Help with last part of the calculus question for integration formulas and the net change theorems

Question

My daughter is having a problem getting the last part of her calculus problem. The problem is listed as:

A particle moves along a straight line and its position at time t is given by s(t)=2t^3−24t^2+42t where s is measured in feet and t in seconds.

She was able to solve for the velocity at time t (42), when it stopped both times (1 and 7), and the position at 16 (2720). The last part of the problem asks to solve for the total distance traveled between 0 and 16. She originally got 5376, which the online assignment said was wrong despite another online source showing the same answer. Another source showed the answer to be 512, which again was wrong. Can someone show us how she needs to work this to get the correct answer? Thanks.

Answer 1

s-s₀ = integral from 0 to a final time of vdt. This is displacement.

distance travelled = integral from 0 to t of |v|dt (or integrating from 0 to first zero of v, from 1st to 2nd zero, and from 2nd zero to the final time. of 16 secs .with the rule that you take the negative of the integral while v is negative.

However, the integral is equal to s(t), so you can take the absolute value of the difference of the values of s between the endpoints

dist = (s(1) - s(0)) - (s(2)-s(1))+(s(16)-s(2)) (sign switches at each 0 +,-,+)

dist = (20-0) - (-196 - 20) + (2720 - (-196)) Be careful with the signs.

As you can see this is the displacement for each interval made positive when the slope (velocity) is negative