Question Number 191052 by otchereabdullai last updated on 17/Apr/23
$${A}\:{particle}\:{Q}\:{is}\:{moving}\:{with}\:{constant} \\ $$$${velocity}\:\left(−\mathrm{5}{i}+\mathrm{3}{j}\right){m}/{s}.\:{At}\:{time}\: \\ $$$${t}=\mathrm{5}{sec}\:,\:{Q}\:{is}\:{at}\:{the}\:{same}\:{point}\:{with} \\ $$$${position}\:{vector}\:\left(−{i}−\mathrm{5}{j}\right){m}.\:{Find}\:{the} \\ $$$${distance}\:{of}\:{Q}\:{from}\:{the}\:{origin}\:{at}\: \\ $$$${time}\:{t}=\mathrm{2}{sec} \\ $$
Answered by mr W last updated on 17/Apr/23
$${position}\:{at}\:{t}=\mathrm{2}: \\ $$$$\left(−\mathrm{1},−\mathrm{5}\right)−\mathrm{3}×\left(−\mathrm{5},\mathrm{3}\right)=\left(\mathrm{14},−\mathrm{14}\right) \\ $$$${distance}\:{to}\:{origin}:\: \\ $$$$\sqrt{\left(\mathrm{14}\right)^{\mathrm{2}} +\left(−\mathrm{14}\right)^{\mathrm{2}} }=\mathrm{14}\sqrt{\mathrm{2}}\approx\mathrm{19}.\mathrm{8}\:{m} \\ $$
Commented by otchereabdullai last updated on 17/Apr/23
$${Am}\:{always}\:{grateful}\:{to}\:{you}\:{for}\:{the}\: \\ $$$${help}\:{but}\:{ProfW}\:{please}\:{i}\:{want}\:{to}\: \\ $$$${understand}\:{it}\:{very}\:{well}\:{expecially}\: \\ $$$${the}\:\mathrm{3}\:{multiplying}\:\left(−\mathrm{5},\mathrm{3}\right) \\ $$
Commented by mr W last updated on 17/Apr/23
$${position}\left({t}=\mathrm{5}\right)={position}\left({t}=\mathrm{2}\right)+\left(\mathrm{5}−\mathrm{2}\right)×{v} \\ $$$$\Rightarrow{position}\left({t}=\mathrm{2}\right)={position}\left({t}=\mathrm{5}\right)−\mathrm{3}×{v} \\ $$
Commented by otchereabdullai last updated on 17/Apr/23
$${God}\:{bless}\:{you}\:{Prof} \\ $$