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I have trouble understanding the following problem's answer:

Which of the following relations are true for any arbitrary motion in space?

(a) $v_{avg}=\frac{r(t_2)-r(t_1)}{t_2-t_1}$

(b) $v_{avg}=\frac{v(t_1)+v(t_2)}{2}$

(c) $a_{avg}=\frac{v(t_2)-v(t_1)}{t_2-t_1}$

Clearly b is wrong.

But my book says a and c are correct:

avg velocity=total distance by total time, so, how is the first part taking out total dist / total time? I would think

$$v_{avg}=\frac{v_1(t_1)+v_2(t_2)}{t_1+t_2}$$

and similarly for avg acceleration. Why isn't this the case?

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  • $\begingroup$ The OP does show some effort, and does tell us the concept at which he or she is stuck. $\endgroup$ – garyp Sep 15 '14 at 12:59
  • $\begingroup$ Remember that $r(t_1)$ and $t_1$ are coordinates. $\endgroup$ – garyp Sep 15 '14 at 12:59
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    $\begingroup$ think units. $v(t_1)$ is velocity at time, $t_1$. So a velocity divided by a time can't give you a velocity. $\endgroup$ – Jim Sep 15 '14 at 13:26
  • $\begingroup$ Related: physics.stackexchange.com/q/55809/2451 $\endgroup$ – Qmechanic Sep 15 '14 at 13:54
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If I understand your question correctly you are asking why you should use $t_2-t_1$ in the denominator instead of $t_2+t_1$. The reason is that both measurements are taken with respect to some initial time. This initial time is arbitrary; think about it as the time when you started your stopwatch. The total time elapsed between the events occurring at $t_1$ and $t_2$ is therefore $t_2-t_1$.

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