Math Help - motion problem and latex prob

1. motion problem and latex prob

actually i already have the ans..but it not correct when i compare it with the actual ans..tq for your time and help

A particle P move along straight line and passes through a fixed point O.It velocity, $

v ms^-1,

$
is given by $

v = 6t^2-4t-2,

$
where t is the time, in seconds, after passing through O.
( my latex wrong here, v ms^-1, how to correct it )

Question (a)
the time interval during which the particle moves towards the left.

Question (b)
the distance, in $

m

$
, travelled by the particle during the first three seconds.

Question (c)
Sketch the velocity-time graph of the motion of the particle for 0 to 3 ( how to write latex for 0 to 3 i.e [0,3]..

2. Hello, nikk!

A particle $P$ moves along straight line and passes through a fixed point $O$.

Its velocity is given by: . $v(t) \:=\: 6t^2-4t-2$
where $t$ is the time, in seconds, after passing through $O$.

(a) The time interval during which the particle moves towards the left.
"Moves towards the left" means that the velocity is negative.

The velocity function is an up-opening parabola.
It is negative between its $x$-intercepts.

$x$-intercepts: . $6t^2 - 4t - 2 \:=\:0 \quad\Rightarrow\quad 3(3t+1)(t-1) \:=\:0 \quad\Rightarrow\quad x \;=\;\text{-}\tfrac{1}{3},\;1$

The particle moves to the left on the interval: . $t \in \left(\text{-}\tfrac{1}{3},\;1\right)$

(b) the distance, in $m$, travelled by the particle during the first 3 seconds.
Let $c$ be the location of point $O$.
This is the position of the particle when $t = 0.$

The position function is given by: . $s(t) \:=\:\int v(t)\,dt$

We have: . $s(t) \;=\;\int(6t^2 - 4t - 2)\,dt \quad\Rightarrow\quad\boxed{s(t) \;=\;2t^3 - 2t^2 - 2t + c}$

The particle stops and changes directions when $v = 0.$

In (a), we found that this happens at: . $t = \text{-}\frac{1}{3}$ and $t = 1.$

At $t = 0\!:\;\;s(0) \:=\:c$

At $t = 1\!:\;\;s(1) \:=\:-2+c$

In the first second, it has moved: . $(-2+c) - c \:=\:-2$ . . . 2 units to the left.

At $t = 3\!:\;\;s(3) \:=\:30+c$

In the next two seconds, it has moved: . $(30+c)-(-2+c) \:=\: +32$ . . . 32 units to the right.

Therefore, in the first 3 seconds, it moves: . $2 + 32 \:=\:34$ units.

(c) Sketch the velocity-time graph of the motion of the particle for 0 to 3
We already know that it is a parabola, but we can plot some points.

. . $\begin{array}{|c|c|}
t & v(t) \\ \hline
0 & \text{-}2 \\ \frac{1}{3} & \text{-}\frac{8}{3} \\ 1 & 0 \\ 2 & 14 \\ 3 & 40 \end{array}$
. . I dropped the "c"

Code:
        |
|                 o (3,40)
|
|
|
|
|           o (2,14)
|
|
- - + - - o - - + - - + - -
|   (1,0)
(0,-2)o
| o (1/3,-8/3)
|

3. heheh tq...i get the same ans for a and c...except b

tq for the nice sketching by for question c..can we plot the line connect to each point....we can i get info on latex for graph plotting????

question (b)
hope u can guide me on this question,...this is my calculation

t = 0 ; s = 0
t = 1 ; s = -2
t = 2 ; s = 4
t = 3 ; s = 30

therefore , total distance = 0 +|-2| + 4 + 30 = 36 m

4. [QUOTE=nikk;408668]actually i already have the ans..but it not correct when i compare it with the actual ans..tq for your time and help

A particle P move along straight line and passes through a fixed point O.It velocity, $

v ms^-1,

$
is given by $

v = 6t^2-4t-2,

$
where t is the time, in seconds, after passing through O.
( my latex wrong here, v ms^-1, how to correct it )[quote]
For more than one symbol in superscript or subscript, enclose in { }: $v ms^{-1}$.

Question (a)
the time interval during which the particle moves towards the left.

Question (b)
the distance, in $

m

$
, travelled by the particle during the first three seconds.

Question (c)
Sketch the velocity-time graph of the motion of the particle for 0 to 3 ( how to write latex for 0 to 3 i.e [0,3]..
You say you got answers. Okay, what answers did you get? We can't tell if you made a mistake if you don't show what you got and how you got it!

5. [quote=HallsofIvy;408698][quote=nikk;408668]actually i already have the ans..but it not correct when i compare it with the actual ans..tq for your time and help

A particle P move along straight line and passes through a fixed point O.It velocity, $

v ms^-1,

$
is given by $

v = 6t^2-4t-2,

$
where t is the time, in seconds, after passing through O.
( my latex wrong here, v ms^-1, how to correct it )
For more than one symbol in superscript or subscript, enclose in { }: $v ms^{-1}$.

You say you got answers. Okay, what answers did you get? We can't tell if you made a mistake if you don't show what you got and how you got it!
tq...i get the same ans for a and c like Soroban...except b

6. Originally Posted by Soroban
Hello, nikk!

"Moves towards the left" means that the velocity is negative.

The velocity function is an up-opening parabola.
It is negative between its $x$-intercepts.

$x$-intercepts: . $6t^2 - 4t - 2 \:=\:0 \quad\Rightarrow\quad 3(3t+1)(t-1) \:=\:0 \quad\Rightarrow\quad x \;=\;\text{-}\tfrac{1}{3},\;1$

The particle moves to the left on the interval: . $t \in \left(\text{-}\tfrac{1}{3},\;1\right)$

Let $c$ be the location of point $O$.
This is the position of the particle when $t = 0.$

The position function is given by: . $s(t) \:=\:\int v(t)\,dt$

We have: . $s(t) \;=\;\int(6t^2 - 4t - 2)\,dt \quad\Rightarrow\quad\boxed{s(t) \;=\;2t^3 - 2t^2 - 2t + c}$

The particle stops and changes directions when $v = 0.$

In (a), we found that this happens at: . $t = \text{-}\frac{1}{3}$ and $t = 1.$

At $t = 0\!:\;\;s(0) \:=\:c$

At $t = 1\!:\;\;s(1) \:=\:-2+c$

In the first second, it has moved: . $(-2+c) - c \:=\:-2$ . . . 2 units to the left.

At $t = 3\!:\;\;s(3) \:=\:30+c$

In the next two seconds, it has moved: . $(30+c)-(-2+c) \:=\: +32$ . . . 32 units to the right.

Therefore, in the first 3 seconds, it moves: . $2 + 32 \:=\:34$ units.

We already know that it is a parabola, but we can plot some points.

. . $\begin{array}{|c|c|}
t & v(t) \\ \hline
0 & \text{-}2 \\ \frac{1}{3} & \text{-}\frac{8}{3} \\ 1 & 0 \\ 2 & 14 \\ 3 & 40 \end{array}$
. . I dropped the "c"

Code:
        |
|                 o (3,40)
|
|
|
|
|           o (2,14)
|
|
- - + - - o - - + - - + - -
|   (1,0)
(0,-2)o
| o (1/3,-8/3)
|
hai every body...i just check the ans for

Question (a)
the time interval during which the particle moves towards the left.

the correct ans is (0,1)

but

my ans and soroban is [-1/3,1]

any one can confirm it??

7. depends on the domain of $t$ ... is it $t \ge 0$ ?

8. Originally Posted by skeeter
depends on the domain of $t$ ... is it $t \ge 0$ ?
the question as below:
The time interval during which the particle moves towards the left.

so , t < 0? am i right?

9. Originally Posted by nikk
the question as below:
The time interval during which the particle moves towards the left.

so , t < 0? am i right?
you did not answer my question.

usually, most motion problems start at t = 0, not for some negative value of time.

10. [quote=HallsofIvy;408698]
For more than one symbol in superscript or subscript, enclose in { }: $v ms^{-1}$.
A better display is the following: $v \text{ m.s}^{-1}$. Units should indeed not be in italic. I typed "v\text{ m.s}^{-1}". (And the dot is to avoid confusion with milliseconds)

When writing physics in LaTeX (not in the forum), there is a package (SIunits) for displaying units more easily.

11. Hello, nikk!

A particle $P$ moves along astraight line
and passes through a fixed point $O$.

Its velocity is given by: . $v \:= \:6t^2-4t-2$
where $t$ is the time, in seconds, after passing through $O.$

(a) Find the time interval during which the particle moves towards the left.
Here's my interpretation of the problem . . .

It does not state anywhere that $t \geq 0.$

Suppose the position function is: . $s(t) \:=\:2t^3 - 2t^2 - 2t + 20$

At $t = 0\!:\;\;s(0) = 20$

Think about it . . . the particle has been moving for perhaps hours.

And the experimenter steps in and starts his/her stopwatch
. . when the particle passes through point $O: +20$
(A purely arbitrary moment.)

We can track the progess of the particle since that time:

. . $\begin{array}{ccc}s(0) &=& 20 \\ s(1) &=& 18 \\ s(2) &=& 24 \\ s(3) &=& 50 \\ s(4) &=& 108 \\ \vdots && \vdots \end{array}$

And we can track the progress of the particle before that time:

. . $\begin{array}{ccc}
\vdots && \vdots \\
s(\text{-}4) &=& \text{-}137 \\ s(\text{-}3) &=& \text{-}46 \\ s(\text{-}2) &=& 0 \\ s(\text{-}1) &=& 18 \end{array}$

So, in this problem, there is such a thing as "negative time."