Need help with calculus problem! 10pts!

A particle moves along a straight line and its position at time t is given by s(t)= 2t^3 - 27 t^2 + 84 t where s is measured in feet and t in seconds.

What is the TOTAL distance the particle travels between time 0 and time 10?

s(t) = 2t³ − 27t² + 84t

s'(t) = 6t² − 54t + 84
s'(t) = 6 (t² − 9t + 14)
s'(t) = 6 (t − 2) (t − 7)

s'(t) = 0 -----> t = 2, t = 7

s''(t) = 12t − 54
s''(2) = 24 − 54 = −30 < 0 ----> local maximum
s''(7) = 84 − 54 = 30 > 0 -----> local minimum

From t = 0 to t = 2, particle moves to the right
From t = 2 to t = 7, particle moves to the left
From t = 7 to t = 10, particle moves to the right

Total distance travelled
= (s(2) − s(0)) + (s(2) − s(7)) + (s(10) − s(7))
= (76 −0) + (76 − (−49)) + (140 − (−49))
= 76 + 125 + 189
= 390 ft

s(t) = 2t^3 -27t^2+84t First find when the particle stops ie v(t) = 0

ds/dt = 6t^2 -54t +84 6t^2 -54t +84 = 0 t^2 -9t + 14 = 0 (t-7)(t-2) = 0 Stopped when t = 2 or 7


Now find how far its position from t = 0 until t = 2 s(0)= 0 s(2) = 76 Distance travelled = 76

Now find position when t = 7 s(7) = -49 distanced travelled from +76 to -49 = 125

s(10) = 140 distance travelled from s(7) to s(10) = 140 +49 = 189

Total distance = 76 + 125 + 189 = 390

compute for s(10)

s(10) = 2(10^3) -27(10^2) + 84(10)
= 2(1000) - 27(100) + 84(10)
=2000-2700+84
= -616

since this is asking for distance, the answer is 616 ft.