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-1 votes
1 answer
103 views

How to Find Trajectory of Particle?

Let’s say I have a particle, and I know all the forces acting on it at every position. (Let’s say the particle is in an electric/gravitational field to simplify the mathematics involved.) Now, is ...
V T Naveen Mugundh's user avatar
-2 votes
1 answer
94 views

What is $V$ in $a$=$V$$dv$/$dx$? [duplicate]

$a$=instantaneous acceleration $V$=instantaneous velocity $x$=position $dx$=small Chang in position $a$=$dv$/$dt$ multiplying numerator and denominator by $dx$,we get $a$=$dv$.$dx$/$dx$.$dt$ now we ...
Aakash's user avatar
  • 161
0 votes
2 answers
268 views

Circular motion equivalent in three dimensions [closed]

Are there equations or even a concept of circular motion/tangential acceleration/centripetal acceleration in three dimensions? Maybe something called "spherical acceleration"? or am I just ...
RhinoPak's user avatar
15 votes
3 answers
4k views

Why does solving the differential equation for circular motion lead to an illogical result?

In uniform circular motion, acceleration is expressed by the equation $$a = \frac{v^2}{r}. $$ But this is a differential equation and solving it gets the result $$v = -\frac{r}{c+t}.$$ This doesn’t ...
Jmh's user avatar
  • 153
0 votes
3 answers
165 views

Power and work contradiction

A body is starting from rest. A force is acting on it for a short period of time. In that given time, power delivered to it at any instance $t$ is given $$P = F \cdot v_1 = ma \cdot v_1 = mv_1^2/t,$$ ...
Priyesha Jha's user avatar
0 votes
1 answer
141 views

Differential Equation & MacLaurin Series for Newton’s Second Law

I am currently working with a differential equation, where I think I need to take the derivative of $ma$ (corrected as per comment). I am trying to write $F = ma$ as a MacLaurin series and eventually ...
Yelena's user avatar
  • 151
1 vote
2 answers
2k views

Acceleration as a function of position and time

I know if you have an acceleration as a function of $t$, $a(t)$, to find the velocity you simply integrate $a(t)$ with respect to $t$. Moreover, if the acceleration was a function of position, $a(x)$, ...
Tim CP's user avatar
  • 33
0 votes
0 answers
370 views

Why don't we define time derivative of acceleration? [duplicate]

When we started the study of kinematics we defined position and its change with respect to time. After that we defined time derivative of velocity which gave us acceleration. These 3 concepts really ...
Shreyansh Pathak's user avatar
-1 votes
1 answer
207 views

Is there any reason why acceleration should not be the first derivative of the absolute value of velocity? [closed]

I ask mainly because I am not familiar enough with newtonian mechanics and higher-level physics in general to know the repercussions of such a change, but on the simpler plane of existence, I have ...
Azorack's user avatar
24 votes
7 answers
12k views

Zero velocity, zero acceleration?

In one dimension, the acceleration of a particle can be written as: $$a = \frac{dv}{dt} = \frac{dv}{dx} \frac{dx}{dt} = v \frac{dv}{dx}$$ Does this equation imply that if: $$v = 0$$ Then, $$\...
7453rfg's user avatar
  • 383