Question Number 10016 by ridwan balatif last updated on 21/Jan/17

Commented byridwan 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...?

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

Commented byridwan balatif last updated on 21/Jan/17

thank you 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.

Commented byridwan balatif last updated on 21/Jan/17

Good Technique  Thank you sir