Re: Units

*To*: mathgroup at smc.vnet.net*Subject*: [mg23810] Re: [mg23718] Units*From*: "Richard Finley" <rfinley at medicine.umsmed.edu>*Date*: Sat, 10 Jun 2000 03:00:19 -0400 (EDT)*Sender*: owner-wri-mathgroup at wolfram.com

Enrique Notice that you have your units wrong...since t is time in seconds, your acceleration should be {3 t Meter/Second^3, t^2 Meter/Second^4} when you enter the ODE you need to be careful of the units: DSolve[{x''[t]==3 t Meter/Second^3,y''[t]==t^2 Meter/Second^4, x[0 Second]==2 Meter,y[0 Second]==5 Meter,x'[1 Second]==2 Meter/Second,y'[1 Second]==3 Meter/Second },{x[t],y[t]},t] This will give the solution: {{ x[t] -> Meter (4 Second^3 + Second^2 t + t^3)/(2 Second^3), y[t]-> Meter (60 Second^4 + 32 Second^3 t + t^4)/{12 Second^4) }} Which gives the result in Meter as it should. When you plug in a value for t...say t -> 3 Second you will get the correctly dimensioned answer. Of course, the easier (and usual) way to do it is to just leave the units out until the end. regards, RF >>> "Enrique Cao" <cao at mundo-r.com> 06/04/00 11:09PM >>> Hi: I need help in order to solve this problem. I have the vector aceleration {3 t Meter/Second^2,t^2 Meter/Second^2}. This is a vector dependent of time (t). I want to get the position vector with initial conditions {x[0]==2 Meter,y[0]==5 Meter,x'[1]==2 Meter/Second,y'[1]==3 Meter/Second}. I do this with the follow expression: << Miscellaneous`Units` << Miscellaneous`SIUnits` DSolve[{x''[t]==3 t Meter/Second^2,y''[t]==t^2 Meter/Second^2, x[0]==2 Meter,y[0]==5 Meter,x'[1]==2 = Meter/Second,y'[1]==3 Meter/Second },{x[t],y[t]},t] The solution is {x[t]->(Meter(4 Second^2+ 8 Second t +t^3 ))/(2 Second^2), y[t]->Mter(3+(6 t)/Second+t^4/(12 Second^2))} The unit of position is Meter but the solution is not correct. How can I get the correct solution ?? Thank you very much Henrique cao at mundo-r.com