Answer to Question #177537 in Mechanical Engineering for Varun

Question

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!

Answer 1

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

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

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