0
$\begingroup$

$\Delta x=\frac 12(v_f+v_i)\Delta t$ because its a linear combination it looks like a linear function, however also as time increases acceleration stays the same therefore velocity changes and that makes it look like a quadratic function at the same time, is it a linear function?

$\endgroup$
3
  • 1
    $\begingroup$ The function is already linear in time if $v_f$ and $v_i$ are assumed to be constant. This equation is to do with the average velocity over time, It doesn't offer much for acceleration, since there is no $\Delta t ^ 2$ terms in the equation. $\endgroup$ Apr 15, 2020 at 1:17
  • $\begingroup$ @Joshua Pasa, i don't think that vf is constant when acceleration is constant since $v_f=a\Delta t$ $\endgroup$
    – Zheer
    Apr 15, 2020 at 10:22
  • $\begingroup$ @Zheer Josh said if $v_f$ is assumed to be constant which it's not as you already know $\endgroup$
    – Andrew
    Apr 15, 2020 at 10:55

2 Answers 2

2
$\begingroup$

The equation you wrote gives the average displacement. It is not linear because $v_f = a\Delta t$ (if $a$ is constant as you said). Average displacement is not the same as total displacement and we are typically interested in average velocity instead of average displacement. Both $v_f$ and $\Delta t$ are growing linearly and the product of $v_f$ and $\Delta t$ grows quadratically as can be seen below:

Substitute $a \Delta t$ for $v_f$ in your equation to see that $\Delta x_{avg} \propto \Delta t^2$

$\Delta x = \frac{1}{2}(a\Delta t + v_i)\Delta t = \frac{1}{2}[a (\Delta t)^2 + v_i\Delta t]$

It is indeed quadratic.

Note: by me writing $(\Delta t)^2$ I'm assuming that $t_i = 0$

Otherwise, we'd need to write $t_f^2 - t_i^2$

$\endgroup$
2
$\begingroup$

Your equation is a representation of $x$ as a linear combination of the speeds. But more practically, you will be considering $t$ to be the independent variable; if the speed $v_f$ depends on time, which it will in any scenario with nonzero acceleration, then one of the terms is itself function of $t$ so you do not have $x$ as a linear combination of your independent variable $t$, you have an equation of the form $x=at+f(t)t$ (where the nonlinear term $f(t)t$ will simplify to a quadratic $t^2$ in the case of constant acceleration).

2nd update pointing out further relationships:

When $a=0$ (no acceleration), $v(t)=v_0$ (a flat horizontal line indicating constant speed), and $x(t)=x_0+v_0t$ (a line where the slope is the speed, position as a linear function of time). The relationship of these graphs to each other will be clear if you are familiar with calculus. To review the more general case, if the acceleration is constant, $a=a_0$ (a flat horizontal line), $v(t)=v_0+a_0t$ (a line where the slope is acceleration, speed as a linear function of time) and $x(t)=x_0+v_0t+\frac{1}{2}a_0t^2$ (a parabola with vertex at $\frac{-v_0}{a_0}$, position as a quadratic function of time).

$\endgroup$
9
  • $\begingroup$ "then one of the terms is a function of t", does that mean it is function inside function? $\endgroup$
    – Zheer
    Apr 15, 2020 at 10:26
  • 2
    $\begingroup$ @Zheer Yes, it would be $\Delta x \, (v_f(t),t)$ which would ultimately be a quadratic function of $t$ after substituting for $v_f$ as I showed $\endgroup$
    – Andrew
    Apr 15, 2020 at 10:53
  • $\begingroup$ @Zheer edited, does that clear everything up for you? $\endgroup$ Apr 15, 2020 at 17:51
  • 1
    $\begingroup$ @Zeer Yes, but then $v_f=v_i$. The velocity is constant and the average displacement becomes equal to the total displacement - but if velocity were constant there'd be no point in using this equation in the first place (although there really never is any for the reason I stated in my comment). $\endgroup$
    – Andrew
    Apr 15, 2020 at 22:37
  • 1
    $\begingroup$ @Zeer You'd get a trivial answer by substituting $v_i$ for $v_f$ that would give you $\Delta x = v_i\Delta t$ which is just the total displacement occurring during $\Delta t $ at a constant velocity $\endgroup$
    – Andrew
    Apr 15, 2020 at 22:44

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.