Distance 11711

The observer sits in a room 2 m from a 50 cm wide window. A road runs parallel at a distance of 500 m. What is the cyclist's average speed on this road when the observer sees him at 15 seconds?

Correct answer:

v =  8.3667 m/s

Step-by-step explanation:

$$\begin{gathered} a=2 \\ b=50\text{cm}\to \text{m}=50:100\text{m}=0.5\text{m} \\ c=500\text{m} \\ t=15\text{m} \\ a:b=(a+c):d \\ d=b/a\cdot (a+c)=0.5/2\cdot (2+500)=\frac{251}{2}=125.5\text{m} \\ d=t\cdot v \\ v=d/t=125.5/15=\frac{251}{30}\doteq 8.3667\text{m/s} \\ {v}_2=v\to \text{km/h}=v\cdot 3.6\text{km/h}=8.3667\cdot 3.6\text{km/h}=30.12\text{km/h} \end{gathered}$$

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