In comparing of result of experiment A and B, the result of A is by 2 factors differ from experiment B.
The given parameters;
in group A, distance traveled by the red block, s = 15 ± 3 feet
in group B, distance traveled by the green block, s = 25 ± 4 feet
To find:
Since both groups performed the experiment at equal times, we assume the time for both motion = t
Also, assume the initial velocity of both blocks = 0
For group A, we set-up the equation of motion as follows;
[tex]s = v_0t + \frac{1}{2} at^2 \\\\15+ 3 = 0 + 0.5\times a_1t^2\\\\18 = 0.5a_1t^2\\\\t^2 = \frac{18}{0.5a_1} \\\\t^2 = \frac{36}{a_1}[/tex]
For group B, we set-up the equation of motion as follows;
[tex]25 + 4 = v_0t + \frac{1}{2} a_2t^2\\\\29 = 0.5\times a_2t^2\\\\t^2 = \frac{29}{0.5a_2} = \frac{58}{a_2}[/tex]
Solve the first equation and the second equation together;
[tex]\frac{36}{a_1} = \frac{58}{a_2} \\\\\frac{a_2}{a_1} = \frac{58}{36} \\\\\frac{a_2}{a_1} = 1.61[/tex]
The ratio of error margin of both experiments;
[tex]\frac{4}{3} = 1.33[/tex]
The resulting parameter for comparison;
[tex]parameter, t' = 1.33 \times 1.61 = 2.14 \approx 2.0[/tex]
Thus, in comparing of result of experiment A and B, the result of A is by 2 factors differ from experiment B.
Learn more here: https://brainly.com/question/12891261
