All Questions
9
questions
-1
votes
1
answer
66
views
Interpretation of velocity-velocity and acceleration-acceleration curves
I am parametrizing equations of motion in the form:
$$x(t) = x_0+v_{0,x}t\\y(t) = y_0+v_{0,y}t+\frac{1}{2}at^2$$
The parametrized equation with respect to time:
$$y(x) = y_0+v_{0,y}\cdot \frac{x - x_0}...
- 135
3
votes
2
answers
156
views
Acceleration in terms of displacement
I am having problems understanding the derivation of acceleration in terms of displacement. The first step is fine:
$$a(x) = \frac{\mathrm dv(x)}{\mathrm dt}
= \frac{\mathrm dv(x)}{\mathrm dx} \frac{\...
- 131
1
vote
2
answers
884
views
Why is position proportional to time squared?
Now I know some of the obvious answers to this, such as if you integrate the acceleration twice, you’ll get time squared, but what I’m really looking for is more of an intuitive answer.
One of the ...
- 11
3
votes
3
answers
2k
views
How do acceleration, velocity, and displacement affect/relate to eachother?
I have been wondering this since learning about position, velocity, and acceleration vs time graphs but can't put numbers/equations to it.
I know that acceleration acts to change velocity, shown by ...
- 169
3
votes
4
answers
2k
views
If displacement is 0, does that mean initial velocity equals final velocity?
For instance, one of the kinematic equations is :
$$v_f^2 = v_i^2 + 2ad$$
where $v_f$ is final velocity, $v_i$ is initial velocity, $a$ is acceleration, and $d$ is displacement.
Say for instance a guy ...
- 33
0
votes
2
answers
85
views
Kinematic displacement: why not represent higher order rates of change?
I understand that the equation for kinematic displacement is:
$x = v_{0x}t+\frac{1}{2}a_xt^2$
Perhaps my understanding is naive, but it seems like this leaves out higher order rates of change. Why ...
- 111
0
votes
0
answers
93
views
Acceleration as the second derivative of displacement function
Let $x$ be displacement as a function of time $t$ and some other physical quantity $k$ such that
$ x = f(t,k) $
Now,
1) Will the acceleration $a$ be $\frac{\partial^2 x}{\partial t^2}$ or $\frac{d^...
- 23
0
votes
2
answers
985
views
Condition of acceleration when using $x=vt$
So my teacher told me that since $v = \delta x/ \delta t$, $\delta x = v • \delta t$ (naturally), and that is equal to the "area under the velocity-time graph", or displacement. This all makes sense ...
- 157
2
votes
2
answers
15k
views
Calculate displacement in position from knowing constant acceleration
I have recently started studying physics at school, and my teacher went over the following equation without explaining about it too much:
$$s=\upsilon_{0}t+\frac{1}{2}a t^2
$$
I have wondered, why ...
- 137