \tag{1}$$
and
$$a^k=v^{\small\beta} \nabla_{\small\beta} v^k.\tag{2}$$...
- 971
3
votes
2
answers
160
views
Acceleration in terms of displacement
I am having problems understanding the derivation of acceleration in terms of displacement. The first step is fine:
$$a(x) = \frac{\mathrm dv(x)}{\mathrm dt}
= \frac{\mathrm dv(x)}{\mathrm dx} \frac{\...
- 131
2
votes
1
answer
435
views
When exactly does velocity increase or decrease on an acceleration time graph? [closed]
How does the acceleration time graph show if and object is speeding up or slowing down?
Is it possible to find the answer without any deep calculations? If yes then how?
Like how can I find the ...
- 137
0
votes
2
answers
353
views
Why isn't tangential acceleration just always 0?
This is probably a very stupid question but I can't help me.
Tangential acceleration is $\vec{a_t}=\frac{dv}{dt}\frac{\vec{v}}{v}=\frac{\vec{v} \cdot \vec{a}}{v} \frac{\vec{v}}{v}$. Since $\vec{a}$ is ...
- 15
0
votes
1
answer
129
views
Why intuitively is the tangent vector the derivative of velocity of position with respect to their modulus?
When trying to find the tangential velocity, many textbooks define the tangent direction as one of the following:
or
Intuitively, why is the tangent vector the derivative of the position with ...
- 849
1
vote
2
answers
319
views
What is the time derivative of the linear velocity vector $\vec{v}\,(t)$?
If $\vec{v}\,(t)$ denotes linear velocity, we can then write $\vec{v}\,(t)$ as $|v(t)|\hat{v}$. My question is what is $\displaystyle\frac{d\vec{v}\,(t)}{dt}?$
The answer I have seen to this question ...
- 39
0
votes
1
answer
42
views
Is such a situation realistically possible where $v$-$t$ graph is continuous but $a$-$t$ graph is not?
Taking for example $v = \cos(t-1)$ from $t \in [0,1]$ and $v = e^{t-1}$ from $t \in (1,\infty)$ and $t \ge 0$. At $t = 1$, the function shifts from cosine to exponential, but remains continuous since ...
3
votes
2
answers
233
views
Generalization of straight line motion under constant acceleration
My question is that, we all know the three equations of straight line motion under constant acceleration,
\begin{align}
x & =x_{\rm o}+v_{\rm o}\,t+\tfrac12 \mathrm a\,t^2
\tag{1d-a}\label{1d-a}\\
...
1
vote
1
answer
459
views
Expressing acceleration in terms of velocity and derivative of velocity with respect to position
we know that
$$a = \dfrac{dv}{dt}$$
dividing numerator and denominator by $dx$, we get $$a=v\dfrac{dv}{dx}$$ provided that $dx$ is not equal to zero or instantaneous velocity not equal to zero
when I ...
- 892
0
votes
3
answers
1k
views
How does gradient give $g$?
How is
$$g=-\nabla V$$
where $V$ is gravitational potential and $g$ is acceleration due to gravity.
I am new to calculus.
0
votes
3
answers
232
views
Are acceleration and velocity simultaneous? [closed]
I would think yes because, if a rope tied to a swinging rock breaks, the rock flies off in the direction that is perpendicular to the direction of the last instant of the acceleration. The ...
- 71
1
vote
2
answers
557
views
In the equation: $a = dv/dt$ , is $dt$ the time taken to achieve that instantaneous acceleration?
If you solve for $dt$ from $a = \frac{dv}{dt}$ , is it the time taken to to achieved that instantaneous acceleration?
$a$ : acceleration
$v$ : velocity
$t$ : time
- 87
9
votes
4
answers
2k
views
Can I find the acceleration or velocity when my displacement-time graph is discontinuous?
Today, I encountered the problem where I was asked to find the velocity and acceleration from displacement-time graph but the displacement-time graph was discontinuous. So I am unable to find the ...
- 123
11
votes
4
answers
3k
views
When the direction of a movement changes, is the object at rest at some time?
The question I asked was disputed amongst XVIIe century physicists (at least before the invention of calculus).
Reference: Spinoza, Principles of Descartes' philosophy ( Part II: Descartes' Physics, ...
user229097
5
votes
2
answers
2k
views
How does instantaneous velocity or acceleration have any other numerical value than 0? [duplicate]
Instantaneous velocity is defined as the limit of average velocity as the time interval ∆t becomes infinitesimally small. Average velocity is defined as the change in position divided by the time ...
- 163
6
votes
6
answers
1k
views
Question about derivation of kinematics equations
Apologies if this has been asked before, but I browsed the sub and couldn't find something specific.
I understand the derivation for one of the equations as follows:
\begin{gather}
\frac{dv}{dt} = a ...
- 69
1
vote
5
answers
158
views
Equation of distance and time
How is this equation derived?
$$r = r_0 + ut + at²/2$$
where $r_0$ is the initial position of particle and $r$ is the position of the particle after all the motion it has undergone, $a$ and $t$ ...
- 131
0
votes
1
answer
58
views
Change of variable in function
Suppose I have a function $h(\theta)$ measuring the height of a piston, with $\theta = \omega t$. I would like to know the vertical acceleration of this piston as $\omega$ changes at the point $\theta ...
- 1,420
11
votes
2
answers
3k
views
Kinematic equation as infinite sum
I'm not sure exactly how to phrase this question, but here it goes:
$v=\dfrac{dx}{dt}$ therefore $x=x_0+vt$
UNLESS there's an acceleration, in which case
$a=\dfrac{dv}{dt}$ therefore $x=x_0+v_0t+\...
