Problem (acceleration/velocity/distance)

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The particle's acceleration is given by a(t) = cos(t) + 3, leading to a velocity function v(t) = sin(t) + 3t + 5 after integrating with initial conditions. The position function is derived as s(t) = -cos(t) + (3/2)t^2 + 5t + 3. To find the total distance traveled from t = π to t = 2π, it's essential to identify points where the velocity changes sign, as integrating directly could yield incorrect results due to negative distances. Properly accounting for these sign changes is crucial for accurately calculating total distance. The discussion emphasizes the importance of using absolute values in integration when dealing with oscillating functions.
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A particle moves with acceleration a(t) = cos(t)+3

Intital velocity v(0)= 5
initial position s(0) =3

a) fin velocity of particle at time t

v(t) = sin(t)+3t+C
plugging 0 for t I get C = 5

s(t) = -cos(t) + 3/2t^2 + 5t + C
plugging 0 for s I get C = 3

answer: v(t) = sin(t)+3t+5

b) What is the total distance traveled from t = pi to t = 2pi

integral pi to 2pi = v(t) = sin(t)+3t+5
= s(2pi)-s(pi)Thanks!
 
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Im not sure what the question is but I am guessing your asking how to do the last part. You can't just intergrate if youve got trig functions all over the place unless you have some well placed absolute value signs. This is because you are subtracting the 'negative' distances which you actually want to add to find the total. Find the points where it turns around (think about what v(t) is at this point) then intergrate between those points.
 
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