Physics lab help, determining acceleration thru V vs T graph?

[papi]

Physics lab help, determining acceleration thru V vs T graph??

the point was to determine the acceleration due to gravity. It was this whole experiment with a ribbon tied to a brass weight. it was put into a machine and youd drop the brass weight and the machine would mark dots on it to show how it feel. each dot represented 1/60, 2/60 seconds etc. I had 28 dots in all. The man in charge said the acceleration was supposed to be 9.8 but I don't know what to do, i graphed and made a table of the displacement and time. (for displacement i measured in cm from origin to the other dots). But how do i find the velocity and graph V vs T. please help!

--- each dot is at its own interval, dot 1 is 1/60, dot 2 is 2/60 dot 3 is 3/30 etc. eaach obviously has its own distance from origin. and its a free falling object...please help, i was told but am very confused to cound 6 dots (to graph it only has to be 1/10 intervals) and count distance from dot before and dot after the 6th like
A.....B....C (with A B &C being dots) so distance from A--->C divided by 1/30 was velocty. but i dnt get that. please explain!
ive been trying so long and am so confused!

 

[tiny-tim]

… i was told but am very confused to cound 6 dots (to graph it only has to be 1/10 intervals) and count distance from dot before and dot after the 6th like
A.....B....C (with A B &C being dots) so distance from A--->C divided by 1/30 was velocty. but i dnt get that. please explain!
ive been trying so long and am so confused!

Hi papi!

You probably know that you can find the speed by drawing the tangent line, and measuring the slope of it.

But that is very difficult to do accurately.

So an alternative experimental method is to draw a chord parallel to the tangent line.

It needs to be about 6 dots away to get reasonable accuracy.

In other words: distance from A--->C divided by 1/30 equals the slope of the chord AC, which equals the slope of the tangent line at B.

(btw, can you prove that the tangent line and the chord are parallel, using v_C - v_A = a(t_C - t_A)? )

 

[papi]

Well is it ok for the velocity to increase every 6 dots?
and from there. once you have let's say 5 velocities. do you do the difference between them or what? help! thanks so much

 

[tiny-tim]

Well is it ok for the velocity to increase every 6 dots?
and from there. once you have let's say 5 velocities. do you do the difference between them or what? help! thanks so much

Hi papi!

You're starting with an x versus t graph, with dots from t = 1 to t = 28.

You want a v versus t graph.

So on the x and t graph, you draw the line joining the dot for t = 1 to the dot for t = 7, and measure the slope …

that gives you the value of v for t = 4.

Repeat, for higher values of t, and you get a v and t graph.

 

[papi]

you are a great help i just still am at a mind block not understanding it. the X vs T graoh and table i was able to do. its just the table and graoh for V vs T where I am at a loss. Right now I have a long thin paper with these multiple dots- so what can i do from that with a cm ruler ?
I was thinking go to every 6th dot. measure distance between 5 and 7th and divide that by 1/30 and that's velocity. but once you did that for every 6 dots, then what? also the velocity gets to be in the 100's so its very big, is that ok? also, do you do the difference between all the velocities? help!

 

[tiny-tim]

I think you're meant to use the ruler by drawing a long line through a pair of dots, and then measuring the slope of the line …*that gives you v.

Do it for 1 and 7 (not 5 and 7 … there's too much room for error there), 2 and 8, 3 and 9, and so on …

 

[papi]

but tim,
we are only supposed to make the chart for every 1/10 seconds meaning every 6 dots... also what slope are you talking about? (im not using the x vs t graoh) I am using the actual data paper looking like dots getting further and further apart. please help again thanks

 

[tiny-tim]

but tim,
we are only supposed to make the chart for every 1/10 seconds meaning every 6 dots... also what slope are you talking about? (im not using the x vs t graoh) I am using the actual data paper looking like dots getting further and further apart. please help again thanks

ok … then do it for 1 and 7, 7 and 13, 13 and 19, …

if you're not doing an x/t graph first, then measure the distance with the ruler, and divide by the time.

 

[papi]

ok so let me get this:
i get distances.. 1-7
7-13
13-19
19-25
and since total of 28 dots id stop there
Id divide each distance by 1/30 (in other words multiply by 30)
and that would be my velocities. once i have those 4 velocities, then what?
do i do the diff between each velocity or what and then how do i find acceleration from there?

 

[tiny-tim]

… then what?
do i do the diff between each velocity or what and then how do i find acceleration from there?

I thought the question asked for a v vs t graph?

You have v for t = 4, 10, 16, and 22.

So draw a graph … it should be a straight line.

 

[papi]

couple q;s:
1. should the velocity be increasing?
--Im pretty sure yes bc its freefall but just wanted to confirm.
2. I am really supposed to do every 6 dots though... like for 1/10 seconds and then 2/10, 3/10 etc

 

[tiny-tim]

couple q;s:
1. should the velocity be increasing?
--Im pretty sure yes bc its freefall but just wanted to confirm.
2. I am really supposed to do every 6 dots though... like for 1/10 seconds and then 2/10, 3/10 etc

1. Yes!

2. oh I see … then do it for 3 to 9, 9 to 15, 15 to 21, …

(or 5 to 7, 11 to 13, 17 to 19, … which is theoretically the same but has a greater experimental error)

 

[papi]

wow youve been an incredible help. I got those velocities now increasing, however, i additonally did the difference between each one and the increaste rate isn't the same, that's ok right? also, as a physicist, do you think i should make the data table and graph keeping it in cm or m? (the X vs T was done in cm) if that matters. thanks so much.
only one more issue though, now that i have all 4 velocities, and i can put that in a table #1 would be 1/10 T and #2 velocity 2/10 T but should i graph it the same way, also MAJOR q, once i have this, how do i find the acceleration??

 

[tiny-tim]

… also, as a physicist, do you think i should make the data table and graph keeping it in cm or m? (the X vs T was done in cm) if that matters. thanks so much.

Hi papi!

I would definitely do it in m, since you'll ultimately be comparing that with g, which is usually given in m/s².

only one more issue though, now that i have all 4 velocities, and i can put that in a table #1 would be 1/10 T and #2 velocity 2/10 T but should i graph it the same way

Sorry, not following you.

also MAJOR q, once i have this, how do i find the acceleration??

You'll have a v vs t graph.

Hint: from an x vs t graph, how would you find v?

 

[papi]

oh through the slope. so id take two points with T being X and Y being velocity. but its ok that the X vs T is in Cm bc still same upward curve? also, why are you so smart? you rock

 

[papi]

oh crap
My X vs T looks likes an upward curve not downward! oh no!
bc X increased as T increased so it looks like

an upward C almost. starts low and slowly curves straight up but isn't it supposed to downward curve, oh no!

 

[papi]

Dot # X T
1 .4cm 1/60
2 .6cm 2/60 (reduced1/30)
3 .9cm 3/60 (reduced 1/20)
4 1.3cm 4/60( reduced 1/15)
5 2.1cm 5/60 (reduced 1/12)
6 3.1cm 6/60 (reduced 1/10)
7 4.1cm 7/60
8 5.5cm 8/60 (reduced 2/15)
9 7.1cm 9/60 ( reduced 3/20)
10 8.8cm 10/60 (reduced 1/6)
11 11.8cm 11/60
12 13.2cm 12/60 (reduced 1/5)
13 15.6cm 13/60
14 18.8cm 14/60 (reduced 7/30)
15 21.6cm 15/60 (reduced¼)
16 24.9cm 16/60 (reduced 4/15)
17 27.8cm 17/60
18 32.5cm 18/60 (reduced 3/10)
19 35.6cm 19/60
20 39.8cm 20/60 (reduced1/3)
21 44.3cm 21/60 (reduced 7/20)
22 48.4cm 22/60 (reduced 11//30)
23 54.0cm 23/60
24 58.8cm 24/60 (reduced 2/5)
25 64.5cm 25/60 (reduced 5/12)
26 68.6cm 26/60 (reduced 13/30)
27 75.7cm 27/60 (reduced 9/20)
28 80.8cm 28/60 (reduced 7/15)


this was my data and it created an upward curve, not downward for X vs T so isn't that wrong? I am so confusedd! because isn't it supposed to be downward? but the data shows increase? also what is the V vs T supposed to look like. please describe and help asap! thanks

 

[tiny-tim]

My X vs T looks likes an upward curve not downward! oh no!
bc X increased as T increased so it looks like

an upward C almost. starts low and slowly curves straight up but isn't it supposed to downward curve, oh no!

uh?

You're supposed to be doing a v vs t graph, not x vs t.

(and anyway, why shouldn't x curve up?)

 

[papi]

I just graphed the v vs T and it came up as a duagonal upward line. Is this ok??

 

[papi]

no i had to do both- and should it all be down bc the object is falling?

 

[tiny-tim]

no i had to do both- and should it all be down bc the object is falling?

If your x is increasing downward (as I assume it is), then the curve should be curving up,

but if your x is increasing upward, then the curve should be curving down.

I just graphed the v vs T and it came up as a duagonal upward line. Is this ok??

Yup!

So what figure did you get for the acceleration?

 

[razored]

The slope between two points on a V vs. T graph is the acceleration.. if you want instantaneous acceleration.. just take the derivative as time approaches 0.

 

[papi]

ok, well i get the V vs T so yaay and I am doing the acceleratio now so don't run away!
but for the X vs T
X increases with time obviously bc it gets further and further away from the initial starting dot.

http://www.glenbrook.k12.il.us/GBSSCI/PHYS/CLASS/1DKin/U1L3a.html
it looks like the positive velocity changing velocity

hope that ok?
doing a now so ill let you know

 

[papi]

TIm!
i got 9.5 for the acceleration yaaay! its supposed to be 9.8 but its close enough!
just one q, what couldve been sources of error to make it not 9.8 aka gravity? thanks

 

[tiny-tim]

TIm!
i got 9.5 for the acceleration yaaay! its supposed to be 9.8 but its close enough!

Excellent!

just one q, what couldve been sources of error to make it not 9.8 aka gravity? thanks

It's only 3% out … I'd call that surprisingly accurate!

hmm … but maybe the marking device was slowing it down?

 

[razored]

If the acceleration is constant... the graph of x vs t should be the right half of a parabola; if the acceleration is 0 then the x v t graph will be a line.

 

[papi]

BEST FIT LINE
ok I am supposed to make these for the graph. but does that mean i draw a line through the points, or have half points above and half below??
please explain (or do i just leave out points that dnt fit>.?)

 

[tiny-tim]

BEST FIT LINE
ok I am supposed to make these for the graph. but does that mean i draw a line through the points, or have half points above and half below??
please explain (or do i just leave out points that dnt fit>.?)

I thought you said you'd already got 9.5?

Sorry, no idea … I guess you draw the two lines with the greatest and least slopes that look reasonable, and then say that the experimental value lies somewhere between them.