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Question-10016




Question Number 10016 by ridwan balatif last updated on 21/Jan/17
Commented by ridwan balatif last updated on 21/Jan/17
a particle moving in x−y plane that fill   the equation y=(5/8)x^2 . Velocity in x−coordinate  is constant (12m/s). When x=1/3 m the velocity is...?
$$\mathrm{a}\:\mathrm{particle}\:\mathrm{moving}\:\mathrm{in}\:{x}−{y}\:\mathrm{plane}\:\mathrm{that}\:\mathrm{fill}\: \\ $$$$\mathrm{the}\:\mathrm{equation}\:\mathrm{y}=\frac{\mathrm{5}}{\mathrm{8}}\mathrm{x}^{\mathrm{2}} .\:\mathrm{Velocity}\:\mathrm{in}\:\mathrm{x}−\mathrm{coordinate} \\ $$$$\mathrm{is}\:\mathrm{constant}\:\left(\mathrm{12m}/\mathrm{s}\right).\:\mathrm{When}\:\mathrm{x}=\mathrm{1}/\mathrm{3}\:\mathrm{m}\:\mathrm{the}\:\mathrm{velocity}\:\mathrm{is}…? \\ $$
Answered by sandy_suhendra last updated on 21/Jan/17
V_x =12 m/s  x=V_x .t=12t ⇒ 12t=(1/3) ⇒ t=(1/(36)) s  y=(5/8)x^2 =(5/8)(12t)^2 =90t^2   V_y =(dy/dt)=180t=180×(1/(36))=5 m/s  V=(√(V_x ^2 +V_y ^2 )) = (√(12^2 +5^2 )) = 13 m/s
$$\mathrm{V}_{\mathrm{x}} =\mathrm{12}\:\mathrm{m}/\mathrm{s} \\ $$$$\mathrm{x}=\mathrm{V}_{\mathrm{x}} .\mathrm{t}=\mathrm{12t}\:\Rightarrow\:\mathrm{12t}=\frac{\mathrm{1}}{\mathrm{3}}\:\Rightarrow\:\mathrm{t}=\frac{\mathrm{1}}{\mathrm{36}}\:\mathrm{s} \\ $$$$\mathrm{y}=\frac{\mathrm{5}}{\mathrm{8}}\mathrm{x}^{\mathrm{2}} =\frac{\mathrm{5}}{\mathrm{8}}\left(\mathrm{12t}\right)^{\mathrm{2}} =\mathrm{90t}^{\mathrm{2}} \\ $$$$\mathrm{V}_{\mathrm{y}} =\frac{\mathrm{dy}}{\mathrm{dt}}=\mathrm{180t}=\mathrm{180}×\frac{\mathrm{1}}{\mathrm{36}}=\mathrm{5}\:\mathrm{m}/\mathrm{s} \\ $$$$\mathrm{V}=\sqrt{\mathrm{V}_{\mathrm{x}} ^{\mathrm{2}} +\mathrm{V}_{\mathrm{y}} ^{\mathrm{2}} }\:=\:\sqrt{\mathrm{12}^{\mathrm{2}} +\mathrm{5}^{\mathrm{2}} }\:=\:\mathrm{13}\:\mathrm{m}/\mathrm{s} \\ $$
Commented by ridwan balatif last updated on 21/Jan/17
thank you sir
$$\mathrm{thank}\:\mathrm{you}\:\mathrm{sir} \\ $$
Answered by mrW1 last updated on 21/Jan/17
there is a way without knowing the time:  y=(5/8)x^2   (dy/dx)=((5×2)/8)x=((5x)/4)  V_x =(dx/dt)    (=12 m/s)  V_y =(dy/dt)=(dy/dx)∙(dx/dt)=(dy/dx)∙V_x =((5x)/4)∙V_x   V=(√(V_x ^2 +V_y ^2 ))=V_x (√(1+(((5x)/4))^2 ))       ...(i)  V_(x=1/3m) =12(√(1+((5/(4×3)))^2 ))=12×((13)/(12))=13 m/s    (i) is also valid when the velocity in  x direction is not constant.
$${there}\:{is}\:{a}\:{way}\:{without}\:{knowing}\:{the}\:{time}: \\ $$$${y}=\frac{\mathrm{5}}{\mathrm{8}}{x}^{\mathrm{2}} \\ $$$$\frac{{dy}}{{dx}}=\frac{\mathrm{5}×\mathrm{2}}{\mathrm{8}}{x}=\frac{\mathrm{5}{x}}{\mathrm{4}} \\ $$$${V}_{{x}} =\frac{{dx}}{{dt}}\:\:\:\:\left(=\mathrm{12}\:{m}/{s}\right) \\ $$$${V}_{{y}} =\frac{{dy}}{{dt}}=\frac{{dy}}{{dx}}\centerdot\frac{{dx}}{{dt}}=\frac{{dy}}{{dx}}\centerdot{V}_{{x}} =\frac{\mathrm{5}{x}}{\mathrm{4}}\centerdot{V}_{{x}} \\ $$$${V}=\sqrt{{V}_{{x}} ^{\mathrm{2}} +{V}_{{y}} ^{\mathrm{2}} }={V}_{{x}} \sqrt{\mathrm{1}+\left(\frac{\mathrm{5}{x}}{\mathrm{4}}\right)^{\mathrm{2}} }\:\:\:\:\:\:\:…\left({i}\right) \\ $$$${V}_{{x}=\mathrm{1}/\mathrm{3}{m}} =\mathrm{12}\sqrt{\mathrm{1}+\left(\frac{\mathrm{5}}{\mathrm{4}×\mathrm{3}}\right)^{\mathrm{2}} }=\mathrm{12}×\frac{\mathrm{13}}{\mathrm{12}}=\mathrm{13}\:{m}/{s} \\ $$$$ \\ $$$$\left({i}\right)\:{is}\:{also}\:{valid}\:{when}\:{the}\:{velocity}\:{in} \\ $$$${x}\:{direction}\:{is}\:{not}\:{constant}. \\ $$
Commented by ridwan balatif last updated on 21/Jan/17
Good Technique  Thank you sir
$$\mathrm{Good}\:\mathrm{Technique} \\ $$$$\mathrm{Thank}\:\mathrm{you}\:\mathrm{sir} \\ $$

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