1. A driver traveled from place $A$ to place $B$, which are $800 \mathrm{~km}$ apart. Instead of driving at the planned speed of $80 \mathrm{~km} / \mathrm{h}$, on the first half of his journey, the driver was forced to drive at a speed of $(80-\mathrm{m})$ km/h. If on the second half of his journey the driver drives at a speed of $(80+m) \mathrm{km} / \mathrm{h}$, will he arrive on time in $B$?
Solution. Let $t_{1}$ be the time the driver took to cover the first half, and $t_{2}$ the time to cover the second half of the journey. Then

$$
400=(80-m) t_{1} \text { and } \quad 400=(80+m) t_{2} .
$$

Since $80-mt_{2}$. Adding the above equations, we get

$$
800=(80-m) t_{1}+(80+m) t_{2}=80\left(t_{1}+t_{2}\right)-m\left(t_{1}-t_{2}\right)
$$

From which it follows that

$$
80\left(t_{1}+t_{2}\right)>800 \text {, i.e. } t_{1}+t_{2}>10 \mathrm{~h}
$$

If the driver had driven at the planned speed, he would have arrived in V in $\frac{800}{80}=10 \mathrm{~h}$. Therefore, the driver will be late.