Question

A particle moves in space along the path z= ax3 + by2 in such a way that dx/dt =c= dy/dt. Where a, b and c are constantsThe acceleration of the particle is-
(A) (6ac2x +2bc2)
(B) (2ax2 + 6by2) k
(C)(4bc2x + 6ac2)
(D) (bc2x + 2by)k​

Answer 1

hopefully it will help u

Answer 2

Answer:

$a=6ac^2x+2bc^2$

Explanation:

Given that,

The path of the particle's motion is given by :

$z=ax^3+by^2$..............(1)

It is given that, $\dfrac{dx}{dt}=\dfrac{dy}{dt}=c$

Velocity of the particle,

$v=\dfrac{dz}{dt}$

$v=\dfrac{d(ax^3+by^2)}{dt}$

$v=3ax^2c+2byc$

Acceleration of the particle is given by :

$a=\dfrac{dv}{dt}$

$a=\dfrac{d(3ax^2c+2byc)}{dt}$

$a=6ac^2x+2bc^2$

So, the acceleration of the particle is $6ac^2x+2bc^2$. Hence, this is the required solution.