At t = 0 s, a car is M position x0 = 19.1 plusminus 0.3 m and traveling with vel

Question

At t = 0 s, a car is M position x0 = 19.1 plusminus 0.3 m and traveling with velocity v0 = 3.3 plusminus 0.2 m/s. It then accelerates with a constant acceleration of a = 4.5 plusminus 0.2 m/s2. what is the distance traveled after 8.9 plusminus 0.1 s? Also, calculate the uncertainty using the simple and the standard deviations methods and enter your values in the second and third boxes respectively. Recall that x = x0 + v0t + 1/2at2. Hint: To find the uncertainty using the standard deviations method, apply the rules of finding the uncertainty for each term first. Then combine using the rule for addition/subtraction. Not sure how to do this problem, click here for an example on calculating the uncertainty. 437 plusminus 14 or plusminus 9 m

Explanation / Answer

X =X0 + Ut + 0.5*a*t2 X = 19.1 + 3.3*8.9 + 0.5*4.5*8.9*8.9 X = 226.7 m using simple method uncertainty in distance / distance = 0.3/19.1 + 0.2/3.3 + 0.1/8.9 + 0.2/4.5 + 2*(0.1/8.9) uncertainty in distance = 35.01 m using standard deviations method uncertainty in distance / distance = sqrt{(0.3/19.1)^2 + (0.2/3.3)^2 + (0.1/8.9)^2 + (0.2/4.5)^2 + 2*(0.1/8.9)^2} uncertainty in distance = 14.2 m

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