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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: 📰 Ws C3650 24Ts S Ios Download 5810174 📰 Inn On Woodlake 9676592 📰 The Truth About Jing Yuans Heartbreak Has You Breathing Hard 8893513 📰 Download Toad For Oracle 3912542 📰 Define Homozygous 6697793 📰 Critical Gains Inside The Mind Blowing Surge In Lics Stock Price Today 430365 📰 When Does Chick Fil A Start Serving Lunch 6774282 📰 Goats In Trees 2804243 📰 Nppes Npi Look Up Unlock Critical Credits Faster Than Ever 4345989 📰 The One Ingredient Kapuzzelle Transforms Every Dish You Make 6701267 📰 Bubba Smith 2784057 📰 Songs For Line Dances 2924705 📰 How To Create Budget 965788 📰 Fire Fire Emblem Heroes 4124105 📰 How Long Is Mackinac Bridge In Michigan 2958605