CCHHAAPPTTEERR1111PROPRIETARY MATERIAL. © 2009 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 3PROBLEM 11.1 The motion of a particle is defined by the relation 4 21.5 30 5 10,x t t t= − + +where x and t are expressed in meters and seconds, respectively. Determine the position, the velocity, and the acceleration of the particle when 4 s.t =SOLUTION Given: 4 2321.5 30 5 106 60 518 60x t t tdxv ttdtdva tdt= − + += = − += =−Evaluate expressions at 4 s.t =4 21.5(4) 30(4) 5(4) 10 66 mx = − + + = − 66.0 mx = −�36(4) 60(4) 5 149 m/sv = − + = 149.0 m/sv =�2 218(4) 60 228 m/sa = − =�2228.0 m/sa = ��3PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.PROPRIETARY MATERIAL. © 2009 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 3PROBLEM 11.1 The motion of a particle is defined by the relation 4 21.5 30 5 10,x t t t= − + +where x and t are expressed in meters and seconds, respectively. Determine the position, the velocity, and the acceleration of the particle when 4 s.t =SOLUTION Given: 4 2321.5 30 5 106 60 518 60x t t tdxv ttdtdva tdt= − + += = − += =−Evaluate expressions at 4 s.t =4 21.5(4) 30(4) 5(4) 10 66 mx = − + + = − 66.0 mx = −�36(4) 60(4) 5 149 m/sv = − + = 149.0 m/sv =�2 218(4) 60 228 m/sa = − =�2228.0 m/sa = ��4PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limiteddistribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.PROBLEM 11.2 The motion of a particle is defined bythe relation 3 212183 2183 22 5,x t12x t12t t3 2t t3 22 5t t2 5= −x t= −x t12x t12= −12x t122 5+ +2 5t t+ +t t2 5t t2 5+ +2 5t t2 5where xand tare expressed in meters and seconds, respectively. Determine the position and the velocity when the acceleration of the particle the velocity when the acceleration of the particle the velocityis equal to zero. SOLUTIONGiven:3 2212183 2183 22 5363627236x t12x t12t t3 2t t3 22 5t t2 5dxv t36v t36tdtdva t72a t72dt= −x t= −x t12x t12= −12x t122 5+ +2 5t t+ +t t2 5t t2 5+ +2 5t t2 5v t= =v tv t= =v t− +36− +36t− +ta t= =a ta t= =a t−Find the time for0.a=72360 0.5st t36t t360 0t t0 0t t− =t t36t t36− =36t t36�0 0t t0 0�0 0t t0 00 0=0 0Substitute into above expressions. 3 212(0.5)18(3 218(3 20.5)3 20.5)3 22(0.5)5 3x= −12= −12(0.5= −(0.5)= −)+ +2(+ +2(0.+ +0.5)+ +5)5 3=5 33.00mx=�236(0.5)36(0.5) 27 m/sv= −36= −36(0.5= −(0.5)= −)) 2+) 2= −7.00m/sv= −��5PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limiteddistribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.PROBLEM 11.3 The motion of a particle is defined bythe relation 3 25 53 25 53 2308 ,3 2xt tt t3 2t t3 2t x8 ,t x8 ,= −= −t t= −t t− +30− +30t x− +t x where xandtare expressed in meters and seconds, respectively. Determine the time, the position, and the acceleration whenv =0.SOLUTION We have 3 25 53 25 53 23083 2xt tt t3 2t t3 2t=− −t t− −t tt t− −t t+Then25 525 5230dxv t5 5v t5 5tdtv t= =v tv t= =v t− −5 5− −5 5t− −tand105dva t10a t10dta t= =a ta t= =a t−When0:v=2 25 52 25 52 2302 2302 25(2 25(2 26)0t t2 2t t2 25 5t t5 52 25 52 2t t2 25 52 2t t2 2t t2 2− −t t− −t t5 5t t5 5− −5 5t t5 5=− −t t− −t t=or3 s and2 s (Reject)t t3 t t3 s at ts andt tnd= =t t= =t t3 t t3 = =3 t t3 s at ts a= =s at ts andt tnd= =ndt tnd−3.00 st=At3 s:t=3 235 53 25 53 2(35 5(35 5) () (3 2) (3 25 5) (5 53 25 53 2) (3 25 53 23)3 23)3 230(3) 83 2(33 2(3) (3 2) (x=− −) (− −) () (− −) (3)− −3)) 8+) 83or59.5mx=−310(3)5a=−3or25.0m/s2a=6PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.PROPRIETARY MATERIAL. © 2009 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 6PROBLEM 11.4 The motion of a particle is defined by the relation 26 8 40 cos ,xt tπ= − +where x and t are expressed in meters and seconds, respectively.