SOLUTION. will destroy the integrity of the work and is not permitted displacement of the particle during the time interval t = 1 s to t = 3s.
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1 74. The velocity of a particle is v = 53i + (6 - t)j6 m>s, where t is in seconds. If r = 0 when t = 0, determine the displacement of the particle during the time interval t = s to t = 3s. Position: The position r of the particle can be determined by integrating the kinematic equation dr = vdt using the initial condition r = 0 at t = 0 as the integration limit. Thus, dr = vdt r t dr = C3i + (6 - t)jddt L0 L 0 r = c3ti + 6t - t Bj dm When t = sand 3 s, r t = s = 3()i + C6() - Dj = [3i + 5j] m>s r t = 3s = 3(3)i + C6(3) - 3 Dj = [9i + 9j] m>s Thus, the displacement of the particle is r = r t = 3s - r t = s = (9i + 9j) - (3i + 5j) = {6i + 4j} m This is work is protected by United States copyright laws and is provided solely for the use of instructors in teaching their courses and assessing student learning. Dissemination or 03 Pearson Education, Inc., Upper Saddle River, NJ. ll rights reserved. This publication is protected by
2 79. car traveling along the straight portions of the road has the velocities indicated in the figure when it arrives at points, B, and C. If it takes 3 s to go from to B, and then 5 s to go from B to C, determine the average acceleration between points and B and between points and C. y x v B 30 m/s v C C 40 m/s B 45 v = 0 i v B =. i +.j v 0 m/s v C = 40i a B = v t =. i +. j - 0 i 3 a B = { i j }m>s a C = v t = 40 i - 0 i 8 a C = {.50 i }m>s This work is protected by United States copyright laws and is provided solely for the use of instructors in teaching their courses and assessing student learning. Dissemination or 03 Pearson Education, Inc., Upper Saddle River, NJ. ll rights reserved. This publication is protected by
3 87. Pegs and B are restricted to move in the elliptical slots due to the motion of the slotted link. If the link moves with a constant speed of 0 m/s, determine the magnitude of the velocity and acceleration of peg when x = m. y Velocity: The x and y components of the peg s velocity can be related by taking the first time derivative of the path s equation. x 4 + y = 4 (xx# ) + yy # = 0 xx# + yy # = 0 x 4 y C B D x v 0 m/s or xv x + yv y = 0 () t x = m, () 4 + y = y = 3 m Here, v x = 0 m>s and x =. Substituting these values into Eq. (), 3 ()(0) + v y = 0 v y = m>s = m>s> T by Thus, the magnitude of the peg s velocity is protected is v = v x + v y = = 0.4 m>s cceleration: The x and y components sof the peg s acceleration ce can be related by taking the second time derivative of the path s equation. (x# x # ## + xx) + (y # y # ## + yy) y) = 0 x# ## + xxb + y # + yy ## B = 0 This work is United States copyright laws and theprovided is solely for the use of instructors in teaching #their ourses and assessing student learning. Dissemination or or v x + xa x B + v y + ya y B = 0 () Since v is constant,. When, y = 3 x a x = 0 x = m, v x = 0 m>s, and m v y = m>s. Substituting these values into Eq. (), 0 + 0B + c(-.887) + 3 a y d = 0 a y = m>s = m>s T Thus, the magnitude of the peg s acceleration is a = a x + a y = 0 + (-38.49) = 38.5 m>s 03 Pearson Education, Inc., Upper Saddle River, NJ. ll rights reserved. This publication is protected by
4 0. It is observed that the skier leaves the ramp at an angle u = 5 with the horizontal. If he strikes the ground at B, determine his initial speed v and the speed at which he strikes the ground. u 4 m v m Coordinate System: x-y coordinate system will be set with its origin to coincide with point as shown in Fig. a. B x-motion: Here, x = 0, x B = 00 a 4 and (v ) x = v cos 5. 5 b = m + : B y-motion: Here, y = 0, y B = -[ a 3 b] = -64 m and (v ) y = v 5 sin n 5 and a y = -g = -9.8 m>s. + c B x B = x + (v ) x t = 0 + (v cos 5 )t t = v cos 5 y B = y + (v ) y t + a yt -64 = 0 + v sin 5 t + (-9.8)t 4.905t - v sin 5 t = 64 Substitute Eq. () into () yields v cos 5 = v sin n 5 y = 64 any cos 5 ro s 5 v cos 5 = 0.65 v cos 5 = () () work is protected by United States copyright laws and is provided solely for the use of instructors in teaching their courses and assessing student learning. Dissemination or sale of cony part of this work (including on the World Wide We b) d. v = 9.4 m>s = 9.4 m>s Substitute this result into Eq. (), t = 9.4 cos 5 = Pearson Education, Inc., Upper Saddle River, NJ. ll rights reserved. This publication is protected by
5 0. continued Using this result, + c B (v B ) y = (v ) y + a y t = 9.4 sin 5 + (- 9.8)(4.5446) = m>s = m>s T nd + : B (v B ) x = (v ) x = v cos 5 = 9.4 cos 5 = 7.60 m>s : Thus, v B = (v B ) x + (v B ) y = = 40.4 m>s This work is protected by United States copyright laws and is provided solely for the use of instructors in teaching their courses and assessing student learning. Dissemination or 03 Pearson Education, Inc., Upper Saddle River, NJ. ll rights reserved. This publication is protected by
6 *. The baseball player hits the baseball at v = 40 ft>s and u = 60 from the horizontal. When the ball is directly overhead of player B he begins to run under it. Determine the constant speed at which B must run and the distance d in order to make the catch at the same elevation at which the ball was hit. v = 40 ft/s θ B v C Vertical Motion: The vertical component of initial velocity for the football is (v 0 ) y = 40 sin 60 = ft>s. The initial and final vertical positions are (s 0 ) y = 0 and s y = 0, respectively. 5 ft d (+ c) s y = (s 0 ) y + (v 0 ) y t + (a c) y t 0 = t + (-3.) t t =.5 s Horizontal Motion: The horizontal component of velocity for the baseball is (v 0 ) x = 40 cos 60 = 0.0 ft>s. The initial and final horizontal positions are (s 0 ) x = 0 and s x = R, respectively. ( : + ) s x = (s 0 ) x + (v 0 ) x t R = (.5) = ft The distance for which player B must travel in order to catch the ebaseball all on is d = R - 5 = = 8.0 ft Player B is required to run at a same speed as the protected horizontal onta component of of velocity of the baseball in order to catch it. v B = 40 cos 60 = 0.0 ft s This work s by United States copyrigh l w and is provided orsolely for the use of instructors in teaching their courses and assessing studen learning. Dissemination or sale of any part of this work (including the World Wide Web) will destroy the integrity onthe work and is not permitted. 03 Pearson Education, Inc., Upper Saddle River, NJ. ll rights reserved. This publication is protected by
= m. 30 m. The angle that the tangent at B makes with the x axis is f = tan-1
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