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

[karush]

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.

before we answer these questions
how do you plot this using tikx

 

[HallsofIvy]

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".​

 

[skeeter]

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 $t=\sqrt{2}$

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

 

[I like Serena]

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}

 

[karush]

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

 

[karush]

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

 

[karush]

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

 

[skeeter]

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 $0 \le t \le \sqrt{6}$