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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

In a space of 20 seconds the school bus covered a distance of d and the ambulance covered a distance of 40+d+60=100+d. Meanwhile speed of ambulance=((100+d)/(20))=30×(5/(18)) (in ms^(−1) ) ⇒100+d=((25×20)/3) ⇒d=((500)/3)−100=((200)/3) m ∴ Speed of bus =(d/(20))=((200)/(3×20))=((10)/3) ms^(−1) =12km/h

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

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