A cheetah

A cheetah was separated from his brothers during a hunt. His brothers were at a watering hole 120 km away. He had to return before sundown, which was 1.5 hours away. His brothers planned to leave the watering hole at sundown.
- For the first 40 km, the cheetah runs at 55% of his maximum speed.
- For the next 50 km, he runs at 70% of his maximum speed.
- For the final 30 km, he runs at full speed.
Does the cheetah reach his brothers before they leave the watering hole?

Final Answer:

v =  80 km/h
v₂ =  116.1039 km/h

Step-by-step explanation:

$$\begin{gathered} s=120\text{km} \\ t=1.5\text{h} \\ v=s/t=120/1.5=80\text{km/h} \end{gathered}$$

$$\begin{gathered} a=40\text{km} \\ b=50\text{km} \\ c=30\text{km} \\ {t}_1=a/(0.55\cdot v_2) \\ {t}_2=b/(0.70\cdot v_2) \\ {t}_3=c/v_2 \\ t=t_1+t_2+t_3 \\ a/(0.55\cdot v_2)+b/(0.70\cdot v_2)+c/v_2=t \\ a/0.55+b/0.70+c=t\cdot v_2 \\ 40/0.55+50/0.70+30=1.5\cdot v_2 \\ 1.5v_2=174.155844 \\ {v}_2=\frac{174.15584416}{1.5}=116.1038961 \\ {v}_2=\frac{8940}{77}\doteq 116.103896=116.1039\text{km/h} \end{gathered}$$

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