Acceleration of a Falling Object from Average Velocities

[ErinSK]

Homework Statement

After gathering data via ticker tape, of a falling object, I had the following information:

Time (s), Displacement (cm)
0.0 - 0.1 , 6.2
0.1 - 0.2 , 16.0
0.2 - 0.3 , 24.5
0.3 - 0.4 , 33.1
0.4 - 0.5 , 43.0
0.5 - 0.6 , 49.7

The Question:

Determine the average acceleration. Is the acceleration due to gravity uniform? Explain your answer using evidence from your investigation.

Homework Equations

Avg. Velocity = Displacement / time

Avg. Acceleration = (Change in Velocity) / time

The Attempt at a Solution

I calculated the average velocities (cm/s) as 62, 160, 245, 331, 430, 497.

I then calculated average velocity as 725cm/s/s :

Avg. Accel. = (497-62) / 0.6​

I was planning on proving that it is uniform acceleration by calculating the acceleration between two consecutive points however when doing so, I always get an acceleration much less than 725cm/s/s.

Accel. = (331 - 245) / 0.2​

= 430cm/s​

I get something similar with all consecutive points and I don't understand why/ I don't know what I'm doing wrong.

Please help!

 

[tms]

I calculated the average velocities (cm/s) as 62, 160, 245, 331, 430, 497.

I assume the object starts at rest. In the first interval the speed goes from 0 to 6.2 cm in 0.1 sec. In the second interval it goes from 6.2 to 16.0 cm in 0.1 sec. What is the average speed in each of those intervals?

 

[ErinSK]

I guess I didn't set up my table properly but by displacement I meant the change for that time interval. As in between 0.1 and 0.2 seconds it traveled from 6.2 to 22.2 for a displacement of 16cm.

 

[tms]

I guess I didn't set up my table properly but by displacement I meant the change for that time interval. As in between 0.1 and 0.2 seconds it traveled from 6.2 to 22.2 for a displacement of 16cm.

That's okay, you should just label it correctly.

 

[tms]

I then calculated average velocity as 725cm/s/s :

Avg. Accel. = (497-62) / 0.6​

There are only 0.5 seconds between those two measurements.

Accel. = (331 - 245) / 0.2​

= 430cm/s​

And there is only 0.1 seconds between those two measurements.

Try doing for acceleration what you did for velocity: At each point put the change in velocity in the table. You might also want to draw graphs of position, velocity and acceleration with respect to time.

 

[ErinSK]

Oh! That makes sense! thank you very much! A few last questions: How would I properly label my chart to indicate the difference in displacement values at each point? Or should I just add them cumulatively to avoid the problem? Also, on a graph I'm assuming the first point would be 0.1sec and 62cm/s even though 62cm/s is the average velocity between 0s and 0.1s. Is this just for simplicity?

 

[tms]

Oh! That makes sense! thank you very much! A few last questions: How would I properly label my chart to indicate the difference in displacement values at each point?

I'd label them t, dt, dx, x, dv, v, and a.

Or should I just add them cumulatively to avoid the problem?

Sweeping the dust under the rug is not a long-term solution.

Also, on a graph I'm assuming the first point would be 0.1sec and 62cm/s even though 62cm/s is the average velocity between 0s and 0.1s. Is this just for simplicity?

I was assuming it started out at t = 0 with x = 0 and v = 0. Or you could leave out the t = 0 point, if the assumption worries you.