Question

Physics help? I know this is math but the physics section is really slow and I need help ASAP
I have to do a lab with the times and distances of which a ball falls on the Earth, moon, and Mars. I have all my values (time in terms of seconds, distance in terms of meters) but I'm supposed to find the average velocity, and average acceleration (using speed) and I'm really confused.

For example, the values for earth are:
0.0s - 0m
0.2s - 0.23m
0.4s - 0.86m
0.6s - 1.87m
0.8s - 3.23m
1.0s - 4.91m

Average velocity = displacement / time, so

4.91m / 1s = 4.91m/s

Average acceleration = change in velocity / change in time.

(4.91 - 0) / (1 - 0) = 4.91 m/s^2

I'm just not sure how all these values could be the same, or what I'm doing wrong?


Average speed = distance/time which in this case would be the same as average velocity. I'm really lost, please help

Answer 1

I'm pretty sure average speed is each interval's velocity divided by the number of intervals..

Answer 2

because each interval would have different speed since the object is accelerating. so calculate the velocity for each interval, add them up and divide by the number of intervals. thatr's your average speed.

Answer 3

@miszery ^avg speed is total distance/time

Answer 4

Do you have the equation of the experiment? Perhaps calculus could be of use.

Answer 5

@LS1088 There is no equation. It was a virtual lab, we were told to write down the distances at those exact time intervals and figure it out from there.

Answer 6

@agent0smith the total distance for all the intervals is still 4.91

Answer 7

I know :)

Answer 8

@agent0smith if an object is accelerating, the speed is not constant......

Answer 9

@miszery neither is the acceleration rate. I feel like the average acceleration shouldn't be 4.91 either. Something here's just not right and I have no idea what it is.

Answer 10

you have the gravitational force which accelerates falling objects.... you need to calculate the speed for each interval, add them up and divide by the number of intervals.

Answer 11

@miszery , average speed is total distance travelled over total time. The keyword is average.

Answer 12

@meegan your acceleration is wrong because you're using average speed. NOT final speed and initial speed for your change in velocity.

Answer 13

Average acceleration = change in velocity / change in time <==== needs to be the total change, not the avg velocity.

Answer 14

to find acceleration use this equation instead \[\Large d = \frac{ 1 }{ 2 }a t^2\]d is distance, t is time.

Answer 15

or use this one, to find final velocity... vf is final, vi is initial (zero isince it's dropped from rest), vav is average velocity \[\huge v_{av} = \frac{ v_f - v_i }{ 2 }\] then once you have final v, you can get acceleration using it.

Answer 16

ah ok.. gonna re-learn this haven't done physics in a while

Answer 17

I think the case might be, acceleration is just gravitational acceleration.

Answer 18

@LS1088 I think you're right.. I've looked at this lab like 6 times and one time I came up with average acceleration being 9.81m/s^2.. Gravitational acceleration is 9.81m/s^2. I'm just sitting here watching this feed progress and getting more and more confused. This was so easy a week ago, I'm not sure what the heck happened haha.

Answer 19

9.81m/s^2 would fit int oall the equations though, right?

Answer 20

@meegan yes a=9.8m/s^2. I'll walk you through it if you need... i teach physics.

Answer 21

@agent0smith @LS1088 Thank you! I was wracking my brain trying to figure this out. I just wish I remembered what equation I used to find that, because whatever it was, it worked.

Answer 22

this is def the easiest way to find a: \[\Large d = \frac{ 1 }{ 2 }a t^2\] plug in distance and time, solve for a.

\[\Large 4.91 = \frac{ 1 }{ 2 }a (1)^2\] gives a=9.8m/s^2

Answer 23

I'm pretty sure it was the equation you typed out, just in terms of a, but I don't remember what the equation was when it's written out in terms of a.

Answer 24

was it this?\[\Large a = \frac{ 2d }{ t^2 }\]

Answer 25

Yes! Because I have 4.91x2 written on a paper here, but of course 1^2 is still 1, so that's why that's not here. Thank you!

Answer 26

No prob :D

Answer 27

what is average speed then? 4.91 m/s?

Answer 28

Yep.

Final speed would be 9.81m/s. Avg is 4.91m/s.

Answer 29

And average velocity is still 4.91m/s? I'm just trying to make sure that 4.91 is still right haha

Answer 30

haha yes of course, if it wasn't i'd have corrected you :)

Answer 31

velocity is just speed with direction (down in this case)

Answer 32

Right, a vector instead of a scalar quantity. Got that part. Thank you!

Answer 33

ok thanks.. for some reason I thought I remeber my physics teacher calculate each interval for speed and add them up and divide to get the average speed. sigh :S

Answer 34

off to khan academy for me haha

Answer 35

You probably could do it that way, but why... but that's way more work than if you know the distance already.

Answer 36

i know... i vageuly remember him using laps as an example.. he said each lap has different speed so have to do it that way.. blach blah, maybe it was a different subject. but anyhoo, that was couple years ago. need to re-learn lol thanks :)

Answer 37

But even with laps, if you want the avg speed of a runner, add up their total distances, divide by time. Perhaps if you don't know the distance of the lap, or something odd like that, you'd need to use the speeds (i'm pretty sure i've never needed to use that method).

But it doesn't matter that the speed is changing... avg speed is ALWAYS equal to total distance travelled/time, no matter what.

Avg velocity is just displacement/time (so it's zero for a runner on a circled track)

Answer 38

/like

Answer 39

Okay, try this one. For the moon, here are the values.
0.0 - 0
0.4 - 0.14
0.8 - 0.54
1.2 - 1.2
1.6 - 2.1
2.0 - 3.28

Gravitational pull on the moon is -1.6m/s^2 (or 1.6m/s^2 downward) so that's my acceleration

3.28/2 = 1.64. This is really messing with me because for the first one, the acceleration was double. I'm sure this is right but I need to make sure my brain is out of "THEY'RE THE SAME" mode.

Answer 40

Use the a=2d/t^2 equation for acceleration.

You didn't double the distance, 3.28 :)

Answer 41

Or square the time... which lucky for you, evened out to give you the correct answer :P

Answer 42

I figured that if you doubled the time and the distance it'd technically come out the same (luckily 2^2 is the same as 2x2 so technically both values doubled)

Answer 43

Wow, pretty smart :)

Answer 44

So you're correct with the accel of 1.6m/s/s

Answer 45

Avg velocity happens to be the same as a this time