ENPH
Seminar3Solutions (1).pdf

C acceleration since v x is constant then a x 0 from

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(c) Acceleration: Since v x is constant, then a x = 0. From a second implicit differentiation, we find 6( y - 20) ˙ y 2 + 3( y - 20) 2 ¨ y = 60 ˙ x 2 , so that a y = ¨ y = 60 ˙ x 2 - 6( y - 20) ˙ y 2 3( y - 20) 2 = 60(10) 2 - 6(40 - 20)(8 . 165) 2 3(40 - 20) 2 = - 1 . 667 m / s 2 1
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so that the acceleration is 1.667 m/s 2 in the - y direction [2 points] Question 3. [8 points] Let us choose the origin of the ( x, y ) coordinate system at the point of departure. We denote by θ 0 the direction of the initial velocity, and φ the slope of the incline. Then the position depends on time as x ( t ) = v 0 cos θ 0 t [1 point] (1) and y ( t ) = v 0 sin θ 0 t - 1 2 gt 2 . [1 point] (2) If the skier lands at a distance R down the slope at time t 0 , then the coordi- nates are x ( t 0 ) = R cos φ, y ( t 0 ) = - R sin φ. [1 . 5 points] (3) From Eqs. (1) and (3), we obtain the landing time, t 0 = R cos φ v 0 cos θ 0 . [1 point] When we substitute this into Eq. (2), together with Eq. (3), we obtain - R sin φ = v 0 sin θ 0 R cos φ v 0 cos θ 0 - 1 2 g R 2 cos 2 φ v 2 0 cos 2 θ 0 , [1 point] which we solve for R , R = 2 v 2 0 cos 2 θ 0 g cos 2 φ (sin φ + cos φ tan θ 0 ) , [1 . 5 points] with θ 0 = 10 , φ = 20 , g = 32 . 2 ft, v 0 = 67 . 1 mph = 67 . 1 × 5280 ft 1 3600 s = 98 . 41 ft/s, we obtain R = 335 ft . [1 point] 2
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