Question Number 139825 by mathdave last updated on 01/May/21 | ||
Commented by mr W last updated on 01/May/21 | ||
$${rate}\:{of}\:{changing}\:{of}\:{distance}\:{is}\:{the} \\ $$$${sum}\:{of}\:{the}\:{speeds}\:{of}\:{the}\:{cars}: \\ $$$${case}\:\mathrm{1}: \\ $$$$−\left(\mathrm{50}+\mathrm{40}\right)=−\mathrm{90}\:{km}/{h}\:{if}\:{A}\:{was}\:{to} \\ $$$${the}\:{north}\:{of}\:{B}.\:{the}\:{distance}\:{decreases}. \\ $$$$ \\ $$$${case}\:\mathrm{2}: \\ $$$$\left(\mathrm{50}+\mathrm{40}\right)=\mathrm{90}\:{km}/{h}\:{if}\:{A}\:{was}\:{to} \\ $$$${the}\:{south}\:{of}\:{B}.\:{the}\:{distance}\:{increases}. \\ $$ | ||