Question

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) \(\left[c \hat{i}+3 A x c^{2} \hat{j}\right] m s^{-2}\)

Short answer

The acceleration of the particle is given by \(\vec{a} = \left(0\hat{i} + 6Axc^2 \hat{j}\right) \, ms^{-2}\). The correct choice is (B) \(6 \mathrm{Axc}^{2} \hat{\mathrm{j}} \mathrm{ms}^{-2}\).

Step-by-step solution

Relation between velocities of x and y components

We know that the path of the particle is given by y = Ax^3 + B. We are also given that its x-component is dx/dt = c, which is a constant. We can calculate the y-component of velocity by differentiating y with respect to x and then multiplying with dx/dt. In short, we have to differentiate y with respect to time.

Calculate the y-component of velocity

To get the y-component of the velocity, we first differentiate y with respect to x, and then multiply the result by dx/dt. \[ \frac{dy}{dx} = \frac{d}{dx}(Ax^3 + B) \] \[ \frac{dy}{dx} = 3 Ax^2 \] Now multiply the result with dx/dt (c): \[ \frac{dy}{dt} = \frac{dy}{dx} \cdot \frac{dx}{dt} = 3 Ax^2 \cdot c \] \[ \frac{dy}{dt} = 3 Acx^2 \]

Calculate the acceleration vector

Now that we have calculated both the x-component and y-component of the velocity, we can calculate their time derivatives to get the components of the acceleration vector. For the x-component, since it is a constant, its time derivative is 0: \[ \frac{d^2x}{dt^2} = 0 \] For the y-component, differentiate again with respect to time: \[ \frac{d^2y}{dt^2} = \frac{d}{dt}(3 Acx^2) \] Now, apply the chain rule, differentiating with respect to x and then multiplying with dx/dt (c): \[ \frac{d^2y}{dt^2} = \frac{d}{dx}(3 Acx^2) \cdot \frac{dx}{dt} = 6 Acx\cdot c \]

Write the vector acceleration in terms of i and j components

By combining the x and y components of acceleration, we can observe the answer matches with one of the given choices. \[ \vec{a} = \left(0\hat{i} + 6Axc^2 \hat{j}\right) \, ms^{-2} \] Comparing the answer with the given options, we can see that the correct choice is: (B) \(6 \mathrm{Axc}^{2} \hat{\mathrm{j}} \mathrm{ms}^{-2}\)

Key concepts

Acceleration in Vector Form

Understanding acceleration in vector form is crucial when analyzing the motion of particles in a 2-dimensional plane. In a simple sense, acceleration describes how fast the velocity of an object changes with time. However, in vector form, acceleration not only has magnitude but also direction.

For our exercise, the particle's motion is along a path determined by the equation \(y = Ax^3 + B\). To capture the full description of acceleration in vector form, both the x and y components of the particle's motion need to be considered. Since the x-component's velocity is constant, its acceleration is zero. Meanwhile, the y-component requires differentiation to uncover its changing velocity and thus allows us to calculate its acceleration.

Ultimately, the vector form of acceleration is expressed as \(\vec{a} = \left(0\hat{i} + 6Axc^2 \hat{j}\right) \) in this scenario. Notice how the i-component is zero, reflecting the constant nature of dx/dt, while the j-component highlights how the function's dependence on x brings about an acceleration dependent on both x and constants like A and c.

Differentiation with Respect to Time

Differentiation with respect to time is a fundamental technique for analyzing dynamic systems. In our particular exercise, as the particle moves along the trajectory, we need to determine how its y-coordinate changes over time to comprehend its velocity, and subsequently, its acceleration.

To achieve this, we differentiate the y function (i.e., \(y = Ax^3 + B\)) with respect to x first, yielding a rate of change in terms of x (\(\frac{dy}{dx} = 3Ax^2\)). However, because the system is dynamic and time-dependent, we multiply this result by the derivative of x with respect to time (\(\frac{dx}{dt} = c\)), effectively transitioning our analysis to the time domain.

Thus, the y-component of the velocity is expressed as \(\frac{dy}{dt} = 3Acx^2\). This process reflects the use of the chain rule and provides a glimpse into the particle's motion along the y-axis, ultimately laying the groundwork to explore how its acceleration is calculated.

Chain Rule in Calculus

The chain rule in calculus is pivotal for deciphering relationships involving composite functions. Essentially, the chain rule assists when differentiating a function with respect to a variable that itself is a function of another variable.

In the context of the exercise, you see the chain rule applied when transitioning from velocity to acceleration along the y-component. Initially, the y-velocity was dependent on x (via \(\frac{dy}{dx} = 3Ax^2\)), but time-dependent analysis required us to apply the chain rule. This rule recognizes that an inner function, x in our pathway \(y = Ax^3 + B\), changes with time.

Differentiating the velocity \(\frac{dy}{dt} = 3Acx^2\) once again with respect to time using the chain rule involves taking the derivative with respect to x (yielding \(6Acx\)), then multiplying by the rate at which x changes over time \(\frac{dx}{dt} = c\).

The complete expression for the acceleration's y-component becomes \(\frac{d^2 y}{dt^2} = 6Acx\cdot c\). The chain rule's critical role ensures accurate differentiation across composite layers of functions even when indirect variables like time affect the derivative calculation. This technique encapsulates how interrelated variables in physics, especially in particle motion, are efficiently unraveled.