Position of Particle over 6 seconds

Question

Suppose the position of a particle is given by

s=f(t)=2t^3-15t^2+24t,

where t is measured in seconds and s is in feet. Find the total distance the particle moves in the first six seconds.

Answer 1

s(t) = 2t³ - 15t²+ 24t = 0;

s'(t) = 6t² - 30t + 24 = 0; t² - 5t + 4 = 0, t = 1 s or t = 4 s

Distance d = ∫₀⁶i6t² - 30t + 24Idt = ∫₀¹(6t² - 30t + 24)dt + ∫₁⁴(- 6t² + 30t - 24)dt + ∫₄⁶(6t² - 30t + 24)dt =

(2t³ - 15t² + 24t)₀¹ + (- 2t³ + 15t² - 24t)₁⁴ + (2t³ - 15t² + 24t)₄⁶ = (2 - 15 + 24) + (- 2·64 + 15·16 - 24·4) -

(- 2 + 15 - 24) + (2·216 - 15·36 + 24·6) - (2·64 - 15·16 + 24·4) = 11 + 16 + 11 + 36 + 16 = 90 feet

Answer 2

s(6)= 2(6)^3 - 15(6)^2 + 24(6) = 432 - 540 + 144 = 36 feet

s(0) = 0

total distance = s(6) - s(0) = 36 - 0 = 36 = total displacement from time t=0 to t=6

36 feet is the difference in positions between time zero and 6 seconds.

unless you're counting negative displacement as positive, then summing all movement. then see Yefin S's answer, with a little modification

the extrema positions are found by taking the derivative & setting = 0

s' = 6t^2 -30t + 24 = 0

t^2 -5t + 4 = 0

(t-4)(t-1) = 0

t= 1, 4

in one second a local maximum position is reached,

s(1) = 2 -15 + 24 = 11. It moved right 11 in one second

at 4 seconds, a negative local and global minimum position is reached

s(4) = 2(64) -240 + 96= -16, left 27, and right 52 (16+36)

add 11 +27 + 52 = 90 feet = total distance moved, left and right.

if you calculated the length of the cubic line, you'd get slightly more, but you don't want the curve's length, you want the left & right distances only.