Homework Help: X versus t graph, simple kinematics?

1. Sep 29, 2004

can someone help me graph this?

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

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...

Last edited by a moderator: May 1, 2017
2. Sep 29, 2004

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.... ???

3. Sep 29, 2004

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...

4. Sep 29, 2004

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

5. Sep 29, 2004

physicsuser

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

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

6. Sep 29, 2004

"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

7. Sep 29, 2004

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

8. Sep 29, 2004

physicsuser

ok so it is distance vs time

distance = speed * time

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

9. Sep 29, 2004

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. Sep 29, 2004

physicsuser

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. Sep 29, 2004

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. Sep 29, 2004

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

13. Sep 29, 2004

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. Sep 29, 2004

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.

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Last edited: Sep 29, 2004