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