Question

A particle is moving in parabolic path y=4x^2 with constant speed vo m/s.the acceleration of the particle when it crosses the origin is

Answer 1

Answer:

$8v_0^2$

Explanation:

1) We have,

a particle moving in a parabolic path $y=4x^2$ with constant speed v_0 m/s.

Then,

$y=4x^2\\ \\=>y'=4*2x=8x\\ \\ =>y''=8*1=8$

2) Radius of curvature is given by

$R=\frac{(1+y'^2)^{3/2}}{y''}\\ \\=\frac{1+(8x)^2}{8}\\ \\=\frac{1}{8}$

3) Now,

Since, this particle moves in parabolic path ,so acceleration will be same radial acceleration .

We know, acceleration of particle is given by

$a=\frac{speed^2}{R}=\frac{v_o^2}{R} =8v_o^2$

Hence, acceleration of particle is given by

$\boxed{8v_o^2}$

Answer 2

Answer:

Hope this will help you

Explanation: