PROBLEM.3 The verticl motion of mss A is defined by the reltion x= 0 sin t+ 5cost+ 00, where x nd t re expressed in mm nd seconds, respectively. Determine () the position, velocity nd ccelertion of A when t = s, (b) the mximum velocity nd ccelertion of A. x = 0sin t+ 5cos t + 00 dx v= = 0cos t 30sin t dt dv = = 40sin t 60cost dt F trigonometric functions set clcult to rdins: () At t = s. x = 0sin + 5cos + 00 = 0.9 x = 0.9 mm v = 0cos 30sin = 35.6 v = 35.6 mm/s (b) Mximum velocity occurs when = 0. = 40sin 60cos =.40 40sin t 60cos t = 0 60 tn t = =.5 40 =.40 mm/s Reject the negtive vlue. t =.588 t = tn (.5) = 0.988 nd 0.988 + π t =.0794 s t =.0794 s f v mx so v mx = 0cos(.588) 30sin(.588) = 36.056 v mx = 36. mm/s Note tht we could hve lso used v mx = 0 + 30 = 36.056 by combining the sine nd cosine terms. F mx we cn tke the derivtive nd set equl to zero just combine the sine nd cosine terms. mx = 40 + 60 = 7. mm/s = 7. mm/s mx PROPRIETARY MATERIAL. 03 The McGrw-Hill Compnies, Inc. All rights reserved. No prt of this Mnul my be displyed, reproduced distributed in ny fm by ny mens, without the pri written permission of the publisher, used beyond the limited 7
PROBLEM.0 The ccelertion of prticle is directly proptionl to the time t. At t = 0, the velocity of the prticle is v = 6 in./s. Knowing tht v = 5 in./s nd tht x = 0 in. when t = s, determine the velocity, the position, nd the totl distnce trveled when t = 7 s. We hve Now At t = 0, v = 6 in./s: At t = s, v= 5 in./s: Also At t = s, x = 0 in.: Then At t = 7 s: When v = 0: v = kt k = constnt dv kt dt = = dv = kt dt 6 0 v 6 = kt t v= 6 + kt (in./s) 5 in./s = 6 in./s + (s) k x 3 k = in./s nd v= 6 t dx v 6 t dt = = dx = (6 t ) dt 0 3 x 0 = 6t t 3 t 3 3 x = t + 6 t + (in.) 3 3 t v 7 = 6 (7) v 7 = 33.0 in./s 3 3 x 7 = (7) + 6(7) + x 7 =.00 in. 3 3 6 t = 0 t = 4 s PROPRIETARY MATERIAL. 03 The McGrw-Hill Compnies, Inc. All rights reserved. No prt of this Mnul my be displyed, reproduced distributed in ny fm by ny mens, without the pri written permission of the publisher, used beyond the limited 4
PROBLEM.0 (Continued) At t = 0: x 0 = t = 4 s: Now observe tht 3 3 0 t < 4 s: v > 0 4 s < t 7 s: v < 0 3 3 x 4 = (4) + 6(4) + = 47 in. 3 3 Then 3 x4 x0 = 47 = 4.67 in. 3 x x = 47 = 45 in. 7 4 Totl distnce trveled = (4.67 + 45) in. Totl distnce = 87.7 in. PROPRIETARY MATERIAL. 03 The McGrw-Hill Compnies, Inc. All rights reserved. No prt of this Mnul my be displyed, reproduced distributed in ny fm by ny mens, without the pri written permission of the publisher, used beyond the limited 5
PROBLEM.5 A piece of electronic equipment tht is surrounded by pcking mteril is dropped so tht it hits the ground with speed of 4 m/s. After contct the equipment experiences n ccelertion of = kx, where k is constnt nd x is the compression of the pcking mteril. If the pcking mteril experiences mximum compression of 0 mm, determine the mximum ccelertion of the equipment. Seprte nd integrte. vf vdv = = kx dx vdv = xf v0 0 kxdx x f f 0 = = f v v kx kx 0 Use v0 = 4 m/s, x = 0.0 m, nd v = 0. Solve f k. f f Mximum ccelertion. 0 (4) = k(0.0) k = 40,000 s = kx : ( 40,000)(0.0) = 800 m/s = 800 m/s mx mx PROPRIETARY MATERIAL. 03 The McGrw-Hill Compnies, Inc. All rights reserved. No prt of this Mnul my be displyed, reproduced distributed in ny fm by ny mens, without the pri written permission of the publisher, used beyond the limited 0
PROBLEM.35 A motist enters freewy t 30 mi/h nd ccelertes unifmly to 60 mi/h. From the odometer in the cr, the motist knows tht she trveled 550 ft while ccelerting. Determine () the ccelertion of the cr, (b) the time required to rech 60 mi/h. () Accelertion of the cr. = 0 + 0 v v ( x x ) v0 v = ( x x ) 0 Dt: v0 = 30 mi/h = 44 ft/s v = 60 mi/h = 88 ft/s x 0 = 0 x = 550 ft (b) Time to rech 60 mi/h. (88) (44) = ()(55 0) = 5.8 ft/s v = v0 + ( t t0) v v0 t t0 = 88 44 = 5.8 = 8.333 s t t0 = 8.33 s PROPRIETARY MATERIAL. 03 The McGrw-Hill Compnies, Inc. All rights reserved. No prt of this Mnul my be displyed, reproduced distributed in ny fm by ny mens, without the pri written permission of the publisher, used beyond the limited 45
PROBLEM.40 In bot rce, bot A is leding bot B by 50 m nd both bots re trveling t constnt speed of 80 km/h. At t = 0, the bots ccelerte t constnt rtes. Knowing tht when B psses A, t = 8 s nd v A = 5 km/h, determine () the ccelertion of A, (b) the ccelertion of B. () We hve va = ( va) 0 + At ( v ) = 80 km/h = 50 m/s At t = 8s: v = 5 km/h = 6.5 m/s Then 6.5 m/s = 50 m/s + A (8 s) A 0 A (b) We hve xa = ( xa) 0 + ( va) 0t + At = 50 m + (50 m/s)(8 s) + (.565 m/s )(8 s) = 500 m nd xb = 0 + ( vb) 0t + Bt ( v B ) 0 = 50 m/s At t = 8s: xa = xb 500 m = (50 m/s)(8 s) + (8 s) B =.563 m/s A = 3.3 m/s B PROPRIETARY MATERIAL. 03 The McGrw-Hill Compnies, Inc. All rights reserved. No prt of this Mnul my be displyed, reproduced distributed in ny fm by ny mens, without the pri written permission of the publisher, used beyond the limited 50
PROBLEM.48 The elevt shown strts from rest nd moves upwrd with constnt ccelertion. If the counterweight W moves through 30 ft in 5 s, determine () the ccelertion of the elevt nd the cble C, (b) the velocity of the elevt fter 5 s. We choose positive direction downwrd f motion of counterweight. yw = Wt At t = 5 s, y = 30 ft W 30 ft = (5 s) W W =.4 ft/s () Accelertions of E nd C. Since y + y = constnt v + v = 0, nd + = 0 W E W E W E W =.4 ft/s Thus: E = = (.4 ft/s ), W =.40 ft/s E Also, y + y = constnt, v + v = 0, nd + = 0 C E C E C E Thus: (b) Velocity of elevt fter 5 s. C = = (.4 ft/s ) =+ 4.8 ft/s, E = 4.80 ft/s C ve = ( ve) 0 + Et = 0 + (.4 ft/s )(5 s) = ft/s ( v E ) 5 =.00 ft/s PROPRIETARY MATERIAL. 03 The McGrw-Hill Compnies, Inc. All rights reserved. No prt of this Mnul my be displyed, reproduced distributed in ny fm by ny mens, without the pri written permission of the publisher, used beyond the limited 60