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If the position of a particle is given by $x(t) = 47t − 3t^3$, where $x$ is in meters and $t$ is in seconds, Graph $x(t), v(t),$ and $a(t).$

I'm not sure how to approach this. Should I differentiate the equation to get $v$, then graph that? But what would be the $x$ and $y$ points?

Alex Ortiz
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user366783
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    Yes you should differentiate and double differentiate for velocity and acceleration. Take $t$ in the X-Axis – N.S.JOHN Sep 15 '16 at 06:30
  • How would I put it on a graph? – user366783 Sep 15 '16 at 06:33
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    You could have one of the directions have axes with different colors that correspond to the different variables (since position, velocity and acceleration have different units this could make things less confusing). – Emil Sep 15 '16 at 06:38
  • How do I even plot x? – user366783 Sep 15 '16 at 06:40
  • You use a program that can plot. Or draw it yourself. If you do not know what a scatterplot is I will not help you. – Emil Sep 15 '16 at 06:42

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You have $v(t)=-9t^2+47$ and $a(t)=-18t$. Then you plot it in the same manner as $x(t)$. You plot for example $y=v(t)$ as a function of $t$.

TheGeekGreek
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  • How would I put it on a graph? – user366783 Sep 15 '16 at 06:32
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    Take $ y = f(x) $ where $x$ is $t$ – N.S.JOHN Sep 15 '16 at 06:34
  • So, I'd put it a random number for t? So, t = 1, then x = 44? And t = 10, so x = -2,530? – user366783 Sep 15 '16 at 06:37
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    You would be plotting 3 separate functions on the same set of axes. Plot each one like you normally would (i.e. pick values of t, solve for x/v/a, put a point on the graph, etc). – Claycrusher Sep 15 '16 at 06:41
  • So, I'd put it a random number for t? – user366783 Sep 15 '16 at 06:47
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    Mostly a range of positive numbers since $t$ denotes time. But why do you want to plot it and which software you use? – TheGeekGreek Sep 15 '16 at 06:49
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    If you access to graphing software, that's the easiest way to graph. Plotting a few individual points on the graph might give you a general idea about it's shape. Additionally, it might help to also reason about what the graph should look like. Notice that $47t-3t^3 = t(47-3t^2) = t(\sqrt{47}-\sqrt{3}t)(\sqrt{47}+\sqrt{3}t)$ so the graph will have zeroes at $t = 0, \pm \sqrt{47/3}$ where $\sqrt{47/3} \approx 3.96$. – benguin Sep 15 '16 at 06:53
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    @user366783 You can use an online sketcher to visualize it when you are lost, like this link: https://graphsketch.com/?eqn1_color=1&eqn1_eqn=47x-3x%5E3&eqn2_color=2&eqn2_eqn=-9x%5E2%2B47&eqn3_color=3&eqn3_eqn=-18x&eqn4_color=4&eqn4_eqn=&eqn5_color=5&eqn5_eqn=&eqn6_color=6&eqn6_eqn=&x_min=-17&x_max=17&y_min=-100&y_max=100&x_tick=1&y_tick=1&x_label_freq=5&y_label_freq=5&do_grid=0&do_grid=1&bold_labeled_lines=0&bold_labeled_lines=1&line_width=4&image_w=850&image_h=525 – iadvd Sep 15 '16 at 06:54