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Unformatted text preview: lim (kl9356) – Homework 2 – Weathers – (17104) 1 This printout should have 10 questions. Multiplechoice questions may continue on the next column or page – find all choices before answering. 001 (part 1 of 2) 10.0 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 . 8 m / s, β = 4 . 46 m / s 3 , and t is in seconds. Find the average acceleration in the time interval from t = 0 to 2 . 59 s. Correct answer: 11 . 5514 m / s 2 . Explanation: The velocities at t i = 0 and t f = 2 . 59 s are found by substituting these values into the expression given for the velocity: v i = ( α β t 2 i ) = (44 . 8 m / s (4 . 46 m / s 3 ) (0 s) 2 ) = 44 . 8 m / s , v f = ( α β t 2 f ) = (44 . 8 m / s (4 . 46 m / s 3 ) (2 . 59 s) 2 ) = 14 . 8819 m / s . Therefore, the average acceleration in the specified time interval Δ t = t f t i = 2 . 59 s is ¯ a = v f v i t f t i = 14 . 8819 m / s 44 . 8 m / s 2 . 59 s 0 s = 11 . 5514 m / s 2 . 002 (part 2 of 2) 10.0 points Determine the acceleration of the particle at t f = 2 . 59 s. Correct answer: 23 . 1028 m / s 2 . Explanation: We can use the rules of the differential calcu lus to find the velocity from the displacement, at t = 2 . 59 s , the acceleration is a ( t ) = d v d t = d d t (44 . 8 m / s 4 . 46 m / s 3 t 2 ) = (2) (4 . 46 m / s 3 ) t = (2) (4 . 46 m / s 3 ) (2 . 59 s) = 23 . 1028 m / s 2 . 003 (part 1 of 2) 10.0 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 51300 m / s to a speed of 4 . 68 × 10 6 m / s in a distance of 2 . 08 cm....
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This note was uploaded on 09/07/2010 for the course PHYS 1710 taught by Professor Weathers during the Fall '08 term at North Texas.
 Fall '08
 Weathers
 Magnetism, Work

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