Problem (acceleration/velocity/distance)

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The discussion focuses on the motion of a particle with acceleration defined by a(t) = cos(t) + 3. The initial conditions are given as v(0) = 5 and s(0) = 3. The velocity function is derived as v(t) = sin(t) + 3t + 5. To calculate the total distance traveled from t = π to t = 2π, it is essential to consider the behavior of the velocity function and identify points where the particle changes direction, necessitating the use of absolute values in the integral calculation.

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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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