2
$\begingroup$

It looks to me that both arrive at the same place, the average velocity. Do I use whichever one based on the data available to me?

$\endgroup$
3
1
$\begingroup$

These formulas correspond to different things (the first hint to which is that they have different units). Under consant acceleration, the former is the average velocity while the latter is the acceleration.

$\endgroup$
1
  • $\begingroup$ I just realized my typo. I meant to say the displacement divided by the change in time. My fault $\endgroup$
    – asdfjkl
    Jun 13 at 21:01
1
$\begingroup$

$x$ can be expressed in terms of either $v_f$ or $v_i$ as follows. \begin{align} x &=v_ft-\tfrac12 at^2\\ x &=v_it+\tfrac12 at^2 \end{align}

Taking the time derivatives, we have \begin{align} \frac{\mathrm{d}x}{\mathrm{d}t} &=v_f-at\\ \frac{\mathrm{d}x}{\mathrm{d}t} &=v_i+at \end{align}

Adding the corresponding sides, we have

$$ \frac{\mathrm{d}x}{\mathrm{d}t} =\frac{v_f+v_i}{2} $$

End of proof!

Conclusion

If an accelerated car is moving with an initial speed $v_i$ and reaches a speed of $v_f$ after $t$, its travelled distance can be calculated in three forms:

\begin{align} x &=v_ft-\tfrac12 at^2\\ x &=v_it+\tfrac12 at^2 \\ x &= \frac{v_f+v_i}{2}t \end{align}

$\endgroup$
0
$\begingroup$

First one is for linear graphs and gives average, for ex, motion with constant acceleration. Second one gives instant velocity, for ex, motion with changing acceleration.

$\endgroup$
0
$\begingroup$

In general for constant acceleration, a, we have x = $x_0$ + $v_0$t + a$t^2$/2 where the subscript 0 denotes initial values. Also, v = $v_0$ +at.

Let's consider times $t_i$ and $t_f$ when v has values $v_i$ and $v_f$:

$v_i$ = $v_0$ + a$t_i$

$v_f$ = $v_0$ + a$t_f$

so ($v_f$ + $v_i$)/2 = $v_0$ + a($t_i$+$t_f$)/2

Now let's use the expression for x to find $\Delta$x:

$x_i$ = $x_0$ + $v_0$$t_i$ + a$t_i^2$/2

$x_f$ = $x_0$ + $v_0$$t_f$ + a$t_f^2$/2

Therefore $\Delta$x = $x_f$-$x_i$ = $v_0$($t_f$-$t_i$) + a($t_f^2$-$t_i^2$)/2

Dividing by $\Delta$t, i.e. ($t_f$-$t_i$):

$\Delta$x/$\Delta$t = $v_0$ + a($t_f$+$t_i$)/2

which is equal to the expression above for ($v_f$ + $v_i$)/2. So yes, you use the expression for which you have the data.

$\endgroup$

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