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I am studying physics equations for constant acceleration and I am having quite a hard time understanding the following.

The average velocity is given by definition as $v_{av-x}=\cfrac{x-x_0}{t-0}$ and this is just the slope of the line connecting points $(x,t)$ and $(x_0,0)$.This is just fine.

Now,another way to calculate $v_{av-x}$ is by taking the average of the $x$-velocities at the beginning and end of the interval (in the $v-t$ diagram),so $v_{av-x}=\cfrac{1}{2} \cdot (v_{0x} + v_x) $ (where the subscript $0x$ indicate the velocity at the beginning of the interval,i.e. when $t=0$, and $x$ at the end of the interval at a given $t$).

My problem is that I don't see how to prove the last statement mathematically,leaving the physics intuition aside.

Please don't give answers of the form "that's obvious intuitively since the rate of change of velocity is constant" because my intuition always follows after I understand it mathematically .

Thanks in advance.

Mr. Y
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3 Answers3

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When acceleration is constant, the velocity has a constant slope, which means the graph of velocity over time is just a line.

If you're willing to accept, as a leap of faith, that the notion of average velocity defined as $v_{av-x} = \dfrac{\Delta x}{\Delta t}$ agrees with the notion of "average height" in the velocity vs time graph, then the result follows as a basic geometric fact: The figure in question is a trapezoid between the times $0$ and $t$, and the average height is indeed simply the average of the two heights at $0$ and $t$. To see this, use the area formula for the trapezoid, and divide by the width, which is $t$.

The fact that I've asked you to accept as a leap of faith is actually the fundamental theorem of calculus; since you haven't been exposed to calculus yet, you're probably not ready to see a rigorous proof. But I think that should give you a taste of the intuition.

Dustan Levenstein
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  • Thank you for this answer.This is less hard to take as a leap of faith ! :)(Actually I've been exposed to calculus,my physics book is calculus based). – Mr. Y Dec 23 '15 at 20:16
  • Ah, well. I assumed you hadn't been, since it's easy to derive these identities with some calculus - this is exactly what Justpassingby has done in their answer. Keep in mind that velocity is the derivative of position, and acceleration is the derivative of velocity. – Dustan Levenstein Dec 23 '15 at 20:36
  • The cute thing is that physics textbooks often derive these kinematics equations for constant acceleration without calculus. It works surprisingly well, because this is the only part that requires a little bit of hand-waving. – Dustan Levenstein Dec 23 '15 at 20:39
  • If it's not too much disturb ,could you tell me where can I find a proof that the average velocity is the averege height in the $v-t$ graph ? – Mr. Y Dec 24 '15 at 18:11
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    @Mr.Y The precise notion of average height here is given by $\dfrac{1}{t}\int_0^t v dt$, abusing the notation of the integral slightly, as physicists are prone to doing. Multiplying the claim that $\dfrac{1}{t} \int_0^t vdt = \dfrac{\Delta v}{\Delta t}$ by $t=\Delta t$ (since the initial time here is defined to be $0$), this claim is precisely the form of the second part of the Fundamental theorem of Calculus seen here. You can find a proof there, or in any standard Calculus text. – Dustan Levenstein Dec 25 '15 at 02:41
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For constant acceleration, ${\Delta x\over\Delta t}={\frac12a{\Delta t}^2 \over \Delta t}=\frac12 a\Delta t=\frac12\Delta v$.

Geometrically, the plot of velocity vs. time is a straight line, so for some some interval $\Delta t$, you end up with a trapezoid. You can think of the average value of $v$ as the average height of this trapezoid, which is the height of a rectangle with the same base and area, i.e., $(v_0\Delta t + \frac12(v-v_0)\Delta t)/\Delta t = \frac12\Delta v$.

amd
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    I don't think this really answers the question; you're using other identities which I believe are typically derived from the basic fact being asked about in this question. – Dustan Levenstein Dec 23 '15 at 20:04
  • @DustanLevenstein Fair enough. I’ve added a geometrical explanation that I see matches your answer. – amd Dec 23 '15 at 20:23
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If the equation of motion is

$$x(t)=at^2+bt+c$$

Then the instantaneous velocity is

$$v(t)=2at+b$$

The average of two velocities is

$$\frac{2at_1+b+2at_2+b}2=a(t_1+t_2)+b$$

The mean velocity is

$$\frac{at_2^2+bt_2+c-at_1^2-bt_1-c}{t_2-t_1}=a(t_2+t_1)+b$$

Justpassingby
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  • My problem is that my book states that fact and then proves this.By going backwards I am just more confused.However thanks for the answer. – Mr. Y Dec 23 '15 at 20:11