Engineering Mechanics - DynamicsChapter 12t21.5 s=Givenvpt2()0=t2Findt2=t22s=vavespt1spt0−t1=vave4ms=vavespeedspt2spt0−spt1spt2−+t1=vavespeed6ms=a1apt1=a118ms2=Problem 12–10A particle is moving along a straight line such that its acceleration is defined as a= −kv. Ifv= v0when d= 0 and t= 0, determine the particle’s velocity as a function of position andthe distance the particle moves before it stops.Given:k2s=v020ms=Solution:apvk−v=vsvddk−v=v0vv1⌠⎮⌡dk−sp=Velocity as a function of positionvv0ksp−=Distance it travels before it stops0v0p−=spv0k
This is the end of the preview. Sign up
access the rest of the document.