1. By the condition

$$
T(t)=\frac{270-s(t)}{s(t) / t}=\frac{t(270-s(t))}{s(t)}=C>1, t \in[0.5 ; 1]
$$

Then $s(t)=\frac{270 t}{t+C}$ on this interval. The speed of movement

$$
\begin{aligned}
& v(t)=s^{\prime}(t)=\frac{270 C}{(t+C)^{2}}=60 \text { when } t=1 \text {, i.e. } \\
& \qquad 2 c^{2}-5 c+2=0 \rightarrow C_{1}=2, C_{2}=\frac{1}{2}
\end{aligned}
$$

The second value of the constant is not allowed by the condition. Thus, $s(t)=\frac{270 t}{t+2}$ and $s(1)=90$.

$$
v(t)=\left.\frac{540}{(t+2)^{2}}\right|_{t=\frac{1}{2}}=\frac{432}{5}=86.4 \text { km } / \text { hour }
$$
Answer: 1) 90 km; 2) 86.4 km/hour