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Question-71051




Question Number 71051 by jagannath19 last updated on 11/Oct/19
Commented by jagannath19 last updated on 11/Oct/19
explain
$${explain} \\ $$
Commented by mr W last updated on 11/Oct/19
v_P =7 m/s  v_Q =17 m/s  v^2 −v_P ^2 =2as  ((v_R ^2 −v_P ^2 )/(v_Q ^2 −v_P ^2 ))=(s_(R−P) /s_(Q−P) )=(1/2)  v_R ^2 =((v_Q ^2 −v_P ^2 )/2)+v_P ^2 =((v_Q ^2 +v_P ^2 )/2)  ⇒v_R =(√((v_Q ^2 +v_P ^2 )/2))=(√((17^2 +7^2 )/2))=13 m/s  v_(ava) =((v_R +v_Q )/2)=((13+17)/2)=15 m/s  ⇒(2) is correct
$${v}_{{P}} =\mathrm{7}\:{m}/{s} \\ $$$${v}_{{Q}} =\mathrm{17}\:{m}/{s} \\ $$$${v}^{\mathrm{2}} −{v}_{{P}} ^{\mathrm{2}} =\mathrm{2}{as} \\ $$$$\frac{{v}_{{R}} ^{\mathrm{2}} −{v}_{{P}} ^{\mathrm{2}} }{{v}_{{Q}} ^{\mathrm{2}} −{v}_{{P}} ^{\mathrm{2}} }=\frac{{s}_{{R}−{P}} }{{s}_{{Q}−{P}} }=\frac{\mathrm{1}}{\mathrm{2}} \\ $$$${v}_{{R}} ^{\mathrm{2}} =\frac{{v}_{{Q}} ^{\mathrm{2}} −{v}_{{P}} ^{\mathrm{2}} }{\mathrm{2}}+{v}_{{P}} ^{\mathrm{2}} =\frac{{v}_{{Q}} ^{\mathrm{2}} +{v}_{{P}} ^{\mathrm{2}} }{\mathrm{2}} \\ $$$$\Rightarrow{v}_{{R}} =\sqrt{\frac{{v}_{{Q}} ^{\mathrm{2}} +{v}_{{P}} ^{\mathrm{2}} }{\mathrm{2}}}=\sqrt{\frac{\mathrm{17}^{\mathrm{2}} +\mathrm{7}^{\mathrm{2}} }{\mathrm{2}}}=\mathrm{13}\:{m}/{s} \\ $$$${v}_{{ava}} =\frac{{v}_{{R}} +{v}_{{Q}} }{\mathrm{2}}=\frac{\mathrm{13}+\mathrm{17}}{\mathrm{2}}=\mathrm{15}\:{m}/{s} \\ $$$$\Rightarrow\left(\mathrm{2}\right)\:{is}\:{correct} \\ $$

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