Peter 16

Peter travels to his uncle's home, 30 km away from his place. He cycles for 2/3 of the journey before the cycle develops a mechanical problem, and he has to push it for the rest of the journey. If he is cycling 10 km per hour faster than his walking speed and completes the journey in 3hrs 20mins, determine his cycling speed.

Final Answer:

v =  15 km/h

Step-by-step explanation:

$$\begin{gathered} s=30\text{km} \\ t=3:20=3hr20min=3+\frac{20}{60}=3.3333hr\doteq 3.3333 \\ {s}_1=\frac{2}{3}\cdot s=\frac{2}{3}\cdot 30=\frac{2\cdot 30}{3}=\frac{60}{3}=20\text{km} \\ {s}_2=s-s_1=30-20=10\text{km} \\ v=10+p \\ v\cdot t_1=s_1 \\ p\cdot t_2=s_2 \\ {t}_1+t_2=t \\ v\cdot t_1=s_1 \\ (v-10)\cdot t_2=s_2 \\ v\cdot (t-t_2)=20\text{km} \\ v\cdot t_2-10\cdot t_2=10\text{km} \\ 20/(t-t_2)\cdot t_2-10\cdot t_2=10\text{km} \\ {s}_1\cdot x-s_2\cdot x\cdot (t-x)=s_2\cdot (t-x) \\ 20\cdot x-10\cdot x\cdot (3.3333333333333-x)=10\cdot (3.3333333333333-x) \\ 10x^2-3.333x-33.333=0 \\ a=10;b=-3.333;c=-33.333 \\ D=b^2-4ac=3.333^2-4\cdot 10\cdot (-33.333)=1344.4444444444 \\ D>0 \\ {x}_{1,2}=\frac{-b\pm \sqrt{D}}{2a}=\frac{3.33\pm \sqrt{1344.44}}{20} \\ {x}_{1,2}=0.166667\pm 1.833333 \\ {x}_1=2 \\ {x}_2=-1.666666667 \\ {t}_2=x_1=2\text{h} \\ {t}_1=t-t_2=3.3333-2=\frac{4}{3}\doteq 1.3333\text{h} \\ p=s_2/t_2=10/2=5\text{km/h} \\ v=s_1/t_1=20/1.3333=15 \\ \text{ Verifying Solution: } \\ {T}_1=\frac{s_1}{v}=\frac{20}{15}=\frac{4}{3}\doteq 1.3333\text{h} \\ {T}_2=\frac{s_2}{p}=\frac{10}{5}=2\text{h} \\ T=T_1+T_2=1.3333+2=\frac{10}{3}\doteq 3.3333\text{h} \end{gathered}$$

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