Solving Position-Dependent Acceleration Problems

1. The Need for a New Kinematic Relation

Let's begin by understanding why our existing tools are insufficient. The definition of acceleration is a = dv/dt. If we have a = a(x), our equation of motion is:

We cannot integrate this equation directly with respect to time because the right-hand side, a(x), depends on position x, and we do not yet know how x varies with t. We are faced with an integral of the form ∫ a(x(t)) dt, which we cannot solve without first finding x(t).

To break this circular problem, we must find a way to eliminate time t as the variable of integration and instead relate velocity v directly to position x. This is achieved with the chain rule.

2. Derivation of the Identity: a = v(dv/dx)

This identity is one of the most powerful tools in kinematics. Its derivation is a beautiful and direct application of the chain rule from calculus.

We start with the definition of acceleration:
$a=\frac{dv}{dt}$
Since we want to introduce the position variable x, we can express the derivative of v with respect to t in terms of derivatives with respect to x. The chain rule states:
$\frac{dv}{dt}=\frac{dv}{dx}\cdot \frac{dx}{dt}$
Now, we simply need to recognize the second term on the right-hand side. The rate of change of position with respect to time, dx/dt, is the definition of instantaneous velocity, v.
$\frac{dx}{dt}=v$
Substituting this back into our chain rule expression gives the desired identity:
$a=v\frac{dv}{dx}$
This elegant result gives us an expression for acceleration that depends on velocity v and the gradient of the velocity with respect to position, dv/dx. It completely bypasses the time variable.

To see this derivation and its application explained in a familiar context, the following video will be helpful.

3. The Integration Method for a(x)

Now that we have the identity a = v(dv/dx), we can use it to solve problems where acceleration is given as a function of position, a(x).

Our equation of motion becomes:
$a(x)=v\frac{dv}{dx}$
Notice that this is a differential equation relating the variables v and x. Just as we did for a(v) problems, we can solve this using the method of separation of variables. We rearrange the equation to group all terms involving x on one side and all terms involving v on the other:
$a(x)dx=vdv$
This form is now ready to be integrated. We integrate both sides between an initial state (position x_i, velocity v_i) and a final state (position x_f, velocity v_f):
${\int }_{x_i}^{x_f}a(x)dx={\int }_{v_i}^{v_f}vdv$
The integral on the right-hand side is always the same, regardless of the physical situation:
${\int }_{v_i}^{v_f}vdv={\left[\frac{1}{2}{v}^2\right]}_{v_i}^{v_f}=\frac{1}{2}{v}_f^2-\frac{1}{2}{v}_i^2=\frac{v_f^2-v_i^2}{2}$
This leads to the general integral form for solving a(x) problems:
${\int }_{x_i}^{x_f}a(x)dx=\frac{v_f^2-v_i^2}{2}$
This equation allows us to find the change in an object's velocity as it moves from one point to another, through a position-dependent acceleration field. You might notice this equation's structure is very similar to the work-energy theorem (W = ΔKE), a deep connection we will explore later in the course when we study dynamics.

4. Worked Example: Acceleration Proportional to Displacement

A classic example of position-dependent acceleration is a mass attached to a spring, where the restoring force (and thus acceleration) is proportional to the displacement from equilibrium, a relationship known as Hooke's Law. Let's analyze this purely from a kinematic perspective.

Problem: A particle starts from rest at x = x_0 and moves in a straight line with an acceleration a = -kx, where k is a positive constant. Find its velocity v as a function of its position x.

Solution:
The acceleration is given as a(x) = -kx. Since we need to relate velocity and position, we use our new tool:
$a=v\frac{dv}{dx}$
$-kx=v\frac{dv}{dx}$
Now, we separate the variables and set up the definite integrals. The particle starts at position x_0 with velocity 0, and we want to find its velocity v at some arbitrary final position x.
${\int }_{x_0}^x-k{x}^{\prime }d{x}^{\prime }={\int }_0^v{v}^{\prime }d{v}^{\prime }$
(Note: Using x' and v' as integration variables is a standard mathematical convention to avoid confusion with the integral's limits.)

Let's evaluate the integrals:
${\left[-\frac{1}{2}k{x}^{\prime 2}\right]}_{x_0}^x={\left[\frac{1}{2}{v}^{\prime 2}\right]}_0^v$
$-\frac{1}{2}k{x}^2-\left(-\frac{1}{2}k{x}_0^2\right)=\frac{1}{2}{v}^2-0$
$\frac{1}{2}k(x_0^2-x^2)=\frac{1}{2}{v}^2$
Solving for v, we get:
$v(x)=\pm \sqrt{k(x_0^2-x^2)}$
This is the velocity of the particle at any position x. The ± sign indicates that the particle will have the same speed at a given position x whether it's moving to the right or to the left. This type of motion is called Simple Harmonic Motion (SHM), a topic you will study in great detail later.

This same problem is discussed in the following resource, which provides a slightly different perspective by relating it to kinetic and potential energy—a preview of concepts to come.

5. Choosing the Right Tool: a(t), a(v), or a(x)?

You now have three distinct methods for handling 1D kinematics with calculus:

1. a = a(t): Use dv = a(t) dt and dx = v(t) dt. Integrate with respect to time.
2. a = a(v): Use dt = dv/a(v) to find t(v), or use dx = v dv / a(v) to find x(v).
3. a = a(x): Use v dv = a(x) dx to find v(x).

The choice of which formula to use depends on the variables involved in the problem statement. The following video solves a complex problem that requires you to make this choice strategically.

The key takeaway from that video is that you must carefully analyze what quantities are given and what quantity is sought.

• If the question involves a, v, and x (or distance), a = v(dv/dx) is your best tool.
• If the question involves a, v, and t, a = dv/dt is the way to go.

Conclusion

In this lesson, we completed our calculus-based toolkit for one-dimensional motion by tackling position-dependent acceleration.

Key Takeaways:

• When acceleration is a function of position, a(x), we cannot directly integrate dv/dt = a(x).
• The correct approach is to use the chain rule identity a = v(dv/dx). This relation is fundamental and must be committed to memory.
• Applying this identity transforms the equation of motion into a separable differential equation in variables x and v: a(x) dx = v dv.
• Integrating this yields the powerful relationship: ∫ a(x) dx = (v_f² - v_i²)/2, which allows you to find the velocity change over a certain displacement.

Preview of the Next Lesson:

We have now developed a sophisticated set of tools for solving complex kinematics problems. However, in a high-stakes exam like the JEE Advanced, solving the problem is only half the battle. You also need to be confident that your answer is correct. In our next lesson, we will focus on verification and error-checking techniques. We will learn how to apply dimensional analysis and limiting case checks to rigorously validate our solutions, a critical skill for building both speed and accuracy.