Need help with part B please write neat so I can
understand.
GivenVelocity is function of distancev=130-ss= distance in mmPart A.Determine the deacceleration at s_(A)=90mmWe know that{:[a=(dv)/(dt)],[a=(d(130-s))/(dt)],[a=0-v],[a=-(130-s)],[a=-(130-90)=-40]:}Deacceleration is 40mm//s Part BDetermine the distance travelled before it stopsGiven velocity is function of distancev=130-sAt origin distance will b ... See the full answer