# Different values of same physical quantity in time domain and space domain OR Physical explanation of Cauchy Schwarz result

I read this example in a popular math book I was browsing, and the author does not give any physical explanation.

The example: Consider a rock falling freely under gravity. There are no viscous or dissipative effects. We will find the average speed at the end of 5 seconds.

Time-domain. $v = 32 t \mbox{feet/sec}$ Average over $5$ seconds $$\langle v\rangle_\text{time} = \frac{1}{5} \int_0^5 32 t \mathrm{d}t = 80\text{ feet/sec}$$

Space Domain. At $t =5$ sec, $y = 400$ feet. $$\langle v\rangle_\text{space} = \frac{1}{400}\int_0^{400} 8\sqrt{y}\; \mathrm{d}y \approx 107\text{ feet/sec}$$

Sorry about the use of imperial units. But I just dont understand the physics of this. The source uses this only as a demonstration of Cauchy Schwarz inequality. Is there something elementary I am missing out here?

Essentially, he derives $$\frac{1}{T}\int_0^T v(t) \mathrm{d}t \leq \frac{1}{L}\int_0^L v(y)\mathrm{d}y$$

The math is ok to follow, but my physical intuition just cant get it somehow.

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I'm not sure I see what you're actually asking about. Are you confused about why the two different averages have different numerical values? (Also, no need to apologize for your units.) – David Z Aug 22 '11 at 20:43
@David they are averaged over the same two events (classically), and represent the same physical quantity. Why are the values different, and in general, why is it an ineuqlaity? – yayu Aug 22 '11 at 20:46

Let us denote by $\langle\cdot\rangle \equiv \int_0^T \cdot \frac{dt}{T}$ the averaging over time. Please notice that:
$$\frac{1}{L}\int_0^Lv\mathrm{d}y=\frac{1}{\langle v\rangle T}\int_0^Tv\frac{\mathrm{d}y}{\mathrm{d}t}\mathrm{d}t = \frac{1}{\langle v\rangle}\int_0^Tv^2\frac{\mathrm{d}t}{T} = \frac{\langle v^2\rangle}{\langle v\rangle}$$
Thus the inequality is given by: $\langle v\rangle \le \frac{\langle v^2\rangle}{\langle v\rangle}$, or $\langle v\rangle^2 \le \langle v^2\rangle$, which is indeed the the Cauchy-Schwarz inequality.