Solução Do Halliday 8.ed.vol.1. Cap.5 - Ch05 - P011

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11. To solve the problem, we note that acceleration is the second time derivative of the position function; it is a vector and can be determined from its components. The net force is related to the acceleration via Newton’s second law. Thus, differentiating x(t ) = −15.0 + 2.00t + 4.00t 3 twice with respect to t, we get dx = 2.00 − 12.0t 2 , dt d2x = −24.0t dt 2 Similarly, differentiating y (t ) = 25.0 + 7.00t − 9.00t 2 twice with respect to t yields dy = 7.00 − 18.0t , dt d2y = −18.0 dt 2 (a) The acceleration is d 2x d2y G a = ax ˆi + a y ˆj = 2 ˆi + 2 ˆj = (−24.0t )iˆ + (−18.0)ˆj. dt dt G At t = 0.700 s , we have a = (−16.8)iˆ + (−18.0)ˆj with a magnitude of G a =| a | = (−16.8) 2 + (−18.0) 2 = 24.6 m/s 2 . Thus, the magnitude of the force is F = ma = (0.34 kg)(24.6 m/s 2 ) = 8.37 N. G G G (b) The angle F or a = F / m makes with + x is 2 § ay · −1 § −18.0 m/s · = 47.0° or − 133°. ¸ = tan ¨ 2 ¸ © −16.8 m/s ¹ © ax ¹ θ = tan −1 ¨ G We choose the latter ( −133° ) since F is in the third quadrant. (c) The direction of travel is the direction of a tangent to the path, which is the direction of the velocity vector: dx dy G v (t ) = vx ˆi + v y ˆj = ˆi + ˆj = (2.00 − 12.0t 2 )iˆ + (7.00 − 18.0t )ˆj. dt dt G At t = 0.700 s , we have v (t = 0.700 s) = (−3.88 m/s)iˆ + (−5.60 m/s)ˆj. Therefore, the angle G v makes with + x is § vy · −1 § −5.60 m/s · ¸ = tan ¨ ¸ = 55.3° or − 125°. © −3.88 m/s ¹ © vx ¹ θ v = tan −1 ¨ G We choose the latter ( −125° ) since v is in the third quadrant.