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Question Number 182716 by sciencestudent last updated on 13/Dec/22

Answered by ARUNG_Brandon_MBU last updated on 13/Dec/22

$$\mathrm{In}\mathrm{a}\mathrm{space}\mathrm{of}20\mathrm{seconds}\mathrm{the}\mathrm{school}\mathrm{bus}\mathrm{covered}\mathrm{a}\mathrm{distance}$$$$\mathrm{of}d\mathrm{and}\mathrm{the}\mathrm{ambulance}\mathrm{covered}\mathrm{a}\mathrm{distance}\mathrm{of}40+d+60=100+d.$$$$$$$$\mathrm{Meanwhile}\mathrm{speed}\mathrm{of}\mathrm{ambulance}=\frac{100+d}{20}=30\times \frac{5}{18}\left(\mathrm{in}{ms}^{-1}\right)$$$$\Rightarrow 100+d=\frac{25\times 20}{3}\Rightarrow d=\frac{500}{3}-100=\frac{200}{3}m$$$$\therefore \mathrm{Speed}\mathrm{of}\mathrm{bus}=\frac{d}{20}=\frac{200}{3\times 20}=\frac{10}{3}{ms}^{-1}=12km/h$$

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