Question

Please help me with this problem!



What is the velocity-time graph if the values for the position-time graph are: position(cm): 0, 6, 14, 25, 40, 60 and the time(s): 0, 0.3, 0.6, 0.9, 1.2, 1.5?

Answer 1

Think you can help me out AccessDenied?

Answer 2

Um, I have an idea of how to do it...
Velocity is just the rate of change of position with respect to time...

Answer 3

Yup

Answer 4

The values I got were: Velocity (cm/s): 0, 20, 26.67, 36.67, 50, 66.67 and the time is the same...Im not sure if this is right because when I graphed it, the first interval had the greatest slope

Answer 5

How about you, what did you get?

Answer 6

I get the same values. I'll have to think about this more...

Answer 7

I was confused to why the slope was so great, and then I moved onto the acceleration-time graph and got: Accelaration(cm/s^2): 66.67, 22.23, 33.33, 44.43, 55.57 and the time is the same... Again, the slope of the first interval is so big

Answer 8

Got anything?

Answer 9

Nope, I'm unsure... I don't know exactly how to get the exact velocity at a point given only points for position. We can only get the average velocity over intervals...

Answer 10

I thought you do y2-y1/x2-x1?

Answer 11

I'll try to find a helpful source on this... I can only think of finding the velocity given a position function. I'm not 100% sure given only points. D:

Answer 12

I see...

Answer 13

What is the velocity-time graph if
the values for the position-time graph are:
position(cm): 0, 6, 14, 25, 40, 60
time: 0, 0.3, 0.6, 0.9, 1.2, 1.5?

you have enough points given to define a function with

Answer 14

Im sorry, but I dont know how to get an equation from a set of data

Answer 15

time is moveing in increments of 1; just notated a bit differently but still the same unit interval each time

Answer 16

Yes, but when I got the values for my velocity-time graph, the first interval had a greater slope than any other interval on the graph

Answer 17

0, 6, 14, 25, 40, 60
+6 +8 +11+15 +20 ....
+2 +3 +4 +5
+1 +1 +1

Answer 18

I dont understand, what are those numbers?

Answer 19

You are taking the difference between the numbers?

Answer 20

those are the common differences between rows

1+2n +6n(n-1) might be a good explicit rule for the sequence

Answer 21

Ok, but can you explain how to make a velocity-time graph from a position-time graph? Isn't it y2-y1/x2-x1?

Answer 22

i looks like im remembering parts of it lol

Answer 23

yes, the slope at each point; but each point on a discrete lattice has no defined slope since it is essentially a corner

Answer 24

it might be the average of the slopes that are on either side of it tho

Answer 25

My pt graph has a curve as does my vt graph and my at graph, I thought that the vt graph was a derivative of the pt and the at graph was the second derivative of the pt

Answer 26

And therefore go down in order

Answer 27

I dont know why they are all the same

Answer 28

Any ideas?

Answer 29

im thinking since velocity = distance/time that we simply do this:

p = 0, 6, 14, 25, 40, 60
d = 0 6 8 11 15 20
t = .3 .........................

Answer 30

well basically, what im doing is, I had to construct a ramp and drop a ball measuring 5 points on the ramp. I did that and with that data I constructed that position-time graph, the question project asks me to make a velocity-time graph and acceleration-time graph from the position time graph...

Answer 31

do you have an angle for your ramp?

Answer 32

No, with the information from the graphs, Im supposed to find the slope of the ramp

Answer 33

.... nothing like working backwards lol

Answer 34

lol yeah

Answer 35

acceleration is going to be constant simply because it mimics gravity; just at an angle

Answer 36

ok

Answer 37

that means velocity is going to be a linear line?

Answer 38

\[a(t) = c_a\]
\[\int a(t) = v(t) = c_at+c_v\]
\[\int v(t) = p(t) = c_at^2+c_vt+c_p\]

pick any velocity point as an inital value to solve for v(t) id spose

Answer 39

is position with regards to height? or with distance traveled down the ramp?

Answer 40

That looks confusing...

Answer 41

distance traveled down the ramp

Answer 42

So, is my position-time graph right atleast?

Answer 43

Its a curve starting from 0 to 1.5, 60

Answer 44

position(cm): 0, 6, 14, 25, 40, 60
time: 0, 0.3, 0.6, 0.9, 1.2, 1.5?

rref{{.3^2,.3,1,6},{.6^2,.6,1,14},{1.5^2,1.5,1,60}}

well, the data isnt quadratic as id expect it to be

Answer 45

Im sorry, I dont understand what you just did :P

Answer 46

i row reduced 3 points of data to find the coeffs of a similat quadratic equation that should have matched it

Answer 47

f(x) = 550/27 x^2 + 25/3 x +5/3

Answer 48

So, thats the equation for the pt graph?

Answer 49

its an "almost" match for the given points

Answer 50

WOW! it works

Answer 51

well; at 0 we aint at 0 so its not an exact fit

Answer 52

hmm, yeah you're right, but that IS the equation for the graph though right?

Answer 53

it an equation for a graph that is close to it in the given interval

Answer 54

So, there is no way to make a vt graph or at graph with the given data?

Answer 55

prolly is, but without knowing what material youve covered as a guide, im just taking a blind stab at this.

im thinking of trying eulers method on it tho; not that itll get me anywhere

Answer 56

Do you think that we could make the vt or at graoh if I changed the data?

Answer 57

Well, the ball takes approximately 1.5s to make it to the bottom from the top, I could prolly extend that a little bit...

Answer 58

Can I email you a picture of the ramp and the project sheet?

Answer 59

(n^3+29n-30)/6

this is the equation that matches all of your points

Answer 60

even zero?

Answer 61

thats starting at n=1

Answer 62

and its assuming intervals of 1 and not 3/10

Answer 63

How did you even derive that equation?

Answer 64

i did what i tried to at the start, but let wolfram iron out the details

Answer 65

How do I do that?

Answer 66

oops nvm :P

Answer 67

Ok, so far, I have the values for a position-time graph, the values and I need to figure out how to make an at graph and a vt graph

Answer 68

Maybe this will help you out, look at the first point,