A particle moves on the x-axis with velocity v(t) = 3t^2 – 15t + 12 for 0 ≤ t ≤ 6. The particle is in position 4 at time 0.

A. Find the Position Function
B. Find the Displacement
C. Write the integrals required to find the total distance traveled and then solve using your calculator

A. s(t) = t^3 - 7.5t^2 + 12t + 4

B. 22

C. Integrals are of the velocity function 0-1 then 4-1 then 4-6. These are all added to equal 45. (the numbers on the left are on the bottom the numbers on the right are on the top of the integral. All should come out positive)

Step-by-step explanation: So the position function will be the integral of the velocity function. t^3 - 7.5t^2 + 12t + C and because the position is 4 at zero C = 4 so t^3 - 7.5t^2 + 12t + 4. To find displacement we just plug in our 6 t value into the position equation and subtract our zero t value s(6) = 22 then - s(0) = 4 = 18. (This is the same as taking the integral of the velocity function from 0 to 6). To find total distance we take the integrals of the negative values of velocity and the positive values of velocity and change the negative values to positive then add. The velocity function is less than zero from x = 1 to x = 4. So we take the integral of the velocity function 0 to 1 then the opposite of 1 to 4 (which is just 4 to 1) and then 4 to 6 and add them all. This becomes 5.5 + 13.5 + 26 = 45