How to find/graph instantaneous speed vs timeby Arooj Tags: find or graph, instantaneous, speed, time 

#1
Aug2512, 02:34 PM

P: 40

1. The problem statement, all variables and given/known data
I have to construct two graphs ( distance vs time and instantaneous speed vs time) based off of a lab in which we made measurements based off of a spark timer and paper tape. I made the following measurements and constructed the D vs T graph based off of them: t / "x" in meters 0.0 / 0.00 0.1 / 0.0360 0.2 / 0.161 0.3 / 0.382 0.4 / 0.697 0.5 / 1.109 0.6 / 1.614 My question is pertaining to making the instantaneous speed vs time graph For this, we had to find the instantaneous speed at the midpoint of each interval. "From your distance vs time graph, find the instantaneous speed at the midpoint of each 0.1 second interval. You can do this by either drawing a tangent line at the midpoint of each interval and determining its slope, or by finding the average speed for each of the intervals." I was a bit confused by this instruction. Drawing a tangent line was and finding the instantaneous speed was impossible, so the teacher told us to find the average speed for each interval. 2. Relevant equations t2t1/s2s1 (I think) 3. The attempt at a solution To find the IS, I first found the midpoint of the time interval (ex: (0.2 + 0.3)/2]= 0.250. Then I divided the displacement of the interval by the midpoint of the time interval (ex: interval 0.20.3) (0.3821.161)/0.250 = 0.884 Time interval /Midpoint of interval /Inst. speed in m's 0.00.1 / 0.0500 / 0.720 0.10.2 / 0.150 / 0.833 0.20.3 / 0.250 / 0.884 ... ... ... I feel like my calculations are incorrect. Also, when plotting the graph would the instantaneous speed be plotted against the time (ex. 0.720/ 0.1) or against the midpoint of the interval (0.0720/0.05)? I plotted form 0.720 / 0.1, and I came up with a graph that looked rose up from zero and leveled off, like a logarithm graph beginning at zero. 



#2
Aug2512, 02:51 PM

Math
Emeritus
Sci Advisor
Thanks
PF Gold
P: 38,898

Dividing " the displacement of the interval by the midpoint of the time interval" is NOT how you find "instantaneous speed". The displacement of the interval is change in distance and you need the corresponding change in time a time interval, not a single time. The theoretical "instaneous speed" would be the limit as the distance interval, and so corresponding time interval, went to 0. If you cannot do that here having only given fixed intervals, the best approximation you can get is the "average speed" over the shortest possible intervals.




#3
Aug2512, 03:12 PM

P: 40

Thanks, I figured out how to do that part. But then how would you graph distance vs time squared? Would you square the time?
t / "x" in meters 0.0 / 0.00 0.1 / 0.0360 0.2 / 0.161 0.3 / 0.382 0.4 / 0.697 0.5 / 1.109 0.6 / 1.614 time squared / distance 0.0 / 0.00 0.01 / 0.0360 0.04 / 0.161 0.09 / 0.382 ... / ... The graph for distance vs time squared is supposed to be constant or close to constant, because any falling body's acceleration is 9.81 m/s^2 And should the graph of instantaneous speed be a line of form y = mx + b? It kind of looks like this (/) and I also included the point zero, which I wasn't sure to include or not because its not an interval of time.( next values after zero are 0.0  0.1, 0.10.2, and so on, so the line from zero to 0.00.1 has a different slope from the rest of the line.) 



#4
Aug2512, 09:22 PM

Sci Advisor
HW Helper
Thanks
PF Gold
P: 4,503

How to find/graph instantaneous speed vs time
In plotting the distance vs time squared, you should plot t^{2} as the abscissa, and d as the ordinate. The slope of the line should be constant, and equal to a/2 (if this is motion down an inclined plane or free fall motion).
The instantaneous velocity at the midpoint (timewise) of each interval can be approximated very accurately by Δd/Δt. This velocity applies to the center of the time interval t_{1/2} = (t_{1}+t_{2})/2. You should plot the center of each time interval as the abscissa, and the ordinate as the velocity at the midpoint. 



#5
Aug2612, 05:43 AM

P: 40

Thank you. So to find the instantaneous velocity, I would just really be dividing a the distance change by the time change.
(So would you always divide the change in distance by 0.1, a constant time change for each interval because 0.50.4 = 0.1, 0.3  0.2 = 0.1, and so on? Correct me if I am wrong. And also for some reason my graph for distance vs time squared rises and levels off; points 0.010.16 will look like a rising line and then points 0.250.36 will level off at a value. 



#6
Aug2612, 12:42 PM

Sci Advisor
HW Helper
Thanks
PF Gold
P: 4,503

The shape of the distance vs time squared plot depends on the geometry of the track. You should be able to look at the instantaneous velocity vs time plot and get a better understanding of what is happening. If a portion of the velocity vs time plot is a straight line, then the acceleration is constant. You should be able to examine the track shape and see why the acceleration is constant in that section of track. Play around with the different plots you have and the corresponding shape of various sections of the track, and see if you can make sense of it. This is what I would do if I were analyzing the system. Chet 



#7
Aug2612, 06:25 PM

Math
Emeritus
Sci Advisor
Thanks
PF Gold
P: 38,898

Also, as I said before, that does NOT give you the "instantaneous velocity". It gives the average velocity over the interval.




#8
Aug2712, 09:48 PM

Sci Advisor
HW Helper
Thanks
PF Gold
P: 4,503

Chet 


Register to reply 
Related Discussions  
How do I find the instantaneous acceleration for a velocitytime graph that is NOT a  Introductory Physics Homework  2  
Instantaneous Velocity with Position v. Time graph  Introductory Physics Homework  1  
[SOLVED] Instantaneous Acceleration on a VelocityTime Graph  Introductory Physics Homework  5  
Finding Instantaneous acceleration from a velocitytime graph  Introductory Physics Homework  4  
velocity time graph &instantaneous acceleration  Introductory Physics Homework  4 