- #1
calculus_jy
- 56
- 0
given acceleration [tex]a = 1 + ln x[/tex]
i can find that [tex]\Delta v^2 = 2xlnx[/tex] and since it is given that when [tex]t = 0, x = 1,v = 0[/tex]
[tex]\therefore v^2 = 2xlnx[/tex]
however i have been asked to prove [tex]v > 0 \; when \;t > 0[/tex] and i have no idea how to explain it in mathematcial terms, can anyone please give any suggestion?
i can find that [tex]\Delta v^2 = 2xlnx[/tex] and since it is given that when [tex]t = 0, x = 1,v = 0[/tex]
[tex]\therefore v^2 = 2xlnx[/tex]
however i have been asked to prove [tex]v > 0 \; when \;t > 0[/tex] and i have no idea how to explain it in mathematcial terms, can anyone please give any suggestion?