Originally Posted by **jackthehat** the correct equation should be ..
v(t) = v_t (1 - e to the power -(k/m)t ) .. |

Yes!

Originally Posted by **jackthehat** My problem with this originally would have still been the same by the way .. that the term for time t forms part of exponent's power .. and how to get a value for t from that equation ..
but now I am thinking (after going away and refreshing myself with some basic maths .. as I left full-time education 40 years ago !) that if you take the log (logarithm function) of each side of the above equation you can take t down from the power as I believe ... log(e to the power x) = x ... that's right isn't it ? |

Yes!

Originally Posted by **jackthehat** my only problem now (I think) is that the equation would then become ..
log (v(t)) = v_t(1 - (-k/m)t) |

Not quite. It's easiest to start by getting the e^(-kt/m) term by itself:

e^(-kt/m) = 1 - v(t)/v_t

Now take the log of both sides:

-kt/m = ln(1-v(t)/v_t)

Rearrange:

t = (-m/k)ln(1-(v(t)/v_t))

Now set v(t)/v_t = 0.9, and solve for t.