position and time, derivative.

**mybrohshi5** · September 20th 2009, 01:00 PM

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

**skeeter** · September 20th 2009, 01:05 PM

Originally Posted by mybrohshi5

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 $\textcolor{red}{= \int_0^8 |v(t)| \, dt}$

...

**mybrohshi5** · September 20th 2009, 01:12 PM

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

**skeeter** · September 20th 2009, 01:29 PM

Originally Posted by mybrohshi5

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

$s(t) = t^4 - 4t + 21<br />$

$v(t) = s'(t) = 4t^3 - 4$

$4t^3 - 4 = 0$ at $t = 1$

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

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

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

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

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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