X versus t graph, simple kinematics?

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The discussion focuses on graphing an x versus t graph for a motion problem involving changing velocity. Participants express confusion about how to represent the motion accurately, noting that the graph should be curved due to acceleration changes. They reference the need for equations related to distance and acceleration to derive the graph, with some suggesting using the formula x = Initial Velocity x Time + 1/2 x Acceleration x Time^2. There is a consensus that the graph will not be a straight line and will include various points based on calculated distances. Overall, the conversation highlights the challenges of visualizing kinematic concepts without a textbook or clear guidance.
WillParadigm
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can someone help me graph this?

http://www.physics.drexel.edu/courses/Physics-152/probgraph.gif

I can understand that, since the velocity is changing, the x versus t graph should not be a straight line, that it is curved, but that's about all I can figure out.

I don't have a textbook, because the bookstore is backordered on all of them, plus this was not covered in the first lecture, so all I can figure out is that it's some curve...

Please help me... :confused:
 
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from 0 to 50 the object speeds up. from 50 to 90 it slows down. from 90 to 105 it is at rest. 105> it speeds up... ?
 
yeahh, some of us were confused too, and did what the instructions on the pic show, but we're supposed to graph an x versus t graph of that picture showing the velocity, so it looks somewhat like an exponential growth graph...
 
that's all I can get, just that it curves up from 0 but then I know there are other dips and peaks, and a level off since the velocity goes to zero, but I don't know how to draw it
 
what does it ask you to find? accelaration or distance?

accelaration is speed/ time and distance is speed * time..
 
the question was this exactly...

"7 Chapter 2, Question 22,and plot an x vs t graph of the motion."

so I mean, the directions aren't that helpful, plot an x vs t graph of the motion
 
the book I have does have a drawing of a x versus t graph, but it derives it from already having acceleration... and used the equation

x=Initial Velocity x Time + 1/2 x Acceleration x Time^2
 
ok so it is distance vs time

distance = speed * time

from that graph speed = y-axis and time= x-axis.
 
wait, now you lost me... how does that help me graph something in

x intervals of Time t (s) and y intervals of Position x (m)
 
  • #10
WillParadigm said:
the book I have does have a drawing of a x versus t graph, but it derives it from already having acceleration... and used the equation

x=Initial Velocity x Time + 1/2 x Acceleration x Time^2

yes that is the equation for distance...

from the graph above acceleration= [V(final)-V(initial)]/[t(final)-t(initial)]
in other words a= (V2-V1)/(t2-t1)
initial velocity= 15m/s

just graph velocity*time from the given graph above
 
  • #11
so it's (10-15)/(120-0) to get acceleration of -1/24

and I just graph

15x + (1/2)(-1/24)x^2? all I got was a straight line
 
  • #12
couldn't the final velocity be any point along the line in the above graph? see that's what's confusing me the most, that there aren't any concrete numbers that I can see to use

I mean, for the first part of the graph, I can see yes, the equation should be

y=1/2x+15

because the slope is 1/2 and picking a point (10,20) plugging that in, got me the y intercept... but that just happened to work out
 
  • #13
but that y intercept equation, I don't see how it helps, it just gives me an equation for a line I already saw, but not even the whole line...
 
  • #14
d= v0xt-0.5 x a x t is good for uniform acceleration

so use d= (v2-v1)* 0.5 (t2-t1)

so
v2=20
v1=12
t2=10
t1=0
d1=0.5*(20-12)*(10-0)=40

so you graph (10,40)

next

v2=25
v1=20
t2=20
t1=10
d2=0.5*(25-20)*(20-10)=25
total distance= d1+d2=40+25=65

graph (20,65)

and so on...

I hope this is right otherwise I'll go insane.
 

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