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# hw 5 - homework 05 HETTIGER CHRISTOF Due 4:00 am at t =...

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homework 05 – HETTIGER, CHRISTOF – Due: Jan 28 2008, 4:00 am 1 Question 1, chap 2, sect 4. part 1 of 2 10 points The velocity of a particle moving along the x axis varies in time according to the expres- sion v ( t ) = ( α β t 2 ) where α = 59 . 9 m / s, β = 4 . 44 m / s 3 , and t is in seconds. Find the average acceleration in the time interval from t = 0 to 2 . 44 s. Correct answer: 10 . 8336 m / s 2 (tolerance ± 1 %). Explanation: The velocities at t i = 0 and t f = 2 . 44 s are found by substituting these values into the expression given for the velocity: v i = ( α β t 2 i ) = (59 . 9 m / s (4 . 44 m / s 3 ) (0 s) 2 ) = 59 . 9 m / s , v f = ( α β t 2 f ) = (59 . 9 m / s (4 . 44 m / s 3 ) (2 . 44 s) 2 ) = 33 . 466 m / s . Therefore, the average acceleration in the specified time interval Δ t = t f t i = 2 . 44 s is ¯ a = v f v i t f t i = 33 . 466 m / s 59 . 9 m / s 2 . 44 s 0 s = 10 . 8336 m / s 2 . Question 2, chap 2, sect 4. part 2 of 2 10 points Determine the acceleration of the particle at t f = 2 . 44 s. Correct answer: 21 . 6672 m / s 2 (tolerance ± 1 %). Explanation: We can use the rules of the differential calcu- lus to find the velocity from the displacement, at t = 2 . 44 s , the acceleration is a ( t ) = d v d t = d d t (59 . 9 m / s 4 . 44 m / s 3 t 2 ) = (2) (4 . 44 m / s 3 ) t = (2) (4 . 44 m / s 3 ) (2 . 44 s) = 21 . 6672 m / s 2 . Question 3, chap 2, sect 7. part 1 of 3 10 points The position of a softball tossed vertically upward is described by the equation y = c 1 t c 2 t 2 , where y is in meters, t in seconds, c 1 = 11 . 2 m / s, and c 2 = 5 . 34 m / s 2 . Find the ball’s initial speed v 0 at t 0 = 0 s. Correct answer: 11 . 2 m / s (tolerance ± 1 %). Explanation: Basic Concepts: v = dx dt a = dv dt = d 2 x dt 2 Solution: The velocity is simply the derivative of y with respect to t : v = dy dt = 11 . 2 m / s 2(5 . 34 m / s 2 ) t, which at t = 0 is v 0 = 11 . 2 m / s .

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