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Two-cars-travel-from-point-A-to-point-B-a-distance-of-200km-Car-X-travels-at-an-average-speed-of-60km-h-with-a-tyre-radius-of-20cm-while-Car-Y-travels-at-an-average-speed-of-50km-h-with-a-tyre-radi




Question Number 228274 by necx122 last updated on 05/Apr/26
Two cars travel from point A to point  B, a distance of 200km. Car X travels  at an average speed of 60km/h with a  tyre radius of 20cm, while Car Y travels  at an average speed of 50km/h with a tyre  radius of 25cm. Assuming all other  conditions are identical, which car  arrives at point B first?  (a) car X  (b) car Y  (c) both arrive at the same time  (d) cannot be determined
$${Two}\:{cars}\:{travel}\:{from}\:{point}\:{A}\:{to}\:{point} \\ $$$${B},\:{a}\:{distance}\:{of}\:\mathrm{200}{km}.\:{Car}\:{X}\:{travels} \\ $$$${at}\:{an}\:{average}\:{speed}\:{of}\:\mathrm{60}{km}/{h}\:{with}\:{a} \\ $$$${tyre}\:{radius}\:{of}\:\mathrm{20}{cm},\:{while}\:{Car}\:{Y}\:{travels} \\ $$$${at}\:{an}\:{average}\:{speed}\:{of}\:\mathrm{50}{km}/{h}\:{with}\:{a}\:{tyre} \\ $$$${radius}\:{of}\:\mathrm{25}{cm}.\:{Assuming}\:{all}\:{other} \\ $$$${conditions}\:{are}\:{identical},\:{which}\:{car} \\ $$$${arrives}\:{at}\:{point}\:{B}\:{first}? \\ $$$$\left({a}\right)\:{car}\:{X} \\ $$$$\left({b}\right)\:{car}\:{Y} \\ $$$$\left({c}\right)\:{both}\:{arrive}\:{at}\:{the}\:{same}\:{time} \\ $$$$\left({d}\right)\:{cannot}\:{be}\:{determined} \\ $$
Answered by TonyCWX last updated on 05/Apr/26
The tyre radius does not affect the Average Speed of the car.    Car X: T = ((200)/(60)) = ((10)/3) h  Car Y: T = ((200)/(50)) = 4 h    T_X  < T_Y , Car X will arrive Point B first.    To be fair, by comparing the average speed, we can  already deduce that Car X will arrive first.
$$\mathrm{The}\:\mathrm{tyre}\:\mathrm{radius}\:\mathrm{does}\:\mathrm{not}\:\mathrm{affect}\:\mathrm{the}\:\mathrm{Average}\:\mathrm{Speed}\:\mathrm{of}\:\mathrm{the}\:\mathrm{car}. \\ $$$$ \\ $$$$\mathrm{Car}\:{X}:\:{T}\:=\:\frac{\mathrm{200}}{\mathrm{60}}\:=\:\frac{\mathrm{10}}{\mathrm{3}}\:\mathrm{h} \\ $$$$\mathrm{Car}\:{Y}:\:{T}\:=\:\frac{\mathrm{200}}{\mathrm{50}}\:=\:\mathrm{4}\:\mathrm{h} \\ $$$$ \\ $$$${T}_{{X}} \:<\:{T}_{{Y}} ,\:\mathrm{Car}\:{X}\:\mathrm{will}\:\mathrm{arrive}\:\mathrm{Point}\:\mathrm{B}\:\mathrm{first}. \\ $$$$ \\ $$$$\mathrm{To}\:\mathrm{be}\:\mathrm{fair},\:\mathrm{by}\:\mathrm{comparing}\:\mathrm{the}\:\mathrm{average}\:\mathrm{speed},\:\mathrm{we}\:\mathrm{can} \\ $$$$\mathrm{already}\:\mathrm{deduce}\:\mathrm{that}\:\mathrm{Car}\:{X}\:\mathrm{will}\:\mathrm{arrive}\:\mathrm{first}. \\ $$

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