Question

a particle moves according to the law a=-ky. find the velocity as a function of distance y, where vnot is initial velocity​

Answer 1

Answer:

The answer will be v^2−v0^2=−ky^2

Explanation:

According to the problem the particle moves a= -ky

Therefore, dv/dt= -ky

=> dv/dt×dy/dy=−ky

=>  v dv =−ky dy

differentiating both side,

=>   v^2/2=−ky^2/2+c    

=> v^2=−ky^2+C

When vnot is initial velocity i.e v=v0   and y=0 thenso C=v0^2

Hence the velocity as a function distance y can be written as  v^2−v0^2=−ky^2