Engineering Mechanics - DynamicsChapter 12t3c−c22bvp0−+b=t37.702 s=sp2b6⎛⎝⎞⎠t23c2⎛⎝⎞⎠t22+vp0t2+sp0+=sp27.127 m=sp3b6⎛⎝⎞⎠t33c2t32+vp0t3+sp0+=sp336.627−m=dsp2sp0−sp2sp3−+sp1sp3−+=d56.009 m=c )The velocityvp1b2⎛⎝⎞⎠t12ct1+vp0+=vp110ms=Problem 12–29A particle is moving along a straight line such that its acceleration is defined as a= ks2. Ifv= v0when s= sp0and t= 0, determine the particle’s velocity as a function of position.Given:k41ms2=v0100−ms=sp010 m=Solution:avspvdd=ksp2=v0vvv⌠⎮⌡dsp0spspp2⌠⎮⌡d=12v2v02−()13p3sp03−=vv0223p3sp03−+=Problem 12–30A car can have an acceleration and a deceleration
This is the end of the preview. Sign up
access the rest of the document.