# Math Help - Displacement of particle motion

1. ## Displacement of particle motion

A particle moves along a line so tat its position at any time t>= 0 is given by the function

$s(t)= t^2-3t+2$

where s is measure in meters and t is measured in seconds

Find the displacement in meters and t is measured in seconds

The book gives this:
The displacement of the object over the time inverval from t to tx (they used delta t but this is easier for the purpose) is

delta S = f(t + tx) - f(t)

I did what I needed, by acting as this is a difference quotient without a limit or denominator, to solve normally and got this:

$tx^2 + 2txt - 3xt$

The answer, however, is 10.
I'm sure there is some reason why the above (as in my answer) = 2t. But, I don't know how.

Thank you!.

2. Originally Posted by Truthbetold
A particle moves along a line so tat its position at any time t>= 0 is given by the function

$s(t)= t^2-3t+2$

where s is measure in meters and t is measured in seconds

Find the displacement in meters and t is measured in seconds

The book gives this:
The displacement of the object over the time inverval from t to tx (they used delta t but this is easier for the purpose) is

delta S = f(t + tx) - f(t)

I did what I needed, by acting as this is a difference quotient without a limit or denominator, to solve normally and got this:

$tx^2 + 2txt - 3xt$

The answer, however, is 10.
I'm sure there is some reason why the above (as in my answer) = 2t. But, I don't know how.

Thank you!.
There seems to be something missing from the problem statement (at least
as far as I can tell). Meybe there are some implied conditions applicable to
the problems in the section of your course this comes from.

RonL

3. Originally Posted by Truthbetold
A particle moves along a line so tat its position at any time t>= 0 is given by the function

$s(t)= t^2-3t+2$

where s is measure in meters and t is measured in seconds

Find the displacement in meters and t is measured in seconds

.
.
.

The answer, however, is 10.
I'm sure there is some reason why the above (as in my answer) = 2t. But, I don't know how.
If the answer is 2t = 10 then obviously you have an upper value of t = 5? (Which you didn't mention.)

If so, then I am assuming the problem is to find the displacement between the times t = 0 s and t = 5 s?

$s(0) = 0^2 - 3 \cdot 0 + 2 = 2$

$s(5) = 5^2 - 3 \cdot 5 + 2 = 12$

so
$s(5) - s(0) = 12 - 2 = 10$

In the future, please write out the whole question.

-Dan