# 2.62 Calculate the objects position and acceleration (using tikx for graph)

• MHB
• karush
In summary, the velocity of an object is given by $v(t)=\alpha-\beta t^2$ where $\alpha=4.00\, m/s$ and $\beta=2.00 \, m/s^3$. The object's position and acceleration can be calculated as functions of time using the derivatives and antiderivatives of the velocity equation. The object will move to the
karush
Gold Member
MHB
2.62 An object's velocity is measured to be
$v(t)=\alpha-\beta t^2$ where $\alpha=4.00\, m/s$ and $\beta=2.00 \, m/s^3$
At $t=0$ the object is an $x=0$.

(a) Calculate the objects position and acceleration as functions of time
(b) What is the object's maximum positive displacement from the origin.

how do you plot this using tikx

Last edited:
I know nothing at all about "tikx". but the math problem is easy to solve. If the velocity at tiime t is $$4- 2t^2$$ and x= 0 when t= 0 then the acceleration at time t is the derivative $$-4t$$ and its position is the anti-derivative, $$4t- \frac{2}{3}t^3$$.

The object, starting at x= 0 will move to the right until its velocity drops to 0 then move back left. Its velocity is 0 when $$4- 2t^2= 0$$ so $$2t^2= 4$$, $$t^2= 2$$, $$t= \pm\sqrt{2}$$. When $$t= \sqrt{2}$$, its position is $$4\sqrt{2}- \frac{16}{3}$$.

However, after that the object moves back to the left without ever stopping there is no "maximum displacement from the origin".

Last edited by a moderator:
Over the interval $0 \le t < \sqrt{2}$, the object has a positive velocity and moves rightward from its initial position at $x=0$. For $t > \sqrt{2}$, the object has a negative velocity and moves left.

Max positive displacement from its initial position at the origin occurs when velocity changes sign from positive to negative at time $t=\sqrt{2}$

$\Delta x = x(\sqrt{2})-x(0) = 4\sqrt{2} - \dfrac{4\sqrt{2}}{3} = \dfrac{8\sqrt{2}}{3}$

We have a 'Live TikZ Editor' on this site. See the topmost link in the MHB Widgets when replying.
It's also explained in the https://mathhelpboards.com/tikz-pictures-63/tikz-announcement-22140.html.

We can for instance do:
[LATEXS]
\begin{tikzpicture}
%preamble \usepackage{pgfplots}
\begin{axis}[
xlabel={$t$}, ylabel={$v$},
grid=both, axis lines=middle,
xmin=0,ymax=12
]
\addplot[blue, smooth, ultra thick] (x, 4-2*x^2);
\addlegendentry{$v(t)$}
\end{axis}
\end{tikzpicture}
[/LATEXS]
\begin{tikzpicture}
%preamble \usepackage{pgfplots}
\begin{axis}[
xlabel={$t$}, ylabel={$v$},
grid=both, axis lines=middle,
xmin=0,ymax=12
]
\addplot[blue, smooth, ultra thick] (x, 4-2*x^2);
\addlegendentry{$v(t)$}
\end{axis}
\end{tikzpicture}

cool...

I want to start using it a lot...

I am just cutting and pasting from desmos (just a lot of steps)
thot the access here was too domineering

HallsofIvy said:
I know nothing at all about "tikx". but the math problem is easy to solve. If the velocity at tiime t is $$4- 2t^2$$ and x= 0 when t= 0 then the acceleration at time t is the derivative $$-4t$$ and its position is the anti-derivative, $$4t- \frac{2}{3}t^3$$.

The object, starting at x= 0 will move to the right until its velocity drops to 0 then move back left. Its velocity is 0 when $$4- 2t^2= 0$$ so $$2t^2= 4$$, $$t^2= 2$$, $$t= \pm\sqrt{2}$$. When $$t= \sqrt{2}$$, its position is $$4\sqrt{2}- \frac{16}{3}$$.

However, after that the object moves back to the left without ever stopping there is no "maximum displacement from the origin".

where does -4t come from I thot the dx of 4 is 4t

skeeter said:
Over the interval $0 \le t < \sqrt{2}$, the object has a positive velocity and moves rightward from its initial position at $x=0$. For $t > \sqrt{2}$, the object has a negative velocity and moves left.

Max positive displacement from its initial position at the origin occurs when velocity changes sign from positive to negative at time $t=\sqrt{2}$

$\Delta x = x(\sqrt{2})-x(0) = 4\sqrt{2} - \dfrac{4\sqrt{2}}{3} = \dfrac{8\sqrt{2}}{3}$

ok so the interval determines the displacement

karush said:
ok so the interval determines the displacement

Area between the velocity curve and the time axis represents displacement ... the graph shown represents zero overall displacement from the origin over the interval $0 \le t \le \sqrt{6}$

## 1. What is the purpose of using tikx for graph in the calculation of an object's position and acceleration?

Tikx is a powerful tool for creating graphs in LaTeX, which allows for precise and customizable representations of mathematical functions. By using tikx for graph, we can visually represent the position and acceleration of an object in a clear and accurate way.

## 2. How is an object's position calculated using tikx for graph?

An object's position can be calculated using the equation x = x0 + v0t + ½at^2, where x is the position at time t, x0 is the initial position, v0 is the initial velocity, a is the acceleration, and t is the time. By plotting this equation on a tikx graph, we can see the object's position at different points in time.

## 3. Can tikx for graph be used to calculate acceleration from a position-time graph?

Yes, tikx for graph can be used to calculate acceleration from a position-time graph by finding the slope of the graph at a specific point. The slope of the graph represents the object's velocity, and the change in velocity over time is the acceleration.

## 4. How can tikx for graph help in understanding an object's motion?

Tikx for graph can help in understanding an object's motion by providing a visual representation of its position and acceleration. By plotting these values on a graph, we can see patterns and relationships that may not be apparent from just looking at the numerical data.

## 5. Is tikx for graph limited to only linear motion calculations?

No, tikx for graph can be used for various types of motion calculations, including linear, circular, and projectile motion. It is a versatile tool that can be customized to fit the specific needs of different types of calculations.

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