What is Position vs. time: Definition and 16 Discussions
In mechanics, the derivative of the position vs. time graph of an object is equal to the velocity of the object. In the International System of Units, the position of the moving object is measured in meters relative to the origin, while the time is measured in seconds. Placing position on the y-axis and time on the x-axis, the slope of the curve is given by:
v
=
Δ
y
Δ
x
=
Δ
s
Δ
t
.
{\displaystyle v={\frac {\Delta y}{\Delta x}}={\frac {\Delta s}{\Delta t}}.}
Here
s
{\displaystyle s}
is the position of the object, and
t
{\displaystyle t}
is the time. Therefore, the slope of the curve gives the change in position divided by the change in time, which is the definition of the average velocity for that interval of time on the graph. If this interval is made to be infinitesimally small, such that
Δ
s
{\displaystyle {\Delta s}}
becomes
d
s
{\displaystyle {ds}}
and
Δ
t
{\displaystyle {\Delta t}}
becomes
d
t
{\displaystyle {dt}}
, the result is the instantaneous velocity at time
t
{\displaystyle t}
, or the derivative of the position with respect to time.
A similar fact also holds true for the velocity vs. time graph. The slope of a velocity vs. time graph is acceleration, this time, placing velocity on the y-axis and time on the x-axis. Again the slope of a line is change in
y
{\displaystyle y}
over change in
x
{\displaystyle x}
:
a
=
Δ
y
Δ
x
=
Δ
v
Δ
t
{\displaystyle a={\frac {\Delta y}{\Delta x}}={\frac {\Delta v}{\Delta t}}}
where
v
{\displaystyle v}
is the velocity, and
t
{\displaystyle t}
is the time. This slope therefore defines the average acceleration over the interval, and reducing the interval infinitesimally gives
d
v
d
t
{\displaystyle {\begin{matrix}{\frac {dv}{dt}}\end{matrix}}}
, the instantaneous acceleration at time
t
{\displaystyle t}
, or the derivative of the velocity with respect to time (or the second derivative of the position with respect to time). In SI, this slope or derivative is expressed in the units of meters per second per second (
m
/
s
2
{\displaystyle \mathrm {m/s^{2}} }
, usually termed "meters per second-squared").
Since the velocity of the object is the derivative of the position graph, the area under the line in the velocity vs. time graph is the displacement of the object. (Velocity is on the y-axis and time on the x-axis. Multiplying the velocity by the time, the time cancels out, and only displacement remains.)
The same multiplication rule holds true for acceleration vs. time graphs. When acceleration is multiplied
So let's assume an object moves along a straight line relative to some fixed origin. Clearly we can study this motion with the help of a position vs. time graph which shows how the position varies as time goes on. Now, as far as I understand, the slope of this graph at any time t gives the...
Homework Statement
Describe how to use the position vs. time graph to determine the numerical value of the particle's acceleration.
Homework Equations
None, these are Lab questions after we released a cart on a track on an incline.
The Attempt at a Solution
I know that you can derive the...
Homework Statement
Given the following graph of displacement vs time for an object moving in a straight line (assume const accel):
Find the acceleration between t=0 and t=4
Homework Equations
A= ((vi-vf)/2)/time
The Attempt at a Solution
I've tried find the area of t=0 to t=4 in order to...
Homework Statement
Three objects can only move along a straight, level path. The graphs below show the po- sition d of each of the objects plotted as a function of time t.
The magnitude of the velocity ∥⃗v∥ of the object increases in which of the cases? (view image/attached file)
Homework...
How could you graph a potential energy vs. time graph only knowing the position vs. time graph and the velocity vs time graph for a hanging object oscillating up and down on a string?
Homework Statement
The graph is of a runner-
time(min) is the x
position(x1000 m) is the y
these are the points
0,0
10,3
20,3.5
30,4.5
The question is- For the time interval between 10 min and 20
min, what is the runner’s displacement?
Answer in units of m
Homework...
And visa versa. And I'm only in 9th grade, so please don't give me any complicated answers or links. I just want to know simple stuff like, "If there is a straight horizontal line in a velocity vs. time graph, how would that look like on a position vs. time graph?" Basically, I just want to know...
Homework Statement
A motion detector transmits a click which is reflected off of a moving car. The echo is received by the motion detector after .0026 seconds. the motion detector sends out a second click .1 seconds after the first click. The echo for the seconds click is received after...
I just completed a lab on the acceleration due to gravity and I have to answer a few questions about my data. Before I share what I am confused about I will give a quick run through of the lab procedures:
--First I used a spark timer to mark a strip of paper with a clamp...
Homework Statement
i am having trouble converting an angular v vs. time to position vs. time graph. i know that the slope of the position is the velocity but the velocity graph has line going up and down.
Homework Equations
The Attempt at a Solution
v v.s t = slope of p vs. t
Homework Statement
http://img364.imageshack.us/my.php?image=90198228dt0.jpg
Image: http://img364.imageshack.us/my.php?image=90198228dt0.jpg
Which of the following statements are correct/incorrect?
1) The velocity is zero at point 4.
2) At point 6 the velocity is negative.
3) At point 3 the...
Homework Statement
http://img370.imageshack.us/img370/9488/96191446sb5.jpg
Evaluate Ax, the x-component of the acceleration between M and S.
Homework Equations
a=x/t^2The Attempt at a Solution
M= 48 m at 8.5 s
S= 40 m at 11.5 s
so change of x=-8 m, and change of t = 3s
so a= -8 / (3^2), so...
The position vs. time graph is wavy and I assume the only point where there is acceleration is where there is a curve, right? It seems like the acceleration is also 0 at the curve though. Is it even possible?
quick question ... if i have constant negative force being applied to an object, what would the curve be on its position-time graph ...
im pretty sure that its got to go upwards since distance can't be reduced, so i was thinking it might be something like the graph of logx, but starting from...
Guys,Please Help me with this question.It would be very very greatful.
Explain what physics quantity the slope of the position vs. time graph represents?:confused:
I tried but i couldn't find it so..Please Help me.i need to write it in my HW.and its due on 17th Jan.
Thank You.