Mechanics Dynamics - (For Complete File, Download link at the end of...

Engineering Mechanics Dynamics 14e Russell Hibbeler

(Solution Manual all Chapters)

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1 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.Solution a=2t-6 dv=a dt L v

dv= L t

(2t-6) dt v=t 2 -6t ds=v

dt L s

ds= L t

(t 2 -6t) dt s= t 3 3 -3t 2 When t=6 s, v=0Ans.When t=11 s, s=80.7 mAns.

12–1.

Starting from rest, a particle moving in a straight line has an acceleration of a = (2t - 6) m>s 2 , where t is in seconds. What is the particle’s velocity when t = 6 s, and what is its position when t = 11 s?

Ans:

s=80.7 m 2 / 3

2 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

12–2.

SOLUTION

1S +2 s=s 0+v

• t+

1 2 a c t 2

=0+12(10)+

1 2

(-2)(10)

2 =20 ft Ans.If a particle has an initial velocity of v 0=12 ft>s to the right, at s 0=0, determine its position when t=10 s, if a=2 ft>s 2 to the left.

Ans:

s=20 ft

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