Question

. I keep getting 108 and rechecking it but it is wrong?! A particle moves according to a law of motion s = f(t) = t^3 - 12t^2 + 21t, t > 0, where t is measured in seconds and s in feet. Find the total distance traveled during the first 8 s.

Answer 1

So we're obviously going to need to know f(0) and f(8). If the particle is always moving in the same direction, then it's displacement over the time period t = 0 to 8 is just
f(8) - f(0)
and the distance travelled is the absolute value of that number.

What you'll want to check for is whether or not the particle changes direction during that time. If so, you'll need to add up the displacement of each of the time segments where it is moving in one direction.

Answer 2

well i got the number 1 and 7. so abs s(1)-s(0), abs s(7)-s(1), abs s(8)-s(7)= 108?

Answer 3

i have also tried 88 and its wrong too

Answer 4

I haven't calculated this myself yet, let's see.
If f(t) = t^3 - 12t^2 + 21t
then f'(t) = 3t^2 - 24t + 21 = 0 <=> t^2 - 8t + 7 = 0 <=> t = 1 or 7
f''(t) = 6t - 24. As f(1) = -18 < 0, t = 1 is a local max; f(7) > 0, t = 7 a local min.
Now,
f(0) = 0
f(1) = 10
f(7) = -98
f(8) = -88

Answer 5

Check my work -- I'm more prone (as I think many of us are) to make more mistakes when we're not using pen and paper.

Answer 6

Now, if that's all correct then the distance travelled is ...

Answer 7

|10 - 0| + |10 -(-98)| + | -98 -(-88)|
= 10 + 108 + 10
= 128

Answer 8

yes thats it! nice i see what i did wrong. could u help me with anoter problem i cant figure out what im doing wrong in?

Answer 9

Post it on the left and I'll see if I can get to it