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1 vote
3 answers
217 views

Why does a particle initially at rest at origin with acceleration as square of its $x$ coordinate ever move?

Consider a particle initially at rest at origin, with acceleration, $a$, such that $ a(x)=x^2$. Since the particle is at origin, initial acceleration would be 0. It's also at rest initially. Its $x$-...
brainfreeze's user avatar
1 vote
5 answers
148 views

The value of $g$ in free fall motion on earth [closed]

When we release a heavy body from a height to earth. We get the value of $g=9.8 \ ms^{-2}$. Now, I'm confused about what it means. For example, does it mean that the body's speed increases to $9.8$ ...
Junaid's user avatar
  • 27
1 vote
3 answers
12k views

Does the SUVAT equations of motion (Kinematics) come from some differential equation?

Wikipedia says about the equations of motion that; "If the dynamics of a system is known, the equations are the solutions for the differential equations describing the motion of the dynamics.&...
Shadman Sakib's user avatar
1 vote
2 answers
546 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
Curious 's user avatar
1 vote
2 answers
111 views

Why isn't tangential acceleration just $a$?

If the tangential acceleration is $\mathrm d|v|/\mathrm dt$ then isn't it just the magnitude of the acceleration of the object because $\mathrm dv/\mathrm dt$ is acceleration?
CatsOnAir's user avatar
  • 139
1 vote
2 answers
129 views

Time derivative of unit velocity vector?

Let's say I have some parametric curve describing the evolution of a particle $\mathbf{r}(t)$. The velocity is $\mathbf{v}(t) = d\mathbf{r}/dt$ of course. I am trying to understand what the expression ...
James Thiamin's user avatar
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
1 vote
5 answers
152 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 $t$ ...
user231094's user avatar
1 vote
2 answers
2k views

Why is there a $\frac{1}{2}$ in the kinematic equation? [duplicate]

In a few of the kinematic equations there is a $2$ or a $0.5$ coefficient. Why is this? For example the kinematic equation for distance is: $$\text{previous velocity} * \text{time} + \frac{1}{2} * \...
Philogy's user avatar
  • 115
1 vote
7 answers
281 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} ...
Kalcifer's user avatar
  • 329
1 vote
3 answers
62 views

Motion in a plane situation

There is something weird I find about the following situation. Suppose a particle has the $X$-coordinate $= 2+2t+4t²$ and $Y$-coordinate $= 4t+8t²$. So it's velocity in $X$ is $2+8t$ and velocity in $...
Qwerty's user avatar
  • 45
1 vote
2 answers
147 views

Determining how long it takes an object to reach a certain speed [closed]

Robotics related. On a linear servo driven rail one can typically set acceleration and maximum move speed. I am trying to determine the amount of seconds it takes the load to accelerate to a certain ...
Eric's user avatar
  • 141
1 vote
3 answers
90 views

How to deal with functions of kinematic quantities not defined in terms of time?

How do I deal with functions of kinematic quantities which are not defined with respect to time? For instance, given acceleration as a function of velocity or displacement, how would I go about ...
Kishor Kunal's user avatar
1 vote
1 answer
431 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 ...
Lalit Tolani's user avatar
0 votes
5 answers
1k views

Why do kinematic equations only work with constant acceleration?

People say that the equations are derived assuming a constant acceleration. I just don't see how this is the case. (I am new to calculus.)
AVS's user avatar
  • 324

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