Question 3: What is the net force on a toy locomotive at t=3s?

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The net force on a toy locomotive, described by the position equation x = t^3 - 6t^2 + 9t, is calculated using derivatives to find velocity and acceleration. At t = 2 seconds, the net force equals zero, as determined by setting the acceleration equation 0 = 6t - 12. The discussion emphasizes the importance of understanding derivatives in physics problems. Additionally, it highlights the community's role in providing guidance rather than performing complete homework checks.

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can someone check my work please thanks, oh btw I am still trying to learn how to use LaTeX mathematical typesetting; I figured itll be slightly faster taking a picture of the problem and my work and posting it here so please bare with me :redface:

1. The position of a toy locamotive moving on a straight track along the x-axis is given by the equation x = t^3 - 6t^2 + 9t, where x is in meters and t is in seconds. The net force on the locomotive is equal to 0 when t is equal to (a) zero (b) 2s (c) 3s (d) 4s (e) 5s


(my work)
x = t^3 - 6t^2 + 9t
v = 3t^2 - 12t -- first derivative
a = 6t - 12 -- second derivative
0 = 6t - 12
12 = 6t
t = 2s


Question 2: http://ourworld.cs.com/LezardV4leth/problem_2.jpg
Work: http://ourworld.cs.com/LezardV4leth/problem_2_work.jpg

Question 3: http://ourworld.cs.com/LezardV4leth/problem_3.jpg
Work: http://ourworld.cs.com/LezardV4leth/problem_3_work.jpg
 
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Looks ok to me...

BTW, don't count too much on us to systematically check your homework :smile:
We want to help with specific difficulties, misunderstandings or so, but we're not here to do the algebra checks for you :wink:
 
oh haha ^^;; Ill make note of that next time thanks a lot though
 

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