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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: 📰 Blackstone Stock Hiding A Secret Heres What Could Make You Rich Overnight 4109737 📰 Call Saul Season 3 Stuns Fansdont Miss These Life Changing Moments 5055137 📰 How A Hidden Mistake Ruins Your Adhesives Full Power 65147 📰 Top 10 Unforgettable Vikings Valhalla Characters That Will Blow Your Mind 5047840 📰 The Big Bang Theory Online Episodes 6025863 📰 Best Cheap In Ear Headphones 1203131 📰 Concra Card Login Erroroon Cause And How To Get Immediate Accessread On 2431331 📰 Whats Hanging In Woodbury Outlet Will Make Your Wallet Queue Up Overflow 1891065 📰 Carolina Lottery Powerball 78306 📰 You Wont Believe What This Minecraft Spiral Staircase Reveals When You Climb It 7977817 📰 62 130 2Ab Rightarrow 36 130 2Ab Rightarrow 2Ab 94 Rightarrow Ab 47 474192 📰 What Time Do The Nfl Draft Start 7142991 📰 Score Big Or Get Eaten The Ultimate Spider Game Playground Now Live 5116267 📰 Rob Riner 5738851 📰 Meaning Ductile 9759381