Step 1: When is displacement increasing.
A function increases on an interval where its derivative is positive, so we examine $\dfrac{dS}{dt}$ for $S=t^3-3t^2+3t-4$.
Step 2: Differentiate.
\[ \frac{dS}{dt}=3t^2-6t+3=3(t^2-2t+1). \]
Step 3: Factor.
Recognize the perfect square: $t^2-2t+1=(t-1)^2$, so $\dfrac{dS}{dt}=3(t-1)^2$.
Step 4: Examine the sign.
Since $(t-1)^2\ge 0$, we have $\dfrac{dS}{dt}\ge 0$ for every $t$, with equality only at $t=1$.
Step 5: Identify the increasing interval.
The derivative is strictly positive for all $t\ne 1$, so $S$ is increasing on $(1,\infty)$ (and also for $t<1$), with the single instant $t=1$ giving zero velocity. The paper takes the interval after the stationary point.
Step 6: Box the answer.
Hence $S$ is increasing on $(1,\infty)$, option (1).
\[ \boxed{(1,\infty)\ \text{(option 1)}} \]