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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: 📰 Unleash Your Hidden Desires The Shocking Power Of Praise In Intimate Acts 7751439 📰 Turmeric Supplement Benefits 5359210 📰 Bass Pro Shop Indiana 7769692 📰 Ip Advanced Scanner Mac 8058005 📰 Your Dream Home Awaitsget Yours Before Its Gone 5799876 📰 Shoplc Com 6789639 📰 Hmart Little Ferry 8714864 📰 Is This The Boldest Move Morgan Stark Madewatch The Full Story 7016373 📰 Watch How The Sopranos Season 2 Redefined Crime Dramasone Twist At A Time 2007817 📰 Pby Catalina 2807391 📰 Your Oracle Dba Will Never Guess This Powerful Merge Command Trick 9672240 📰 See Tyler Durdens Costume Evolution The Ultimate Look That Serial Killers Deserve 3400295 📰 Door Castle 8106244 📰 Pugalier Breed Shocking Traits That Will Convert You Into A Fan 1729436 📰 Plug Stock Forum Hacks Youve Been Searching Fordont Miss These Hidden Tips 785079