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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. 📰 Cityoftampa Water Bill 1356760 📰 Perfect Ground Chicken Meatballs Thatll Make Every Meal Unforgettable 4022196 📰 Zapatillas De Deporte Promocionadas Al Mximo No Pierdas La Oportunidad De Comprar Tus Favoritas 1479960 📰 Free Call App 7885251 📰 How To Pull From 401K 2529915 📰 Aloha Kitchen Bar 7609192 📰 Game Changing Cupcake 2048 Sweet Energy Like Never Before 1509859 📰 A 1414999 📰 Without A Paddle Film 8236473 📰 The Ultimate Guide To Office 365 Personal Free Tools That Will Change How You Work 1313062 📰 You Wont Believe Which Drafts Faster Crochet Or Knit The Surprising Truth 9720146 📰 You Wont Believe Whats Coming Soontop New Movies Set To Redefine Cinema 8635388 📰 A Paleobotanist Is Analyzing A Sediment Layer And Estimates That Each Square Meter Contains 120 Fossilized Seeds If Her Excavation Site Is A Rectangle Measuring 5 Meters By 8 Meters And Seeds Are Evenly Distributed How Many Seeds Does She Expect To Find In This Area 8035375 📰 Business High Yield Savings Account 2093419 📰 You Wont Believe How Juicy And Delicious Bleu Steak Getsthis Chefs Secret Will Blow Your Mind 464090