Train delay

Due to a breakdown, the train lost 16 minutes of standing on the track behind Brno. He "eliminated" this delay so that after the start, the 80 km long section went at a speed 10 km/h higher than originally planned. What speed was it and what was it supposed to be?

Correct answer:

v₁ =  50 km/h
v₂ =  60 km/h

Step-by-step explanation:

$$\begin{gathered} s=80\text{ km } \\ {v}_0=10\text{ km/h } \\ {t}_0=16\text{ min}\to \text{ h}=16\mathrm{/}60\text{ h}=0.26667\text{ h } \\ s=v_1\cdot t \\ s=v_2\cdot (t-t_0) \\ {v}_2=v_1+v_0 \\ s=v_1\cdot t \\ s=v_2\cdot t-v_2\cdot t_0 \\ {v}_2=v_1+v_0 \\ s=(v_2-v_0)\cdot t \\ s=v_2\cdot t-v_2\cdot t_0 \\ s=v_2\cdot t-v_0\cdot t \\ s=v_2\cdot t-v_2\cdot t_0 \\ 80=v_1\cdot t \\ 80=v_2\cdot (t-16\mathrm{/}60) \\ {v}_2=v_1+10 \\ 80=v_1\cdot t; \\ 80=(v_1+v_0)\cdot (t-16\mathrm{/}60) \\ 80=v_1\cdot t+v_0\cdot t-16\mathrm{/}60\cdot v_1-16\mathrm{/}60\cdot v_0 \\ t=1.6\text{ h } \\ {v}_1=s\mathrm{/}t=80\mathrm{/}1.6=50\text{ km/h} \end{gathered}$$

$v_2=v_1+v_0=50+10=60\text{ km/h}$

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