x versus t graph, simple kinematics?

[WillParadigm]

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:

[physicsuser]

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.... ???

[WillParadigm]

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...

[WillParadigm]

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

[physicsuser]

what does it ask you to find? accelaration or distance?

accelaration is speed/ time and distance is speed * time..

[WillParadigm]

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

[WillParadigm]

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

[physicsuser]

ok so it is distance vs time

distance = speed * time

from that graph speed = y-axis and time= x-axis.

[WillParadigm]

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)

[physicsuser]

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

[WillParadigm]

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

[WillParadigm]

couldn't the final velocity be any point along the line in the above graph? see that's whats 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

[WillParadigm]

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...

[physicsuser]

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.