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📰 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: 📰 At $ t = 3 $: $ s(3) = 9a + 3b + c $ 📰 Jugar Doomsday Last Survivors En Pc 5055763 📰 Lifting Simulator 4239686 📰 Can Infold Games Outgame Every Smartphone Game Find Out Now 6611185 📰 Why This Mlp Card Is The Ultimate Must Have For Rare Collectors See Why 8485003 📰 System Alert The Reserved Partition Refused To Updatediscover What Happened And How To Recover 8598453 📰 How Many Days Until June 7Th 4495290 📰 Youll Never Guess What These Valentines Nails Secretly Look Like 1311693 📰 The Shocking Truth Behind Herofx That Will Change Everything 7652033 📰 Snow Totals In Indiana 2301988 📰 Frontwar Secrets The Epic Showdown No Ones Talking About Revealed 1523225 📰 How Maiff Stock Blasted Past Predictions The Full Story Of Its Explosive Rise 438095 📰 From Hidden Fortresses To Floating Castlesthese Minecraft Houses Will Leave You Speechless 3050968 📰 Cr Champions Reveal The Secret Decks Ready To Crush Every Match 5877510 📰 Hdmovie2 Hd Quality You Havent Seen This One Is Uncensored 7422645 📰 Viprostol 2657951