position and time, derivative.

Sep 2009
230
7
A particle moves along a straight line and its position at time
is given by
where
is measured in feet and
in seconds.

1. Use interval notation to indicate when the particle is moving in the positive direction.
-----i thought this was [0,infinity) because it is always increasing?
but its wrong.

2. Find the total distance traveled during the first 8 seconds.
-----i thought i just plugged in 8 into the original equation but thats wrong too.

could i get some help on these two parts please =)

thank you

chad
 
Jun 2008
16,216
6,764
North Texas
A particle moves along a straight line and its position at time
is given by
where
is measured in feet and
in seconds.

1. Use interval notation to indicate when the particle is moving in the positive direction.
-----i thought this was [0,infinity) because it is always increasing?
but its wrong.

particle is moving in the (+) direction when v(t) > 0

2. Find the total distance traveled during the first 8 seconds.
-----i thought i just plugged in 8 into the original equation but thats wrong too.

total distance \(\displaystyle \textcolor{red}{= \int_0^8 |v(t)| \, dt}\)
...
 
Sep 2009
230
7
ok so if its >0 wouldnt the interval notation be (0,infinity)?

i just plugged that into my online math homework and it said that was wrong too?

and for the second part wouldnt it be 4064?
that was wrong too?

what am i doing wrong with these two?

thank you
 
Jun 2008
16,216
6,764
North Texas
ok so if its >0 wouldnt the interval notation be (0,infinity)?

i just plugged that into my online math homework and it said that was wrong too?

and for the second part wouldnt it be 4064?
that was wrong too?

what am i doing wrong with these two?

thank you
\(\displaystyle s(t) = t^4 - 4t + 21
\)

\(\displaystyle v(t) = s'(t) = 4t^3 - 4\)

\(\displaystyle 4t^3 - 4 = 0\) at \(\displaystyle t = 1\)

for \(\displaystyle 0 \le t < 1\) , \(\displaystyle v(t) < 0\) ... the particle is moving in the negative direction.

for \(\displaystyle t > 1\) , \(\displaystyle v(t) > 0\) ... the particle is moving in the positive direction.

\(\displaystyle |s(1) - s(0)|\) = distance particle moves between \(\displaystyle t = 0\) and \(\displaystyle t = 1\)

\(\displaystyle |s(8) - s(1)|\) = distance particle moves between \(\displaystyle t = 1\) and \(\displaystyle t = 8\)


your knowledge of the basics of rectilinear motion is lacking, particularly the difference between displacement and distance traveled ... recommend that you research the topic.
 
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