Problem (acceleration/velocity/distance)

[sapiental]

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!

 

[FunkyDwarf]

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.

 

[kyle8921]

Your response looks correct. The total distance traveled from t = pi to t = 2pi would be the difference between the position at t = 2pi and the position at t = pi, which is s(2pi) - s(pi), as you mentioned. This can also be calculated by taking the integral of the velocity function from t = pi to t = 2pi. Great job!