Homework 1_Solutions_Graded - PROBLEM 11.8 The motion o f a...

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PROBLEM 11.8 The motion of a particle is defined by the relation x := 2t 3 18t 2 + 48t 16, where x and t are expressed in millimeters and seconds, respectively. Determine (a) when the velocity is zero, (b) the position and the total distance traveled when the acceleration is zero. SOLUTION Position: x = 2t 3 - 18t 2 + 48t - 16 mm dx 2 mmI Velocity: v =- = 6t 36t + 48 s dt dv Acceleration: 12t - 36 mmls 2 a = dt (a) When v = 0, 6t 2 - 36t + 48 = 0 Solving the quadratic equation, t = 2 s and t 4 s . ... For 0 ~ t < 2 s, v is positive, and x is increasing. 2s~t~4s, v is negative, and x is decreasing. t:2:4s, v is positive, and x is increasing. (b) When a = 0, 12t - 36 = 0 t = 3 s When t = 3 s, X3 (2)(3)3 - (18)(3)2 + (48)(3) 16 X3 = 20mm . ... When t = 0, Xo -16 mm When t = 2 s, X2 = (2)(2)3 (18)(2)2 + (48)(2) -16 = 24 mm Distance traveled over 0 ~ t ~ 2 s: d l = I x2 Xo I = 124 - (-16) I := 40 mm Distance traveled over 2 s ~ t ~ 3 s: d 2 := I X3 - x21 = 120 241:= 4 mm Total distance traveled: d l + d 2 = 40 + 4 d=44mm . ...
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Pil I r I 1 I , PROBLEM 11.29 The acceleration due to gravity of a particle falling toward the earth is
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Homework 1_Solutions_Graded - PROBLEM 11.8 The motion o f a...

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