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Question Number 88963 by Zainal Arifin last updated on 14/Apr/20

How far can a cyclist travel in 4 h  if his average speed is 11.5 km/h ?

$$\mathrm{How}\:\mathrm{far}\:\mathrm{can}\:\mathrm{a}\:\mathrm{cyclist}\:\mathrm{travel}\:\mathrm{in}\:\mathrm{4}\:\mathrm{h} \\ $$$$\mathrm{if}\:\mathrm{his}\:\mathrm{average}\:\mathrm{speed}\:\mathrm{is}\:\mathrm{11}.\mathrm{5}\:\mathrm{km}/\mathrm{h}\:? \\ $$

Commented by MJS last updated on 14/Apr/20

seriously?  ...ok:  he travels 11.5km in the 1^(st)  hour  he travels 11.5km in the 2^(nd)  hour  he travels 11.5km in the 3^(rd)  hour  he travels 11.5km in the 4^(th)  hour  11.5+11.5+11.5+11.5=4×11.5=46  he travels 46km in 4 hours

$$\mathrm{seriously}? \\ $$$$...\mathrm{ok}: \\ $$$$\mathrm{he}\:\mathrm{travels}\:\mathrm{11}.\mathrm{5km}\:\mathrm{in}\:\mathrm{the}\:\mathrm{1}^{\mathrm{st}} \:\mathrm{hour} \\ $$$$\mathrm{he}\:\mathrm{travels}\:\mathrm{11}.\mathrm{5km}\:\mathrm{in}\:\mathrm{the}\:\mathrm{2}^{\mathrm{nd}} \:\mathrm{hour} \\ $$$$\mathrm{he}\:\mathrm{travels}\:\mathrm{11}.\mathrm{5km}\:\mathrm{in}\:\mathrm{the}\:\mathrm{3}^{\mathrm{rd}} \:\mathrm{hour} \\ $$$$\mathrm{he}\:\mathrm{travels}\:\mathrm{11}.\mathrm{5km}\:\mathrm{in}\:\mathrm{the}\:\mathrm{4}^{\mathrm{th}} \:\mathrm{hour} \\ $$$$\mathrm{11}.\mathrm{5}+\mathrm{11}.\mathrm{5}+\mathrm{11}.\mathrm{5}+\mathrm{11}.\mathrm{5}=\mathrm{4}×\mathrm{11}.\mathrm{5}=\mathrm{46} \\ $$$$\mathrm{he}\:\mathrm{travels}\:\mathrm{46km}\:\mathrm{in}\:\mathrm{4}\:\mathrm{hours} \\ $$

Answered by Rio Michael last updated on 14/Apr/20

hahaha ,what the hell  use the formula   distance = speed × time

$$\mathrm{hahaha}\:,\mathrm{what}\:\mathrm{the}\:\mathrm{hell} \\ $$$$\mathrm{use}\:\mathrm{the}\:\mathrm{formula} \\ $$$$\:\mathrm{distance}\:=\:\mathrm{speed}\:×\:\mathrm{time} \\ $$

Commented by Zainal Arifin last updated on 01/May/20

yes , you are right, simple solution,  not too long.

$$\mathrm{yes}\:,\:\mathrm{you}\:\mathrm{are}\:\mathrm{right},\:\mathrm{simple}\:\mathrm{solution}, \\ $$$$\mathrm{not}\:\mathrm{too}\:\mathrm{long}. \\ $$

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