Question

Help! How does displacement change with time for a falling object? How might you describe the mathematical relationship between the velocity and time of a falling object? I have a picture of the data table if you need it

Answer 1

The kinematic equations are a perfect example of how free fall relates with displacement, velocity, time and acceleration.
is this a physics class or a calculus class?

Answer 2

physics

Answer 3

ok,
then the use of the kinematic equations will really help you. also looking at the units of each variable:
like, displacement is a unit of meters
velocity is a unit of meters/second
acceleraion is a unit of meters/seconds^2
and time is a unit of seconds

is this helping any?

Answer 4

Yes, I have a data table though you probably need to help

Answer 5

post it up ^_^

Answer 6

ok, lets work on the first question, "How does displacement change with time for a falling object"
so looking at the table, and only looking at the column of displacement and time. what can you contrast between them? does displacement increase or decrease? does it increase the same amount each second passes or is it less or is it more?

Answer 7

I'm confused on what the column for displacement is? Would it be the distance column or..?

Answer 8

oh yes yes, displacement is the 'general word' for distance ^_^

Answer 9

Ok! The displacement is increasing and is increasing the same amount while the seconds are increasing also

Answer 10

yes! ^_^
now, does distance increase the same amount from one second to the next second to the next second?

Answer 11

Well, the seconds aren't increasing on a constant basis, but yes, the distance is

Answer 12

very good
so seconds increase more and more as distance increase only 0.1
do you feel that answers your first question, or should we talk about it more?

Answer 13

Oh alright, I got the first question

Answer 14

next question, "How might you describe the mathematical relationship between the velocity and time of a falling object"

Answer 15

any ideas on this one, or should I ask you some questions to help?

Answer 16

Well velocity seems to be increasing for the most part although it decreases a couple times while seconds continues to only increase. I'm not sure how to describe that relationship mathematically though

Answer 17

good, you are right. normally velocity would continue to increase as time increase, but the way this chart is done, its almost like the object is allowed to 'free-fall' because the velocity grows, then shrinks, then grows again.
anyways (that was all a side note)
see in the top of the column where it says, "Velocity (m/s)" the m stands for meters and the s stands for seconds, so its saying velocity = meters/seconds. now pick any row and get the distance and divide it by the time. Do you get the same velocity as it shows in the table for that row?

Answer 18

Yes, I do. So I would say that the relationship is v=m/s? That sounds like a little bit too short of an answer haha

Answer 19

From what I can understand what the question is asking, "v=m/s" is a great answer. I might also to "velocity=displacement/time"

Answer 20

do, not "to" sorry

Answer 21

Oh ok, please don't go anywhere. I only have two more questions left. I really appreicate the help and not just solely the answer. Do you want me to post the questions here or on a new topic?

Answer 22

we can do it here, this way other people wont get involved ^_^

Answer 23

Haha good idea. Ok, so the question is:How does the velocity of a falling object change with time? Would I just answer that as time continues to increase, velocity increases for the most part, but decreases a few times or..?

Answer 24

is this question for the same table?

Answer 25

Yes, all questions use the same table

Answer 26

hmm... i need to think about this one...

Answer 27

i'm not gonna lie to you, this table is a bit weird, it says it's 'free-fall' but the numbers do match an object that falls at 'free-fall'...

Answer 28

I think your answer, "that as time continues to increase, velocity increases for the most part" is good, and you'll have a gut feeling of what your teach is looking for more than I can.
I would just add, "when velocity decreases, it is due to a force resisting free-fall"

Answer 29

Ok because I think that she actually did this lab herself, so maybe there are human errors unaccounted for. I don't know

Answer 30

Last question:Compare the slope of the velocity-time graph to the average of all your acceleration values. Are they close? What does the slope of a velocity (or speed) vs. time graph mean? Explain the answer using your data. How does the value of g that you calculated compare to the accepted value of 9.80 m/s2? What is your percent error? Remember that the value of g can be calculated by using g = a/sinθ and percent error can be calculated using the following equation: (it's just the percent error calculation equation)

Answer 31

I'm gonna attach a graph of what 'free-fall' should look like to the table you sent me, and you'll see what I mean.

Answer 32

the top chart is what the data looks like of the labs 'free-fall'
and the bottom chart is an example of what 'free-fall' should look like.
do you see how the example curve is smooth and the lab's curve is jagged? thats why I'm saying the numbers are weird for the lab, because it should look like the example chart.

Answer 33

I have a feeling that my teacher will be fine with the #2 question, so I think I'll leave it at that, but I do understand what you're saying. Can you help me on the last question, #3? I know the slope of the velocity time graph which is 0.1846

Answer 34

":Compare the slope of the velocity-time graph to the average of all your acceleration values"
0.1846 is correct
what do you get as the average of the acceleration column?

Answer 35

The average is .39

Answer 36

remember that slope is: rise/run, so if we have velocity/second (remember its rise/run) then it looks like this:
\[\frac{ \frac{ meters }{ second } }{ second}\]
which turns out to be
meters/seconds^2 which is acceleration

sooo... just like how we did meters/ seconds = velocity
velocity/seconds = acceleration

^_^

Answer 37

So the slope of the velocity vs. time graph would compare to the average acceleration because to get the slope you divide velocity/seconds, which turns out to be meters/s^2 (acceleration). Is that what I would say to compare?

Answer 38

I'm running late, and should have left 10min ago... so i gotta leave super quick here...

Answer 39

Yes! you got it ^_^

Answer 40

that is how they are similar, or how you can compare them ^_^

Answer 41

How does the value of g that you calculated compare to the accepted value of 9.80 m/s2? What is your percent error

Answer 42

well we got 0.1846 and the acutall value should be 9.8, so what was calculated from the lab is waaay off from what g should be.
as for percent error... i haven't touched that stuff in years....

Answer 43

Percent error is: (measure value-accepted value/accepted value) x 100

Answer 44

I gotta run though! hope everythiing works out ok!

(measure value-accepted value/accepted value) x 100 so,
((0.1846 - 9.8)/9.8) * 100 = percent error

^_^

Answer 45

Ok! Thank you so much. I really appreciate it