Question

A particle is moving in parabolic path x2 = y, with constant speed v. when particle crosses origin, the radius of curvature at origin will be

Answer 1

Answer:

$\frac{1}{2}$

Explanation:

1) We have,

Particle moving in parabolic path $y=x^2$ with constant speed 'v'.

We have,

$y'(0)=\frac{d(x^2)}{dx} =2x=2*0=0$

$y''(0)=\frac{d(y')}{dx}=\frac{d(2x)}{dx}=2$

2) Radius of curvature at origin ,

$R(0,0)=\frac{(1+y'^2)^{3/2}}{y''} =\frac{1+0}{2}=\frac{1}{2}$

That is , radius of curvature at origin will be 0.5 units.

Answer 2

Explanation:

for ans see above ans is 1/2 or 0.5