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Unformatted text preview: homework 02 FRENNEA, KYLE Due: Jan 31 2007, 4:00 am 1 Question 1 part 1 of 4 10 points Consider the plot below describing the ac celeration of a particle along a straight line with an initial position of 32 m and an ini tial velocity of 5 m / s. 5 4 3 2 1 1 2 3 1 2 3 4 5 6 7 8 9 time (s) acceleration(m/s 2 ) What is the velocity at 2 s? Correct answer: 1 m / s (tolerance 1 %). Explanation: In order to use the above graph, let x = x , 1 = 32 m , v = v , 1 = 5 m / s , ( t , a ) = ( t , 1 , a ) = (0 s , 0 m / s 2 ) , ( t 1 , a 1 ) = ( t , 1 , a 1 , 2 ) = (0 s , 2 m / s 2 ) , ( t 2 , a 2 ) = ( t 2 , 3 , a 1 , 2 ) = (3 s , 2 m / s 2 ) , ( t 3 , a 3 ) = ( t 2 , 3 , a 3 , 4 ) = (3 s , 4 m / s 2 ) , ( t 4 , a 4 ) = ( t 4 , 5 , a 3 , 4 ) = (9 s , 4 m / s 2 ) , and ( t 5 , a 5 ) = ( t 4 , 5 , a 5 ) = (9 s , 0 m / s 2 ) . Basic Concepts: The plot shows a curve of acceleration versus time . The change in velocity is the area ( a 1 , 2 t ) between the acceleration curve and the time axis v = v , 1 + a 1 , 2 t , where the acceleration is constant. Solution: With constant acceleration ( a 1 , 2 = 2 m / s 2 ), v = v , 1 + a 1 , 2 t (1) = ( 5 m / s) + (2 m / s 2 ) (2 s) = 1 m / s . Equations 1 and 3 are plotted below. 3 2 1 1 1 2 3 4 5 6 7 8 9 time (s) velocity(m/s)[ . 1] Question 2 part 2 of 4 10 points What is the position at 2 s? Correct answer: 38 m (tolerance 1 %). Explanation: Basic Concepts: The change in position is the area ( v , 1 t + 1 2 a t 2 ) between the veloc ity curve and the time axis x = x , 1 + v , 1 t + 1 2 a t 2 . Solution: With constant acceleration ( a 1 , 2 = 2 m / s 2 ), x = x , 1 + v , 1 t + 1 2 a 1 , 2 t 2 (2) = ( 32 m) + ( 5 m / s) (2 s) + 1 2 (2 m / s 2 ) (2 s) 2 = 38 m . Equations 2 and 4 are plotted below. 2 1 1 1 2 3 4 5 6 7 8 9 time (s) position(m)[ . 01] homework 02 FRENNEA, KYLE Due: Jan 31 2007, 4:00 am 2 Question 3 part 3 of 4 10 points What is the velocity at 7 s? Correct answer: 15 m / s (tolerance 1 %). Explanation: The calculation is done in two parts, each with constant acceleration ( a 1 , 2 = 2 m / s 2 ) and ( a 3 , 4 = 4 m / s 2 ). v = v , 1 + a 1 , 2 t 2 , 3 + a 3 , 4 [ t t 2 , 3 ] (3) = ( 5 m / s) + (2 m / s 2 ) (3 s) + ( 4 m / s 2 ) [(7 s) (3 s)] = 15 m / s , where v 2 , 3 = v , 1 + a 1 , 2 t 2 , 3 = ( 5 m / s) + (2 m / s 2 ) (3 s) = 1 m / s . Question 4 part 4 of 4 10 points What is the position at 7 s? Correct answer: 66 m (tolerance 1 %). Explanation: The calculation is done in two parts, each with constant acceleration ( a 1 , 2 = 2 m / s 2 ) and ( a 3 , 4 = 4 m / s 2 )....
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 Spring '08
 Turner
 Acceleration, Work

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