Distance

If Dana increases her walking speed by 1km/h, she will cover a distance greater than 20 km in 4 hours. If she decreases her speed by 1km/h, she will not cover more than 20 km in 5 hours. At what speed did Dana walk? (s=v. t;s- distance, v- speed, t- time). Solve both inequalities and check for correctness for both; write down the resulting intervals

Final Answer:

v₁ =  4 km/h
v₂ =  5 km/h

Step-by-step explanation:

$$\begin{gathered} v_1<v<v_2 \\ {t}_1=4\text{h} \\ {t}_2=5\text{h} \\ s=20\text{km} \\ \Delta =1\text{km/h} \\ s=v\cdot t \\ (v+\Delta )\cdot t_1>s \\ (v-\Delta )\cdot t_2<s \\ {t}_1\cdot v>s-\Delta \cdot t_1 \\ {t}_2\cdot v<s+\Delta \cdot t_2 \\ v>(s-\Delta \cdot t_1)/t_1 \\ v<(s+\Delta \cdot t_2)/t_2 \\ {v}_1<v<v_2 \\ {v}_1=(s-\Delta \cdot t_1)/t_1=(20-1\cdot 4)/4=4\text{km/h} \end{gathered}$$

$v_2=(s+\Delta \cdot t_2)/t_2=(20+1\cdot 5)/5=5\text{km/h}$

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