Question about the average acceleration

[Emily0203]

Hello everyone,
I have to find the average acceleration in the intervals 0m - 30m, 30m - 60m, and 60m - 90m. Here is the table:

Position (m)	0	10	20	30	40	50	60	70	80	90	100
Time (s)	0	1,89	2,88	3,78	4,64	5,47	6,29	7,10	7,92	8,75	9,58

My teacher has given us the answers, but we have to come up with the solution ourselves. The answers are supposedly 3,0 m/s^2, 0,36 m/s^2 and -0,090 m/s^2, but my results come out wrong every time I try to solve this.

I would really appreciate it, if someone would be able to help me!

 

[berkeman]

Show us what you've done already, even if you aren't getting the correct answers.

 

[rudransh verma]

intervals 0m - 30m, 30m - 60m, and 60m - 90m.

What are the intervals in minute or second?
I guess you need to be given the position function. The data is not that helpful.
Without instantaneous velocity how will you calculate average acceleration?

 

[valenumr]

Hello everyone,
I have to find the average acceleration in the intervals 0m - 30m, 30m - 60m, and 60m - 90m. Here is the table:

Position (m)	0	10	20	30	40	50	60	70	80	90	100
Time (s)	0	1,89	2,88	3,78	4,64	5,47	6,29	7,10	7,92	8,75	9,58

My teacher has given us the answers, but we have to come up with the solution ourselves. The answers are supposedly 3,0 m/s^2, 0,36 m/s^2 and -0,090 m/s^2, but my results come out wrong every time I try to solve this.

I would really appreciate it, if someone would be able to help me!

I might suggest drawing a plot of position versus time, and then just integrating. Then use the result to work backwards to average acceleration.

 

[kuruman]

I would add to this list the values of the average velocity at the end of each time interval. . That will give an idea of how the velocity is changing with time. Then I would try the definition for average acceleration (what is it?)

 

[haruspex]

I would add to this list the values of the average velocity at the end of each time interval. . That will give an idea of how the velocity is changing with time. Then I would try the definition for average acceleration (what is it?)

Yes, but how to calculate those velocities?
I used the neighbouring data, e.g. for velocity at 30m I used the data for 20m and 40m, etc. This is suggested by the fact that we have data for 100m, but the last acceleration we're asked for is at 90m.
Got pretty much the official answers.

 

[kuruman]

Yes, but how to calculate those velocities?

I did it the straightforwrd way, $\bar v_i=\dfrac{x_i-x_{i-10}}{t_i-t_{i-10}},~~~~i=10,20~\dots~100.$ Then for the average acceleration, I used the bounding values of average velocities and times at each interval. "Pretty much the official answers" is what I got. The sparsity of data at the beginning (0-30 m) where the velocity changes most rapidly makes refinement unwarranted.

 

[haruspex]

I did it the straightforwrd way, $\bar v_i=\dfrac{x_i-x_{i-10}}{t_i-t_{i-10}},~~~~i=10,20~\dots~100.$ Then for the average acceleration, I used the bounding values of average velocities and times at each interval. "Pretty much the official answers" is what I got. The sparsity of data at the beginning (0-30 m) where the velocity changes most rapidly makes refinement unwarranted.

I just tried that way and got rather different results.

Your way:
2.94, 0.43, -0.06
My way:
3.01, 0.36, -0.09
Given answers:
3.0, 0.36, -0.09

 

[kuruman]

I just tried that way and got rather different results.

Your way:
2.94, 0.43, -0.06
My way:
3.01, 0.36, -0.09
Given answers:
3.0, 0.36, -0.09

I agree, if the exercise is to get the teacher's answers, your way works better than mine. However, if this is an exercise in data analysis, the way to go would be to do a polynomial fit, take derivatives to find the instantaneous velocities and use them to find the average acceleration at the stated intervals. I did just that out of curiosity and fitted a polynomial in $t$ of order 3, which is the minimum needed for non-constant acceleration.

I found $x(t) = -0.0747~t^3+1.4423~t^2+3.422~t.$ The interesting feature of this fit is that it implies an initial velocity $v_0=3.42~\text{m/s}.$ Neither method takes that into account when doing the 0-30 m interval and for good reason: it is an initial condition that should have been provided by the teacher. My recalculated average accelerations using the fit are 2.03 m/s², 0.628 m/s², and -0.486 m/s². That is why I considered my crude calculation that you duplicated close enough to the teacher's and deemed that any further refinement is unwarranted.

 

[haruspex]

the way to go would be to do a polynomial fit

I disagree. I chose the method I used because it seemed the most reasonable, not because it matched the teacher's answers.

 

[rudransh verma]

I used the bounding values of average velocities and times at each interval. "Pretty much the official answers" is what I got.

I don’t understand this. What are bounding values. How did @haruspex got that result very close to actual answer?
Either the function should be given or there should be infinite entries of data to calculate.

 

[haruspex]

What are bounding values.

@kuruman considered three ranges, 0 to 30m, 30m-60m, 60m-90m. For each, he calculated the velocity at the end of the range based on the last 10m of it. Then he used the change in velocity from start to end of range to get the acceleration.
As he notes, we have to assume it starts at rest.
The only difference in my calculation is that I found the average velocities by combining the last 10m of one range with the first 10m of the next.

Either the function should be given or there should be infinite entries of data to calculate.

There is no algebraic function. You just have to assume the velocity varies fairly smoothly. If you wanted to improve on my approach you would use some kind of spline fit to the data.

 

[rudransh verma]

For each, he calculated the velocity at the end of the range based on the last 10m of it. Then he used the change in velocity from start to end of range to get the acceleration.

I don’t get it. We need to find the instantaneous velocity at the start and end of each interval.
So he calculated the average velocity for the last 10m and assume it as velocity at the end of the interval and calculated the average acceleration

 

[haruspex]

I don’t get it. We need to find the instantaneous velocity at the start and end of each interval.
So he calculated the average velocity for the last 10m and assume it as velocity at the end of the interval and calculated the average acceleration

Yes, whereas I used the average velocity over the 10m either side.

 

[haruspex]

Further thoughts...
The flaw in my method is that the given time intervals are not quite equal.
My post #12 comment about assuming the motion is fairly smooth made me realize a more general approach: for any two consecutive periods, find the velocity at the boundary by presuming acceleration constant.
If the displacements and durations of the two intervals are $s_1, s_2, t_1, t_2$ this leads to
$v=\frac{s_1t_2^2+s_2t_1^2}{t_1t_2(t_1+t_2)}$.
Note that if $t_1=t_2=t$ this reduces to $\frac{s_1+s_2}{2t}$, as I used before.
In the question in this thread it make no more than a 1% difference, but it would be much better in general.

 

[kuruman]

If the displacements and durations of the two intervals are $s_1, s_2, t_1, t_2$ this leads to
$v=\frac{s_1t_2^2+s_2t_1^2}{t_1t_2(t_1+t_2)}$.

Yes, and the same equations that yield this relation also yields the assumed constant acceleration as $a=\frac{s_2t_1-s_1t_2}{t_1t_2(t_1+t_2)}.$ Logically, that would be the average acceleration over the two adjacent intervals.
For the first two intervals (0-10-20) this gives 3.34 m/s².
For the next pair of intervals (10-20-30) this gives 1.07 m/s².
My question is, how one can extract the average acceleration from 0 to 30 out of these two? Take a weighted average? Also, OP stated

I have to find the average acceleration in the intervals 0m - 30m, 30m - 60m, and 60m - 90m.

I automatically assumed that this would be a time-averaged acceleration. Could it be an average over distance?

 

[haruspex]

For the first two intervals (0-10-20) this gives 3.34 m/s².
For the next pair of intervals (10-20-30) this gives 1.07 m/s².
My question is, how one can extract the average acceleration from 0 to 30 out of these two? Take a weighted average?

Having found estimates for the instantaneous velocities at the given boundaries (30m, 60m, 90m), and assuming stationary at 0m, why not just velocity increase/elapsed time for each average acceleration?