Determine safety factor

You will see that at time $1.32$ the friction angle and cohesion have become such low that the slope becomes unstable. At that time the cohesion has decreased from $10.0 \frac{kN}{m^2}$ to $(1.-0.32)*10.0 = 6.8 \frac{kN}{m^2}$. The safety factor thus is $\frac{10}{6.8} = 1.47$.