mt1_review_friday_soln - Midterm 1 Review Solution Aritoki...

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Midterm 1 Review Solution Aritoki Suzuki September 28th 2007 1 Mathematical Representation of Position v.s. Ve- locity v.s. Acceleration 1. vectorv ( t ) = dvectorx ( t ) dt = 2 At + B 2. vectora ( t ) = d 2 vectorx ( t ) dt 2 = dvectorv ( t ) dt = 2 A 3. t = parenleftBig - B ± B 2 - 4 AC 2 A parenrightBig 4. Since vectora ( t ) is a constant (2A), we can use v ( t 2 )+ v ( t 1 ) 2 = A ( t 2 + t 1 ) + B 5. vectorv ( t ) = 2( - 10) t + 50 , t = 2 and 3 6. (a) vectorv ( t ) = dvectorx ( t ) dt = 3 At 2 + 2 Bt + C (b) vectora ( t ) = d 2 vectorx ( t ) dt 2 = dvectorv ( t ) dt = 6 At + 2 B (c) t = 0 and t = parenleftBig - B ± B 2 - 4 AC 2 A parenrightBig (d) R t 2 t 1 3 At 2 +2 Bt + C dt t 2 - t 1 = A ( t 2 2 + t 2 t 1 + t 2 1 ) + B ( t 2 + t 1 ) + C 2 Graphical Representation of Position v.s. Velocity v.s. Acceleration 1. I will post them later 1
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2. During t 1 to t 2 , vectorv ( t ) is positive, thus particle a is traveling in a positive direction 3. During t 3 to t 4 , vectorv ( t ) is negative, thus particle a is travelign in a negative direction 4. Since vectora ( t ) = Δ vectorv ( t ) Δ t t 0 to t 1 : a t 1 - t 0 t 1 to t 3 : b - a t 3 - t 1 t 3 to t 4 : 0 t 4 to t 5 : - b t 5 - t 4 5. Since at t 0 and t 5 velocity is 0, average velocity is ¯ a = Δ vectorv t = 0 6. Displacement is a sum of area under the curve of vectorv ( t ) . vectorx ( t ) = 1 2 a ( t 1 - t 0 )+ 1 2 a ( t 2 - t 1 ) + 1 2 b ( t 3 - t 2 ) + b ( t 4 - t 3 ) + 1 2 b ( t 5 - t 4 ) 7. With given condition, sum of area under v ( t ) becomes negative (work this out graphically by yourself). Displacement is a sum of area under the curve of vectorv ( t ) so total displacement is negative. Also average velocity is defined by ¯ v = Δ vectorx t thus if total displacement is negative, average velocity is also negative.
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