B-sol - Chapter B - Problems Blinn College - Physics 2425 -...

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Unformatted text preview: Chapter B - Problems Blinn College - Physics 2425 - Terry Honan Problem B.1 x H m L t H s L 2 3 7 2 4 8 (a) What is the average velocity between 0 s and 2 s, between 3 s and 7 s, between 2 s and 7 s? (b) What is the instantaneous velocity at 1 s and and at 4 s? (c) What is the average acceleration between 1 s and 4 s? (b) What is the instantaneous acceleration at 1 s and at 4 s? Solution to B.1 (a) The average velocity is defined as v = D x D t = x f- x i t f- t i . Between 0 s and 2 s: v = x H 2 L- x H L 2- = 4- 2 2 = 1 m s Between 3 s and 7 s: v = x H 7 L- x H 3 L 7- 3 = 4- 8 4 = - 1 m s Between 2 s and 7 s: v = x H 7 L- x H 2 L 7- 2 = 4- 4 5 = (b) The instantaneous velocity is the slope of the x vs. t graph. The instantaneous velocity of a straight-line segment is the average velocity for that segment. The instantaneous velocity at t = 1 s is the same as the average velocity between 0 and 2 s, which is 1 m s . The instantaneous velocity at t = 4 s is the same as the average velocity between 3 s and 7 s, which is - 1 m s . (c) The average acceleration is defined as a = D v D t = v f- v i t f- t i . Between 1 s and 4 s: a = v H 4 L- v H 1 L 4- 1 =- 1- 1 3 = - 2 3 m s 2 (d) During periods of constant velocity the acceleration is zero, so at both 1 s and 4 s, a = 0. Problem B.2 A particle moves in one dimension. It's position as a function of time is given, in SI units, by: x H t L = 2 t 4- 5 t 2 + 18. (a) What is the average velocity between 2s and 4s? (b) What is the instantaneous velocity at 3s? (c) What is the average acceleration between 2s and 4s? (d) What is the instantaneous acceleration at 3s? Solution to B.2 (a) x H 2 L = 2 2 4- 5 2 2 + 18 = 30 and x H 4 L = 2 4 4- 5 4 2 + 18 = 450 v = D x D t = x H 4 L- x H 2 L 4- 2 = 450- 30 4- 2 = 210 m s (b) v = t I 2 t 4- 5 t 2 + 18 M = 2 4 t 3- 5 2 t + = 8 t 3- 10 t v H 3 L = 8 3 3- 10 3 = 186 m s (c) v H 2 L = 8 2 3- 10 2 = 44 and v H 4 L = 8 4 3- 10 4 = 472 a = D v D t = v H 4 L- v H 2 L 4- 2 = 472- 44 4- 2 = 214 m s 2 (d) a = t I 8 t 3- 10 t M = 8 3 t 2- 10 1 = 24 t 2- 10 a H 3 L = 24 3 2- 10 = 206 m s 2 Problem B.3 x t t 1 t 2 t 3 t 4 The above diagram is a graph of position versus time. For t < t 1 the graph is a straight line. For t > t 4 the graph is a horizontal line. I t 3 is a point of inflection.) (a) Where is the velocity zero? Where is it positive? Where is it negative? (b) Where is the acceleration zero? Where is it positive? Where is it negative? Solution to B.3 (a) Velocity is the slope of the x vs. t graph....
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B-sol - Chapter B - Problems Blinn College - Physics 2425 -...

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