Question

4. To find the acceleration of a glider moving down a sloping air track, you measure its velocity at two points (????1 and ????2) and the time ???? it takes between them: ????1 = 0.21 ± 0.05 m/???? ????2 = 0.85 ± 0.05 m/2 ???? = 8.0 ± 0.1 ???? a. Assuming all uncertainties are independent and random, and acceleration is calculated using ???? = ????2−????1 ???? , what should you report for ???? and its uncertainty? b. You calculate using an air resistance model that the acceleration should be 0.13 ± 0.01 m/???? 2 . Does your measurement agree with this prediction?

Answer 1

a = 0.08 +/- 0.025

Does not agree

Explanation:

Given:

- V_1 = 0.21 +/- 0.05 m/s

- V_2 = 0.85 +/- 0.05 m/s

- t = 8 +/- 0.1 s

- a = ( V_2 - V_1 ) / t

Find:

a) The average acceleration a and its uncertainty

b) Compare the results with theoretical model a = 0.13 +/- 0.01 m/s^2

Solution:

a) The average acceleration a is:

a = ( 0.85 - 0.21 ) / 8 = 0.08 m / s^2

- The uncertainty can be calculated by the following equation:

da / a = dV_1 / V_1 + dV_2 / V_2 + dt / t

da = (dV_1 / V_1 + dV_2 / V_2 + dt / t)*a

da = (0.05 / 0.21 + 0.05 /0.85 + 0.1/8) * 0.08

da = 0.30941*0.08 = 0.02475 m/s^2

- Hence, report a = 0.08 +/- 0.025

b)

The acceleration predicted by model a = 0.13 +/- 0.01 m/s^2.

The experimental acceleration a = 0.08 +/- 0.025

-Evaluate a relative error in mean value:

R.E = |(0.08 - 0.13)| / 0.13 * 100

R.E = 38.5%

- As per standard practices the results obtained experimentally and thematically must lie within 10% of deviation. However, in our case its 38.5% from which it can be concluded that the measurement does not agree with this prediction.