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# hw5 - TAM212 Introductory Dynamics Spring 2005 S...

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Unformatted text preview: TAM212 Introductory Dynamics Spring 2005 S. Balachandar Homework 5 (1.74, 1.96) 1.74 At any instance of time, the position of the object is given by (x, y). We are given that y = sinh x. So the displacement vector is given by ˆ ˆ r p = xi + sinh x j ˆ = xi + e x − e− x ˆ j 2 Now the velocity and accceleration are given by ˙ x x ˆ e + e− x j 2 ˆ = 0.08i + 0.04 e x + e−x jˆ ˙ˆ v p = xi + ∴ vp x =0 . 2 ˆ ˆ = 0.08i + 0.0816 j m/s 0 ! ¡ ˆ ¨ ¨ a p = ¡ i + 0.04 e x − e−x x jˆ (x p = 0 as the speed is constant) xˆ = 0.04(0.08) e x − e−x jˆ ∴ ap x =0 . 2 ˆ = 0.00129 j m/s2 1.96 The velocity is given in terms of it components as ˙ˆ ˙ ˆ v p = xi + y j We are given 2π x ft 3000 2π x 100π ˙ ˙ sin x ∴y = − 3000 3000 y = 50 cos 1 The speed is given to be Speed = 52 mph = 52(88/60) ft/s = v p 80.7 = ˙ ∴ x|x=2500 = ∴ vp x=2500 2 π 2π x sin 30 3000 80.7 = 80.3 ft/s π2 1 + 900 sin2 5000π 3000 ˙ ˙ x2 + x2 −π 5π ˆ ˆ = 80.3i + 80.3 j ft/s sin 30 3 ˆ = 80.3i + 7.29 jˆ ft/s The x and y components of the velocity are 80.3 ft/s and 7.29 ft/s respectively. 2 ...
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hw5 - TAM212 Introductory Dynamics Spring 2005 S...

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