Placement And Displacement

**metallica007** · January 8th 2010, 07:10 PM

Hi my friends ,, again
Sorry for asking to many question in short time but what can i do , i am studying calculus all by myself ( wish me luck

)
in the following question

what the different when it comes to Placement and Displacement
should i divide the integration when i try find the displacement ??
i dont need u to solve the question for me , i just want to understand the general rule for this case

Thanks in advanced

**TKHunny** · January 8th 2010, 07:51 PM

"Placement" isn't normally encountered in this sort of a problem. I suppose it might mean "starting somewhere". In the absence of other information, one might assume that one starts at the Origin.

Displacement is simply how far one has travelled.

Always, and the notation is ubiquitous, is this relationship for constant acceleration.

$s(t) = -½at^{2} + v_{0}t + h_{0}$

h_{0} is the place to start - your "placement", I suppose. If you start at zero, $s(0) = 0 = h_{0}$ .

This leaves: $s(t) = -½at^{2} + v_{0}t$

The first derivative: $s'(t) = v(t) = at + v_{0}$ gives the velocity.

The second derivative: $s"(t) = v'(t) = a(t) = a$ gives the acceletation.

It is this last equation that will fill your head with solutions.

$s"(t) = v'(t) = a(t)$

Given v(t), find v'(t) to get the acceleration.
Given v(t), find the antiderivative to get the displacement. You will also need to know where to start.

Givenm then v(t) = t^2 - t, we have $s(t) = (1/3)t^3 - (1/2)t^2 + h_{0}$ . In this case, since we are asked only how far things move, not where they land, we can ignore the initial displacement so that $h_{0} = 0$ .

Simple enough, then, s(5) = (1/3)125 - (1/2)25 = 25((1/3)5 - (1/2))

That's enough of that. You tell me why Part B is even a different question. They look rather similar, don't they?

Hint: Is our friendly particla ALWAYS moving in the same direction?

**Soroban** · January 8th 2010, 08:15 PM

Hello, metallica007!

Who uses the term "placement"?
The location of the particle is usually called its Position.

A particle moves along a line with the velocity function: . $v(t) \:=\:t^2-t$
. . where $v$ is measured in meters per second.

Find:

(a) the displacement during the time interval [0, 5]

The position function is the integral of the velocity function.

$s(t) \;=\;\int (t^2 - t)\,dt \;=\;\tfrac{1}{3}t^3 - \tfrac{1}{2}t^2 + s_o$

. . where $s_o$ is the particle's initial position (at $t = 0$ ).

At $t=0\!:\;\;s(0) \:=\:\tfrac{1}{3}(0^3) - \tfrac{1}{2}(0^2) + s_o \;=\;s_o$

At $t = 5\!:\;\;s(5) \;=\;\tfrac{1}{3}(5^3) - \tfrac{1}{2}(5^2) + s_o \;=\;\frac{175}{6} + s_o$

During that time interval, its displacement is:
. . $s(5) - s(0) \;=\;\left(\frac{175}{6}+s_o\right) - s_o \;=\;\frac{175}{6} \;=\;29\tfrac{1}{6}\text{ meters}$

(b) the distance traveled during the time interval [0, 5].

The particle changes direction when $v = 0.$

. . $t^2-t \:=\:0 \quad\Rightarrow\quad t(t-1) \:=\:0 \quad\Rightarrow\quad t \;=\;0,\;1$

At $t=0$ , its position is: . $s(0) = s_o$

At $t = 1$ , its position is: . $s(1) \:=\:\tfrac{1}{3}(1^3) - \tfrac{1}{2}(1^2) + s_o \;=\;s_o-\tfrac{1}{6}$

Its displacement is: . $s(1) - s(0) \;=\;\left(s_o - \tfrac{1}{6}\right) - s_o \;=\;-\tfrac{1}{6}$
. . In the first second, the particle moved $\tfrac{1}{6}$ meter to the left.

At $t = 5$ , its position is: . $s(5) \;=\;\tfrac{1}{3}(5^3) - \tfrac{1}{2}(5^2) + s_o \;=\;\tfrac{175}{6} +s_o$

Its displacement is: . $s(5) - s(1) \;=\;\left(\tfrac{175}{6} + s_o\right) - \left(s_o - \tfrac{1}{6}\right) \;=\;\tfrac{88}{3}$
. . In the next 4 seconds, it moved $29\tfrac{1}{3}$ meters to the right.

Therefore, it moved a total distance of: . $\tfrac{1}{6} + 29\tfrac{1}{3} \;=\;29\tfrac{1}{2}$ meters.

**metallica007** · January 8th 2010, 08:48 PM

Originally Posted by Soroban

Hello, metallica007!

Who uses the term "placement"?
The location of the particle is usually called its Position.

The position function is the integral of the velocity function.

$s(t) \;=\;\int (t^2 - t)\,dt \;=\;\tfrac{1}{3}t^3 - \tfrac{1}{2}t^2 + s_o$

. . where $s_o$ is the particle's initial position (at $t = 0$ ).

At $t=0\!:\;\;s(0) \:=\:\tfrac{1}{3}(0^3) - \tfrac{1}{2}(0^2) + s_o \;=\;s_o$

At $t = 5\!:\;\;s(5) \;=\;\tfrac{1}{3}(5^3) - \tfrac{1}{2}(5^2) + s_o \;=\;\frac{175}{6} + s_o$

During that time interval, its displacement is:
. . $s(5) - s(0) \;=\;\left(\frac{175}{6}+s_o\right) - s_o \;=\;\frac{175}{6} \;=\;29\tfrac{1}{6}\text{ meters}$

The particle changes direction when $v = 0.$

. . $t^2-t \:=\:0 \quad\Rightarrow\quad t(t-1) \:=\:0 \quad\Rightarrow\quad t \;=\;0,\;1$

At $t=0$ , its position is: . $s(0) = s_o$

At $t = 1$ , its position is: . $s(1) \:=\:\tfrac{1}{3}(1^3) - \tfrac{1}{2}(1^2) + s_o \;=\;s_o-\tfrac{1}{6}$

Its displacement is: . $s(1) - s(0) \;=\;\left(s_o - \tfrac{1}{6}\right) - s_o \;=\;-\tfrac{1}{6}$
. . In the first second, the particle moved $\tfrac{1}{6}$ meter to the left.

At $t = 5$ , its position is: . $s(5) \;=\;\tfrac{1}{3}(5^3) - \tfrac{1}{2}(5^2) + s_o \;=\;\tfrac{175}{6} +s_o$

Its displacement is: . $s(5) - s(1) \;=\;\left(\tfrac{175}{6} + s_o\right) - \left(s_o - \tfrac{1}{6}\right) \;=\;\tfrac{88}{3}$
. . In the next 4 seconds, it moved $29\tfrac{1}{3}$ meters to the right.

Therefore, it moved a total distance of: . $\tfrac{1}{6} + 29\tfrac{1}{3} \;=\;29\tfrac{1}{2}$ meters.

Thank u i get it now
if i integrated from 0 to 5 i get the total displacement and if i divide the integration i get the total distance
distance>=displacement

Thank you

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