A particle moves on a plane along the path y=Ax^3+B in such a way that (dx / dt)=c . c, A, B are

step 1 Hello students, our question is a particle moves on a plane along the path y is equal to a x, Q,

A particle moves on a plane along the path y=Ax^3+B in such...

A particle moves on a plane along the path $\mathrm{y}=\mathrm{Ax}^{3}+\mathrm{B}$ in such a way that $(\mathrm{dx} / \mathrm{dt})=\mathrm{c} . \mathrm{c}, \mathrm{A}, \mathrm{B}$ are constant. Calculate the acceleration of the particle. (A) $3 \mathrm{Ax} \mathrm{cj} \mathrm{ms}^{-2}$ (B) $6 \mathrm{Axc}^{2} \hat{\mathrm{j}} \mathrm{ms}^{-2}$ (C) $3 \mathrm{Axc}^{2} \hat{\mathrm{j}} \mathrm{ms}^{-2}$ (D) $\left[c \hat{i}+3 A x c^{2} \hat{j}\right] m s^{-2}$

A particle moves on a plane along the path $\mathrm{y}=\mathrm{Ax}^{3}+\mathrm{B}$ in such a way that $(\mathrm{dx} / \mathrm{dt})=\mathrm{c} . \mathrm{c}, \mathrm{A}, \mathrm{B}$ are constant. Calculate the acceleration of the particle.
(A) $3 \mathrm{Ax} \mathrm{cj} \mathrm{ms}^{-2}$
(B) $6 \mathrm{Axc}^{2} \hat{\mathrm{j}} \mathrm{ms}^{-2}$
(C) $3 \mathrm{Axc}^{2} \hat{\mathrm{j}} \mathrm{ms}^{-2}$
(D) $\left[c \hat{i}+3 A x c^{2} \hat{j}\right] m s^{-2}$

Key Concepts

Parametric Equations in Kinematics

When an object’s motion is described by different functions for each spatial coordinate in terms of a common parameter (usually time), the path of the object is expressed in parametric form. This approach is fundamental in analyzing motions along curves, where derivatives with respect to time yield velocity and acceleration components in each coordinate direction.

Chain Rule in Differentiation

The chain rule is a critical concept in calculus used to differentiate composite functions. In kinematics, when a coordinate is expressed as a function of another variable (such as y as a function of x) that itself depends on time, the chain rule allows us to link the derivatives and compute rates of change in time, such as converting dy/dx to dy/dt.

Velocity and Acceleration Vectors

Understanding the calculation of velocity and acceleration from position functions is essential in kinematics. The velocity vector is obtained by taking the first derivative of the position vector with respect to time, while the acceleration vector is the derivative of the velocity vector. Each component of these vectors represents the rate of change in the corresponding spatial direction.

Constant Rate of Change

When a derivative such as dx/dt is constant, it simplifies the analysis of motion. This constant rate leads to a linear component in motion, which can be directly differentiated, reducing complexity when computing higher derivatives like acceleration, and allowing for easier application of the chain rule.

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