Depending on the structure of your date/time values, there are a few ways to do this.
If the dates will always be 10 digits, you could simply extract 10 characters from the left: Left(yourvalue,10). I presume #'s will not be part of your date.
If not, you need to determine the length of the entire date (remember for this suggestion the date would sometimes be 01/01/2015 and sometimes 1/1/2015 or even 1/1/15 [not good]). So from 26 (2015/01/01 01:03:15 PM EST) you subtract 16 (the " 01:03:15 PM EST" portion), thus running the function: Left(Len(yourDate)-16).
If you want a format that differs from the value you have, or different from your system settings
(e.g. dd/mm/yyyy vs yyyy,mm/dd) you'll have to wrap the example within the Format function.
Maybe as: Format(Left(Len(yourDate)-16),"mm/dd/yyyy")
For the date serial numbers, Format(42238, "dd/mm/yyyy") will return 22/08/2015 as noted by cooper645