Free fall prac help ...?

Question

So we did a prac at school it involved a 0.5kg mass being dropped 2m, we used a ticker timer to calculate the speed. We were then given questions to complete.

I have done a histogram of the tape - 10 dots per piece.

1st tape is 14cm long, the 5th dot is 3.3 cm from the first (0) dot.

2nd tape is 43.5 cm long, the 15th dot is 18.3 cm from dot 10.

3rd tape is 75.5cm long dot 25 is 32.8cm from dot 20.

Distance between dots:

4 & 5 - 1.3 cm

14 & 15 - 4.3cm

24 & 25 - 7.3 cm

Frecuency of the dots is 50 per second ~ 0.02 seconds apart

We were given the eq.

a= change in speed between dots 5&15 / Time between dot 15 & 25

Heres my question:

What is the actual speed at

i) dot 5

ii) dot 15

iii) dot 25

Then, calculate the acceleration between:

i) dots 5 and 15

ii) dots 15 and 25

My answers can't be right as I didn't end up with an acceleration even close to 980cms^-2.

Please any help would be excellent

Thank you

Answer 1

This is why g-d, Dan Brinklin, and Bob Frankston invented interactive spread-sheet programs.

The right way to solve this is to pull out your favorite (Open Office, Excel, Apple's new Numbers, etc.); plug in the data, generate a scatter plot, and ask it for a polynomial fit.

Since the equation under constant acceleration is:

D = Vi T + (1/2) A T^2

a quadratic best fit would give you the acceleration directly.

Otherwise you are almost guaranteed to have all sorts of algebra and arithmetic errors, even you use a calculator.

still, as you don't give us all your date, we have to do things the hard way:

- distance between 4 & 5 = 1.3 cm so velocity is approximately 1.3/0.02 or 65 cm/sec

- distance between 14 & 15 - 4.3cm so velocity is

approximately 215 cm/sec

- difference in velocity  is approximately 150 cm/sec. and the difference in time is about 10 x 0.02 or 0.2 seconds so

acceleration looks like around 750 cm/sec, pretty close to what it should be, given the quality of your data and the fact that we are using so few points.