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You have graphed position s as a function of time: $s(t) $in Physics notation. But you call time $t$ your independent variable $x$, and your $f$ refers to the mathematical shape of the function rather than the resulting Physical quantity position $s$, which is the dependent variable, a.k.a. $ y$ in maths.

The derivative with respect to t would be the velocity $v(t) = 5t$, a straight line. The second derivative would be the constant change of velocity, i.e. the acceleration $a(t) = 5.$

You "get" only what you draw in the first place (you said it: position). From there, it depends on what you do next with it (derivation, looking for zeros etc.) to get some other information.

You have graphed position s as a function of time: $s(t) $in Physics notation. But you call time $t$ your independent variable $x$, and your $f$ refers to the mathematical shape of the function rather than the resulting Physical quantity position $s$, which is the dependent variable, a.k.a. $ y$ in maths.

The derivative with respect to t would be the velocity $v(t) = 5t$, a straight line. The second derivative would be the constant change of velocity, i.e. the acceleration $a(t) = 5$

You "get" only what you draw in the first place (you said it: position). From there, it depends on what you do next with it (derivation, looking for zeros etc.) to get some other information.

And once you know that the acceleration $a(t)$ doesn't actually depend on the time here, you can define the constant a = 5 to rewrite $s(t) = \frac{1}{2}at^2$ or with your variables $f(x) = \frac{1}{2}ax^2$ which should look familiar.

You have graphed position s as a function of time: $s(t) $in Physics notation. But you call time $t$ your independent variable $x$, and your $f$ refers to the mathematical shape of the function rather than the resulting Physical quantity position $s$, which is the dependent variable, a.k.a.$ y$ in maths.

The derivative with respect to t would be the velocity $v(t) = 5t$, a straight line. The second derivative would be the constant change of velocity, i.e. the acceleration $a(t) = 5.$

You "get" only what you draw in the first place (you said it: position). From there, it depends on what you do next with it (derivation, looking for zeros etc.) to get some other information.

You have graphed position s as a function of time: $s(t) $in Physics notation. But you call time $t$ your independent variable $x$, and your $f$ refers to the mathematical shape of the function rather than the resulting Physical quantity position $s$, which is the dependent variable, a.k.a.  $ y$ in maths.

The derivative with respect to t would be the velocity $v(t) = 5t$, a straight line. The second derivative would be the constant change of velocity, i.e. the acceleration $a(t) = 5.$

You "get" only what you draw in the first place (you said it: position). From there, it depends on what you do next with it (derivation, looking for zeros etc.) to get some other information.

You have graphed position s as a function of time: $s(t) $in Physics notation. But you call time $t$ your independent variable$x$ and your $f$ refers to the mathematical shape of the function rather then the resulting Physical quantity position $s$, which is the dependent variable, a.k.a.$ y$in maths.

The derivative with respect to t would be the velocity $v(t) = 5t$, a straight line. The second derivative would be the constant change of velocity, i.e. the acceleration $a(t) = 5.$

You "get" only what you draw in the first place (you said it: position). Then it depends on what you do next with it, like derivation, looking for zeros etc. to be able to get some other information.

You have graphed position s as a function of time: $s(t) $in Physics notation. But you call time $t$ your independent variable  $x$, and your $f$ refers to the mathematical shape of the function rather than the resulting Physical quantity position $s$, which is the dependent variable, a.k.a.$ y$ in maths.

The derivative with respect to t would be the velocity $v(t) = 5t$, a straight line. The second derivative would be the constant change of velocity, i.e. the acceleration $a(t) = 5.$

You "get" only what you draw in the first place (you said it: position). From there, it depends on what you do next with it (derivation, looking for zeros etc.) to get some other information.