Solução Do Halliday 8.ed.vol.1. Cap.5 - Ch05 - P031

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31. The acceleration vector as a function of time is G G dv d a= = 8.00t ˆi + 3.00t 2 ˆj m/s = (8.00 ˆi + 6.00t ˆj) m/s 2 . dt dt ( ) (a) The magnitude of the force acting on the particle is G F = ma = m | a | = (3.00) (8.00) 2 + (6.00t ) 2 = (3.00) 64.0 + 36.0 t 2 N. Thus, F = 35.0 N corresponds to t = 1.415 s, and the acceleration vector at this instant is G a = [8.00 ˆi + 6.00(1.415) ˆj] m/s 2 = (8.00 m/s 2 ) ˆi + (8.49 m/s2 )ˆj. G The angle a makes with +x is § ay © ax θ a = tan −1 ¨ 2 · −1 § 8.49 m/s · = = 46.7°. tan ¸ ¨ 2 ¸ 8.00 m/s © ¹ ¹ (b) The velocity vector at t = 1.415 s is G v = ª¬8.00(1.415) ˆi + 3.00(1.415)2 ˆjº¼ m/s = (11.3 m/s) ˆi + (6.01 m/s)ˆj. G Therefore, the angle v makes with +x is § vy © vx θ v = tan −1 ¨ · −1 § 6.01 m/s · ¸ = tan ¨ ¸ = 28.0°. © 11.3 m/s ¹ ¹