Answer to Question #177537 in Mechanical Engineering for Varun

Question #177537

A point moves along the path Y= (X/3) with a constant speed of Sm/s. What are the X and Y components of velocity when X=3m? What is the acceleration of point when x=3m!


1
Expert's answer
2021-04-12T01:47:46-0400

"y= \\frac{x}{3}"

Differentiating "\\frac{dy}{dt}=3 \\frac{dx}{dt}"

"\\frac{dy}{dt}=v_y; \\frac{dx}{dt}=v_x"

Also from "x^2+y^2=s^2" we know that "\\sqrt{v_x^2+v_y^2}=s= const"

At x=3, "\\frac{dy}{dt}=3 \\frac{dx}{dt} \\implies v_y=3v_x"

1) "v_x^2+9v_x^2=s"

"10v_x^2=s"

"v_x= \\sqrt{\\frac{s}{10}}m\/s"

"v_y= 3\\sqrt{\\frac{s}{10}} m\/s"

And acceleration, "a= \\frac{v^2}{r}"

Radius of curvature ,"r=\\frac{(1+ (\\frac{dy}{dx})^2)^3}{\\frac{d^2y}{dx^2}}"

r="\\frac{(1+3^2)^3}{0}=0"

"a= \\frac{s^2}{0}=0 m\/s^2"


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