If we were to form a carbocation, the tertiary carbocation is the most stable by hyperconjugation. However, there is no carbocation formed in E2 (no intermediate formed in E2), and I still don't get the idea of bond length
Time is a variable in reactions. Carbocations are not an either/or proposition. If a bond breaks, it does so over time. It begins at the measured bond length and ends at some point in which the separation is so large that a bond no longer exists. Therefore, hyperconjugation should also participate in a proportional degree.
As a bond breaks, the neighboring electrons can be donated to the growing carbocation. This electron donation also increases the acidity of the hyperconjugated proton.
We should see two effects here, first, neighboring electrons can compete with a nucleophile in an attack to the backside of a C-X bond. This will retard an attack. Second, this attack by neighboring electrons should increase as a bond breaks.
How do we know a bond is breaking? Check the distance. Let us
assume the methyl chloride bond length represents no bond breakage and no polarization. Therefore, any increase in bond length represents a weaker bond, a greater degree of polarization, a greater amount of hyperconjugation, a greater amount of backside interference, and a greater acidity of the beta hydrogens. If that were so, then I would argue a tertiary halide should react faster via an E2 reaction than a secondary or primary.
Those are assumptions for completely synchronous reactions. Since we know tertiary halides can break faster than nucleophiles can attack, if bond actually stretched rather than remain fixed and motionless, this stretching would increase the amount of bond breakage, hyperconjugation, etc.. The net effect would actually be increased for a tertiary halide in an E2 reaction.
I don't know if that helps or not, but that is how I envision these elimination reactions taking place. I also use this model to predict Zaitsev versus Hoffmann elimination. More bond breakage is more carbocation-like, more hyperconjugation, and more Zaitzev products. The reverse is Hoffmann products. Ethanol/ethoxide with bromide or iodide favor Zaitzev products as rationalized above. Fluorides and ammonium salts have less bond polarization, hence depend on CH acidity and favor Hoffmann products.