Solving Problems with Time-Dependent Acceleration

Hello! Let's dive into the next topic in our exploration of one-dimensional kinematics.

In our previous lessons, you derived and applied the kinematic equations for constant acceleration. We established that these equations are a special case, valid only when $a$ is constant. This led us to a crucial question: what happens when acceleration itself changes?

Today, we address that question head-on. The learning outcome for this lesson is to solve problems with time-dependent acceleration, a(t), using direct integration. We will return to the fundamental definitions of velocity and acceleration and use calculus as our primary tool. This moves us from applying pre-derived formulas to a more general and powerful problem-solving methodology, which is essential for the advanced problems you're aiming to master.

1. The Breakdown of Constant Acceleration Formulas

First, let's be explicit about why the familiar kinematic equations fail when acceleration is not constant. You derived equations like $v = u + a t$ by integrating $d v = a d t$ , assuming $a$ could be pulled out of the integral as a constant. If $a$ is a function of time, $a (t)$ , this is no longer mathematically valid.
$v_{f} - v_{i} = \int_{t_{i}}^{t_{f}} a (t) d t$
The integral of $a (t)$ is generally not $a (t_{f} - t_{i})$ . Therefore, we must abandon the constant acceleration formulas and work directly with the definitions.

2. The General Method: Direct Integration

Our entire approach is built on the two fundamental differential relationships you've already mastered:

Velocity from Acceleration:
$a (t) = \frac{d v}{d t} ⟹ d v = a (t) d t$
Position from Velocity:
$v (t) = \frac{d x}{d t} ⟹ d x = v (t) d t$

To find the velocity or position at a certain time, we integrate these differential expressions between appropriate limits (initial and final states).

Step 1: Finding Velocity from $a (t)$

Given $a (t)$ and an initial velocity $v_{i}$ at time $t_{i}$ , we can find the velocity $v_{f}$ at a later time $t_{f}$ by integrating:
$\int_{v_{i}}^{v_{f}} d v = \int_{t_{i}}^{t_{f}} a (t) d t$
$v_{f} - v_{i} = \int_{t_{i}}^{t_{f}} a (t) d t$
Solving for $v_{f}$ , we get the velocity as a function of time:
$v_{f} (t_{f}) = v_{i} + \int_{t_{i}}^{t_{f}} a (t) d t$
It is crucial to remember that integration yields the change in velocity. You must know the initial velocity to find the final velocity.

Step 2: Finding Position from $v (t)$

Once we have the velocity function $v (t)$ from Step 1, we can find the change in position (displacement) over the same time interval by integrating again:
$\int_{x_{i}}^{x_{f}} d x = \int_{t_{i}}^{t_{f}} v (t) d t$
$Δ x = x_{f} - x_{i} = \int_{t_{i}}^{t_{f}} v (t) d t$
The final position is then $x_{f} = x_{i} + Δ x$ .

This two-step integration process is the core technique for handling any problem where acceleration is given as a function of time.

Integration in Kinematics of Linear Motion — This diagram illustrates the sequential process of integration. Integrating acceleration with respect to time gives velocity, and integrating velocity with respect to time gives displacement. This is the fundamental calculus-based framework we will use.

3. Formalizing the Procedure

The following resource from Resonance provides a concise summary of this method and a clear, worked example. Your mathematical background should make this section feel very natural.

Rectilinear Motion

Please read the following sections from the 'Rectilinear Motion' study material by Resonance. This will formalize the integration technique we just discussed.

Navigate to page 20, section '13. MOTION WITH NON-UNIFORM ACCELERATION'. Read subsection '(i) Acceleration depending on velocity v or time t' to see the integral formulas laid out formally. Study 'Example 19' carefully. This is a perfect, simple application of our two-step integration method for the case where a(t) = 6t. Notice how they first integrate 'a' to find 'v(t)' and then integrate 'v(t)' to find the displacement \Delta x.

As you saw in the reading, the procedure is systematic:

Set up the integral for velocity: $\int d v = \int a (t) d t$ .
Solve for $v (t)$ , using the initial conditions to find the constant of integration (or by using definite integrals).
Set up the integral for position: $\int d x = \int v (t) d t$ .
Solve for $x (t)$ , again using initial conditions.

4. Worked Examples in Action

Let's watch a couple of short video clips where this method is applied to typical JEE-level problems.

First, let's see an example of the first step: finding velocity by integrating acceleration.

This clip from Vedantu JEE's lesson on non-uniformly accelerated motion solves a problem where acceleration is a quadratic function of time.

Watch from 36:54 to 38:44. The problem gives a(t) = 3t^2 + 2t + 2 and an initial velocity. Pay close attention to how the instructor sets up the definite integral for the change in velocity, \Delta v = \int a(t) dt, and uses the initial condition v(0) = 2 m/s to find the final velocity.

Now, let's see the second step: finding position by integrating velocity.

This next clip from the same video demonstrates integrating a velocity function to find position.

Watch from 28:20 to 32:08. Here, the velocity is given as v(t) = 10 + 5t - t^2, and the goal is to find the position. Observe how the definite integral \Delta x = \int_0^3 v(t) dt is used to calculate the x-coordinate at t=3 seconds, given the particle starts at the origin.

Combining these two processes allows you to solve for the complete motion of a particle starting from a given $a (t)$ .

Test your understanding!

A particle starts from rest at the origin at $t = 0$ . Its acceleration is given by the function $a (t) = (6 t - 4) m/s^{2}$ .

Derive the expression for the particle's velocity, $v (t)$ .
Find the time(s) at which the particle is momentarily at rest.
Calculate the particle's position and the total distance traveled at $t = 3 s$ .

Show answer

1. Velocity Function $v (t)$ :
We start with $d v = a (t) d t$ and integrate. The particle starts from rest ( $v (0) = 0$ ) at $t = 0$ .
$\int_{0}^{v (t)} d v^{'} = \int_{0}^{t} (6 t^{'} - 4) d t^{'}$
$v (t) - 0 = [3 t^{'2} - 4 t^{'}]_{0}^{t}$
$v (t) = 3 t^{2} - 4 t$

2. Time(s) at Rest:
The particle is at rest when $v (t) = 0$ .
$3 t^{2} - 4 t = 0$
$t (3 t - 4) = 0$
The solutions are $t = 0$ (the initial moment) and $t = 4/3 s$ .

3. Position and Distance at $t = 3 s$ :
First, we find the position function, $x (t)$ . The particle starts at the origin ( $x (0) = 0$ ).
$\int_{0}^{x (t)} d x^{'} = \int_{0}^{t} (3 t^{'2} - 4 t^{'}) d t^{'}$
$x (t) - 0 = [t^{'3} - 2 t^{'2}]_{0}^{t}$
$x (t) = t^{3} - 2 t^{2}$
The position at $t = 3 s$ is:
$x (3) = (3)^{3} - 2 (3)^{2} = 27 - 18 = 9 m$
This is the displacement. To find the distance, we must check for any change in direction. We know the velocity is zero at $t = 4/3 s$ .

For $0 < t < 4/3$ , let's check $t = 1$ : $v (1) = 3 (1)^{2} - 4 (1) = - 1 m/s$ . The particle moves in the negative direction.
For $t > 4/3$ , let's check $t = 2$ : $v (2) = 3 (2)^{2} - 4 (2) = 12 - 8 = 4 m/s$ . The particle moves in the positive direction.

The particle turns around at $t = 4/3 s$ . So, we must calculate the path length in two parts.

Position at the turning point ( $t = 4/3 s$ ):
$x (4/3) = (4/3)^{3} - 2 (4/3)^{2} = \frac{64}{27} - 2 (\frac{16}{9}) = \frac{64}{27} - \frac{32}{9} = \frac{64 - 96}{27} = - \frac{32}{27} m$
Distance traveled from $t = 0$ to $t = 4/3 s$ is $∣ x (4/3) - x (0) ∣ = ∣ - \frac{32}{27} - 0∣ = \frac{32}{27} m$ .
Distance traveled from $t = 4/3 s$ to $t = 3 s$ is $∣ x (3) - x (4/3) ∣ = ∣9 - (- \frac{32}{27}) ∣ = ∣9 + \frac{32}{27} ∣ = \frac{243 + 32}{27} = \frac{275}{27} m$ .

Total distance = $\frac{32}{27} + \frac{275}{27} = \frac{307}{27} \approx 11.37 m$ .

Conclusion

This lesson has equipped you with the universal method for solving 1D kinematics problems where acceleration is a function of time. This is a significant step beyond relying on the specialized constant-acceleration equations.

Key Takeaways:

When acceleration is not constant, the standard kinematic equations ( $v = u + a t$ , etc.) are invalid.
The fundamental relationships $a = d v / d t$ and $v = d x / d t$ are always true.
For time-dependent acceleration, $a (t)$ , we find velocity and then position through a two-step process of direct integration with respect to time.
1. $v (t) = v_{0} + \int_{t_{0}}^{t} a (t^{'}) d t^{'}$
2. $x (t) = x_{0} + \int_{t_{0}}^{t} v (t^{'}) d t^{'}$
Always be mindful of initial conditions ( $x_{0}, v_{0}$ ), as they are essential for finding the unique solution.

Preview of the Next Lesson:

Time dependence is just one way acceleration can be non-uniform. What if acceleration depends on how fast an object is moving? In our next lesson, we will tackle problems with velocity-dependent acceleration, a(v). This will require a different integration strategy—the method of separation of variables—and will allow us to analyze realistic physical situations like motion with air resistance.

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