Using a modified Arrhenius plot to find activation energy
The use of $t$ only makes sense if $t$ is defined as the time it takes for the reaction to progress a certain relative amount (and if we are talking about first-order reactions).  For example if $t$ is the time it takes for half (or 90% or 99% or whatever, as long as it is specified) the reactant to be consumed, then the idea makes sense.  In that case it would be more helpful and clear to say $t_{1/2}$ (or $t_{90\%}$ or whatever) instead of just $t$.

Assuming $t_{1/2}$ is what is meant by $t$:

Arrhenius equation:

$k = A_0 e^{\frac{-E_a}{RT}}$

$k = A_0 e^{\frac{-E_a}{R}\frac{1}{T}}$

$\ln k = \ln A_0 - \frac{E_a}{R}\frac{1}{T}$

First-order kinetics:

$C = C_0 e^{-k t}$

$\ln \frac{C_0/2}{C_0} = - k t_{1/2}$

$k = \frac{1}{t_{1/2}} \ln 2$

$\ln k = - \ln t_{1/2} + \ln \ln 2 = -\ln{t_{1/2}} + \ln \ln 2$

Combining:

$\ln k = \ln A_0 - \frac{E_a}{R}\frac{1}{T} = -\ln{t_{1/2}} + \ln \ln 2$

$\ln t_{1/2} - \ln \ln 2 = \frac{E_a}{R} T^{-1} - \ln A_0$

$\ln t_{1/2} = \frac{E_a}{R} T^{-1} - \ln A_0 + \ln \ln 2$

So if $t$ means $t_{1/2}$, then a plot of $\ln t_{1/2}$ vs. inverse temperature has a slope of $+\frac{E_a}{R}$, instead of $-\frac{E_a}{R}$ if you plot $\ln k$ vs. inverse temperature.  

N.B. If $t$ just means time, then the idea doesn't make any sense, because then instead of $\ln \ln 2$ you would have $\ln \ln \frac{C(t)}{C_0}$, which is a term that depends on time, so instead of being an explicit equation that is clearly a line, you would have an implicit equation, non-linear in $t$, that would need to be solved before the plot was made, and the result would not be linear and would be impossible to use to determine $E_a$.