Trial 2 12 021 06 s 07 s 07 s 06 s 01 ms2 Trial 3 104 018 07 s 06 s 07 s 06 s

Trial 2 12 021 06 s 07 s 07 s 06 s 01 ms2 trial 3 104

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Trial 2 12 ° 0.21 0.6 s 0.7 s 0.7 s 0.6 s 0.1 m/s^2 Trial 3 10.4 ° 0.18 0.7 s 0.6 s 0.7 s 0.6 s 0.1 m/s^2 Trial 4 8.6 ° 0.15 0.8 s 0.7 s 0.8 s 0.8 s 0.07 m/s^2 Trial 5 5.6 ° 0.10 1.0 s 1.0 s 1.0 s 1.0 s 0 m/s^2 a net δ a net a g % Error a net vs. a g Trial 1 2.2 m/s^2 0.58 m/s^2 2.5 m/s^2 12% Trial 2 2.2 m/s^2 0.58 m/s^2 2.1 m/s^2 4.8% Trial 3 2.2 m/s^2 0.58 m/s^2 1.8 m/s^2 22% Trial 4 1.3 m/s^2 0.30 m/s^2 1.5 m/s^2 13% Trial 5 0.8 m/s^2 2.2*10^-5 m/s^2 0.98 m/s^2 18% Calculation Trial 1 displacement = 0.8 m ( constant ) ´ t 1 = 0.5 s + 0.5 s + 0.7 s 3 = 0.6 s ´ v 1 = ∆ x ∆t = 0.8 m 0.6 s = 1.3 m / s ´ a net 1 = v ∆t = 1.3 m / s 0.6 s = 2.2 m / s 2 ´ d anet 1 = 2.2 1.7 = 0.3 m / s 2
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PHYS 2211L Name: Jonghyun Lee October 2, 2019 Motion on an Incline t Σ ( ¿ ¿ i t avg ) 2 N = ( 0.5 0.6 ) 2 +( 0.5 0.6 ) 2 +( 0.7 0.6 ) 2 5 = 0.1 s σ t 1 = ¿ δ x = 0.01 m = 0.00001 δ a 1 = a avg ( δ x x + 2 δ t t ) ¿ a avg ( 0.00001 0.8 + 2 σ t t avg ) = 0.58 m / s 2 % Error of a net 1 vs .a g 1 = | a net a g a g | 100 = | 2.2 2.5 2.5 | 100 = 11.2% Trial 2 ´ t 2 = 0.6 s ´ v 2 = 1.3 m / s ´ a net 2 = 2.2 m / s 2 ´ d a 2 = a −´ a = 2.2 1.7 = 0.3 m / s 2 σ t 2 = 0.1 s δ a 2 = 0.58 m / s 2 % Error of a net 2 vs .a g 2 = | 2.2 2.1 2.1 | 100 = 5.7%
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PHYS 2211L Name: Jonghyun Lee October 2, 2019 Motion on an Incline Trial 3 ´ t 3 = 0.6 s ´ v 3 = 1.3 m / s ´ a net 3 = 2.2 m / s 2 ´ d a 3 = a −´ a = 2.2 1.7 = 0.3 m / s 2 σ t 3 = 0.1 s δ a 3 = 0.58 m / s 2 % Error of a net 3 vs .a g 3 = | 2.2 1.8 1.8 | 100 = 23.3% Trial 4 ´ t 4 = 0.8 s ´ v 4 = 1 m / s ´ a net 4 = 1.3 m / s 2 ´ d a 4 = a −´ a = 1.3 1.7 =− 0.4 m / s 2 σ t 4 = 0.07 s δ a 4 = 0.30 m / s 2 % Error of a net 4 vs.a g 4 = | 1.3 1.5 1.5 | 100 = 16.6%
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PHYS 2211L Name: Jonghyun Lee October 2, 2019 Motion on an Incline Trial 5 ´ t 5 = 1 s ´ v 5 = 0.8 m / s ´ a net 5 = 0.8 m / s 2 ´ d a 5 = a −´ a = 0.8 1.7 =− 0. 9 m s 2 σ t 5 = 0 s δ a 5 = 2.2 10 5 m / s 2 % Error of a net 5 vs .a g 5 = | 0.8 0.98 0.98 | 100 = 18.4 % Average of the net acceleration for five trials ´ a net = a net 1 + a net 2 + a net 3 + a met 4 + a net 5 5 ¿ 0.005 0.005 + 0.007 + 0.001 + 0.001 5 =− 0.0002 m / s 2 Standard deviation of acceleration
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PHYS 2211L Name: Jonghyun Lee October 2, 2019 Motion on an Incline ´ d δ δ δ δ δ ( ¿¿ a 3 ) 2 +( ¿¿ a 4 ) 2 + ( ¿¿ a 5 ) 2 5 ( ¿¿ a 2 ) 2 + ¿ ( ¿¿ a 1 ) 2 + ¿ ¿ ¿ Σ ( ¿¿ anet ) 2 N = ¿ σ anet = ¿ ¿ (− 0.005 ) 2 +(− 0.005 ) 2 +( 0.007 ) 2 +( 0.001 ) 2 +( 0.001 ) 2 5 = 0.004 m / s 2 Results Part A, Displacement vs. Time For the first part of the experiment, we focus on changing displacement and then recording the time for figuring out that the displacement affects the acceleration. The results of this part are if reducing traveled distance is not related to acceleration. Only the first trial contains different acceleration-value. I guess that the displacement is much short, so we push the timer early. That is why the first trial`s acceleration has a bigger value than others. The most interest fact from part A is that the average and net acceleration are the same. The reason why they are the same is the ignoration of friction. The ‘net’ means the sum of all acceleration, but we ignore the friction. Therefore, the average acceleration and net acceleration are the same in this case: also, the average and final velocity are equal because of the same reason.
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PHYS 2211L Name: Jonghyun Lee October 2, 2019 Motion on an Incline v / t v / t^2 v avg v f a avg a net 1 0.3 m/s 0.5 m/(s^2) 0.3 m/s 0.3 m/s 0.5m/(s^2) 0.5 m/(s^2) 2 0.2 m/s 0.1 m/(s^2)
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