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10:26 PM S ¸ −6 ± √(6)² − (4)(0.375)(-120) (2)(0.375) 41.1 -6±14.697 0.75 11.596 s and -27.6 s 7-11.60 s Reject the negative root. Corresponding values of x and xp. Example 6 x-(0.375) (11.596) 50.4 m x 120 (6)(11.596) = 50.4 m x 50.4 m◄ The elevator shown in the figure moves downward with a constant velocity of 4 m/s. Determine (a) the velocity of the cable C, (b) the velocity of the counterweight W, (c) the relative velocity of the cable C with respect to the elevator, (d) the relative velocity of the counterweight W with respect to the elevator. SOLUTION Choose the positive direction downward. (a) Velocity of cable C. 11 Yw Je + 2y constant Ve+2vg=0 Уе But, V-4 m/s or Ve-2v-8 m/s Ve -8.00 m/s (b) Velocity of counterweight W yu+yg constant V+V-0---4 m/s Va -4.00 m/s (c) Relative velocity of C with respect to E. Ver Ve-VE (-8 m/s)-(+4 m/s)=-12 m/s Yor-12.00 m's (d) Relative velocity of W with respect to E. VV-V=(-4 m/s)-(4 m/s)=-8 m/s "WE 8.00 m/s Examples 7 A particle moves in a straight line with a constant acceleration of -4 ft/s2 for 6 s, zero acceleration for the next 4 s, and a constant acceleration of +4 ft/s2 for the next 4 s. Knowing that the particle starts from the origin and that its velocity is -8 ft/s during the zero-acceleration time interval, (a) construct the v-t and x-t curves for 0sts 14 s, (b) determine the position and the velocity of the particle and the total distance traveled when t = 14 s. a (ft/s²) 6 0 10 14 -4 ||| = t(s) כ Մ

10:26 PM S ¸ −6 ± √(6)² − (4)(0.375)(-120) (2)(0.375) 41.1 -6±14.697 0.75 11.596 s and -27.6 s 7-11.60 s Reject the negative root. Corresponding values of x and xp. Example 6 x-(0.375) (11.596) 50.4 m x 120 (6)(11.596) = 50.4 m x 50.4 m◄ The elevator shown in the figure moves downward with a constant velocity of 4 m/s. Determine (a) the velocity of the cable C, (b) the velocity of the counterweight W, (c) the relative velocity of the cable C with respect to the elevator, (d) the relative velocity of the counterweight W with respect to the elevator. SOLUTION Choose the positive direction downward. (a) Velocity of cable C. 11 Yw Je + 2y constant Ve+2vg=0 Уе But, V-4 m/s or Ve-2v-8 m/s Ve -8.00 m/s (b) Velocity of counterweight W yu+yg constant V+V-0---4 m/s Va -4.00 m/s (c) Relative velocity of C with respect to E. Ver Ve-VE (-8 m/s)-(+4 m/s)=-12 m/s Yor-12.00 m's (d) Relative velocity of W with respect to E. VV-V=(-4 m/s)-(4 m/s)=-8 m/s "WE 8.00 m/s Examples 7 A particle moves in a straight line with a constant acceleration of -4 ft/s2 for 6 s, zero acceleration for the next 4 s, and a constant acceleration of +4 ft/s2 for the next 4 s. Knowing that the particle starts from the origin and that its velocity is -8 ft/s during the zero-acceleration time interval, (a) construct the v-t and x-t curves for 0sts 14 s, (b) determine the position and the velocity of the particle and the total distance traveled when t = 14 s. a (ft/s²) 6 0 10 14 -4 ||| = t(s) כ Մ

Elements Of Electromagnetics
Elements Of Electromagnetics
7th Edition
ISBN:9780190698614
Author:Sadiku, Matthew N. O.
Publisher:Sadiku, Matthew N. O.
ChapterMA: Math Assessment
Section: Chapter Questions
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10:26 PM S ¸ −6 ± √(6)² − (4)(0.375)(-120) (2)(0.375) 41.1 -6±14.697 0.75 11.596 s and -27.6 s 7-11.60 s Reject the negative root. Corresponding values of x and xp. Example 6 x-(0.375) (11.596) 50.4 m x 120 (6)(11.596) = 50.4 m x 50.4 m◄ The elevator shown in the figure moves downward with a constant velocity of 4 m/s. Determine (a) the velocity of the cable C, (b) the velocity of the counterweight W, (c) the relative velocity of the cable C with respect to the elevator, (d) the relative velocity of the counterweight W with respect to the elevator. SOLUTION Choose the positive direction downward. (a) Velocity of cable C. 11 Yw Je + 2y constant Ve+2vg=0 Уе But, V-4 m/s or Ve-2v-8 m/s Ve -8.00 m/s (b) Velocity of counterweight W yu+yg constant V+V-0---4 m/s Va -4.00 m/s (c) Relative velocity of C with respect to E. Ver Ve-VE (-8 m/s)-(+4 m/s)=-12 m/s Yor-12.00 m's (d) Relative velocity of W with respect to E. VV-V=(-4 m/s)-(4 m/s)=-8 m/s "WE 8.00 m/s Examples 7 A particle moves in a straight line with a constant acceleration of -4 ft/s2 for 6 s, zero acceleration for the next 4 s, and a constant acceleration of +4 ft/s2 for the next 4 s. Knowing that the particle starts from the origin and that its velocity is -8 ft/s during the zero-acceleration time interval, (a) construct the v-t and x-t curves for 0sts 14 s, (b) determine the position and the velocity of the particle and the total distance traveled when t = 14 s. a (ft/s²) 6 0 10 14 -4 ||| = t(s) כ Մ
Transcribed Image Text:10:26 PM S ¸ −6 ± √(6)² − (4)(0.375)(-120) (2)(0.375) 41.1 -6±14.697 0.75 11.596 s and -27.6 s 7-11.60 s Reject the negative root. Corresponding values of x and xp. Example 6 x-(0.375) (11.596) 50.4 m x 120 (6)(11.596) = 50.4 m x 50.4 m◄ The elevator shown in the figure moves downward with a constant velocity of 4 m/s. Determine (a) the velocity of the cable C, (b) the velocity of the counterweight W, (c) the relative velocity of the cable C with respect to the elevator, (d) the relative velocity of the counterweight W with respect to the elevator. SOLUTION Choose the positive direction downward. (a) Velocity of cable C. 11 Yw Je + 2y constant Ve+2vg=0 Уе But, V-4 m/s or Ve-2v-8 m/s Ve -8.00 m/s (b) Velocity of counterweight W yu+yg constant V+V-0---4 m/s Va -4.00 m/s (c) Relative velocity of C with respect to E. Ver Ve-VE (-8 m/s)-(+4 m/s)=-12 m/s Yor-12.00 m's (d) Relative velocity of W with respect to E. VV-V=(-4 m/s)-(4 m/s)=-8 m/s "WE 8.00 m/s Examples 7 A particle moves in a straight line with a constant acceleration of -4 ft/s2 for 6 s, zero acceleration for the next 4 s, and a constant acceleration of +4 ft/s2 for the next 4 s. Knowing that the particle starts from the origin and that its velocity is -8 ft/s during the zero-acceleration time interval, (a) construct the v-t and x-t curves for 0sts 14 s, (b) determine the position and the velocity of the particle and the total distance traveled when t = 14 s. a (ft/s²) 6 0 10 14 -4 ||| = t(s) כ Մ
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