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📰 Since average speed equals speed at $ t = 2 $, the condition is satisfied for all $ a $, but we must ensure consistency in the model. However, the equality holds precisely due to the quadratic nature and linear derivative — no restriction on $ a $ otherwise. But since the condition is identically satisfied under $ b = 4a $, and no additional constraints are given, the relation defines $ b $ in terms of $ a $, and $ a $ remains arbitrary unless more data is provided. But the problem implies a unique answer, so reconsider: the equality always holds, meaning the condition does not constrain $ a $, but the setup expects a specific value. This suggests a misinterpretation — actually, the average speed is $ 8a $, speed at $ t=2 $ is $ 8a $, so the condition is always true. Hence, unless additional physical constraints (e.g., zero velocity at vertex) are implied, $ a $ is not uniquely determined. But suppose the question intends for the average speed to equal the speed at $ t=2 $, which it always does under $ b = 4a $. Thus, the condition holds for any $ a $, but since the problem asks to find the value, likely a misstatement has occurred. However, if we assume the only way this universal identity holds (and is non-trivial) is when the acceleration is consistent, perhaps the only way the identity is meaningful is if $ a $ is determined by normalization. But given no magnitude condition, re-express: since the equality $ 8a + b = 4a + b $ reduces to $ 8a = 8a $, it holds identically under $ b = 4a $. Thus, no unique $ a $ exists unless additional normalization (e.g., $ s(0) = 0 $) is imposed. But without such, the equation is satisfied for any real $ a $. But the problem asks to find the value, suggesting a unique answer. Re-express the condition: perhaps the average speed equals the speed at $ t=2 $ is always true under $ b = 4a $, so the condition gives no new info — unless interpreted differently. Alternatively, suppose the professor defines speed as magnitude, and acceleration is constant. But still, no constraint. To resolve, assume the only way the equality is plausible is if $ a $ cancels, which it does. Hence, the condition is satisfied for all $ a $, but the problem likely intends a specific value — perhaps a missing condition. However, if we suppose the average speed equals $ v(2) $, and both are $ 8a + b $, with $ b = 4a $, then $ 8a + 4a = 12a $? Wait — correction: 📰 At $ t = 3 $: $ s(3) = 9a + 3b + c $ 📰 $ v(2) = 2a(2) + b = 4a + 4a = 8a $ — equal. 📰 Youll Be Astounded These Painting Games Will Transform Your Art Skills Overnight 5857884 📰 17 Hidden Meanings Behind Butterfly Tattoos You Need To Know Now 4944559 📰 Stop Searchingthis Simple Windows Shortcut For Screenshots Works In Seconds 8822702 📰 Watch How Movies Are Controling What You Feelmind Blowing Truth Exposed 3202623 📰 Action Unlock Boot Into Safe Mode Now And Fix Your Windows 10 Sluggish Startup 2052234 📰 City Of St Petersburg Water Department 7494533 📰 How Long Is March Madness Halftime 6463211 📰 Gladdihoppers 7551769 📰 In The Stars Benson Boone 6713344 📰 Cryoversia Unleashed Justin Biebers Divorce Shakes Fanbase To The Core 5438177 📰 No Healthy Upstream 3835355 📰 Year Round School 780317 📰 The Law Of Reflection States That The Angle Between The Incoming Ray And The Normal Equals The Angle Between The Outgoing Ray And The Normal 4587186 📰 Iccpr 179305 📰 The Shocking Truth About Pa Dockets Standard File System 4121109