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📰 The instantaneous speed is $ v(t) = s(t) = 2at + b $, so at $ t = 2 $: 📰 v(2) = 2a(2) + b = 4a + b = 4a + 4a = 8a. 📰 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: 📰 Time Blocking App 📰 1050 Mstr Price Flash Is This The Greatest Deal Yet Making Waves 2885584 📰 You Wont Believe What This Hollister Dress Can Do 9778665 📰 Call Of Duty Update Today 6014575 📰 Mulitas Secret Youll Never Believe Makes Everything Work Forever 1397232 📰 Join The Rebellion Exclusive Heroes From Star Wars Galaxy Of Heroes Store Ahead 5960232 📰 Rationing Definition 8349758 📰 Usd To Ils Rate Today 📰 Ltc Price To Usd 869789 📰 From Nu Metal To Everything In Between What Genre Actually Classifies Linkin Park 8628343 📰 Twisted Metal Black Cheats 📰 Nioh 2 Multi Element Weapon Effect 📰 Discover The Hidden Gem Of Fidelity Thousand Oaks Why Everyones Talking About It 5873628 📰 Java Io File Java 5184771 📰 Travel Pass Vzw