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solutions 2

# solutions 2 - Patel Kinal Homework 2 Due Sep 7 2007 7:00 pm...

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Patel, Kinal – Homework 2 – Due: Sep 7 2007, 7:00 pm – Inst: D Weathers 1 This print-out should have 10 questions. Multiple-choice questions may continue on the next column or page – fnd all choices beFore answering. The due time is Central time. 001 (part 1 oF 2) 10 points The velocity oF a particle moving along the x axis varies in time according to the expression v ( t ) = ( α - β t 2 ) where α = 44 . 2 m / s, β = 5 . 09 m / s 3 , and t is in seconds. ±ind the average acceleration in the time interval From t = 0 to 2 . 68 s. Correct answer: - 13 . 6412 m / s 2 . Explanation: The velocities at t i = 0 and t f = 2 . 68 s are Found by substituting these values into the expression given For the velocity: v i = ( α - β t 2 i ) = (44 . 2 m / s - (5 . 09 m / s 3 ) (0 s) 2 ) = 44 . 2 m / s , v f = ( α - β t 2 f ) = (44 . 2 m / s - (5 . 09 m / s 3 ) (2 . 68 s) 2 ) = 7 . 64158 m / s . ThereFore, the average acceleration in the specifed time interval Δ t = t f - t i = 2 . 68 s is ¯ a = v f - v i t f - t i = 7 . 64158 m / s - 44 . 2 m / s 2 . 68 s - 0 s = - 13 . 6412 m / s 2 . 002 (part 2 oF 2) 10 points Determine the acceleration oF the particle at t f = 2 . 68 s. Correct answer: - 27 . 2824 m / s 2 . Explanation: We can use the rules oF the di²erential calcu- lus to fnd the velocity From the displacement, at t = 2 . 68 s , the acceleration is a ( t ) = dv dt = d dt (44 . 2 m / s - 5 . 09 m / s 3 t 2 ) = - (2) (5 . 09 m / s 3 ) t = - (2) (5 . 09 m / s 3 ) (2 . 68 s) = - 27 . 2824 m / s 2 . keywords: 003 (part 1 oF 2) 10 points An electron in the cathode ray tube oF a tele- vision set enters a region where it acceler- ates uniFormly From a speed oF 34600 m / s to a speed oF 4 . 67 × 10 6 m / s in a distance oF 2 . 95 cm. What is its acceleration?

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solutions 2 - Patel Kinal Homework 2 Due Sep 7 2007 7:00 pm...

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