Skip to content# Variable Acceleration: An Easy and Enjoyable Journey to Mastery

## Variable Acceleration

## Velocity

## Conclusion

## Practice Questions

### Application 1: Free-Falling Object

### Accelerating Car with Variable Acceleration

**s = ****t3 ****/ 3**

** ****At t = 5 sec**** ****s = 5****3 ****/ 3**

** ****s = 41.67 m**

Today, we’re diving into an exciting topic in kinematics: Variable Acceleration

We’ve already covered the applications of SUVAT equations with constant acceleration.

If you haven’t watched that video, you can click on this link. SUVAT Equations

If you’re seeking a more personalized and tailored approach to mastering Variable Acceleration and tackling SUVAT Equations with confidence, look no further. My one-to-one tutoring is designed to cater to your unique learning style, pace, and academic goals. Together, we’ll conquer the challenges of variable acceleration and excel in A-level kinematics.

Whether you’re preparing for your GCSE or A-Level exams, you can click on this link: Physics Private Tutoring

Delve deeper into the world of SUVAT Equations by clicking on the link for an in-depth exploration and a comprehensive understanding: SUVAT Equations

If you want to watch a Video tutorial, please watch this video:

Contents

Chapter 1

Now, let’s take things to the next level! Real-world scenarios often involve variable acceleration, where the velocity changes at different rates throughout motion due to factors like air resistance and friction.

But fear not, as we have a few formulas with differentiation and integration that can help us solve all questions related to variable acceleration.

Let’s start with the basic definition of acceleration. We know acceleration is the change in velocity divided by time taken.

The same formula, when written in the form of a derivative, will give us “a = dv/dt.” That means by differentiating the velocity function, we can obtain acceleration.

Also, if we multiply “dt” on both sides, we get the integral of “a times dt = dv” Therefore, by integrating acceleration with respect to time, we get the change in velocity.

Chapter 2

Next, lets move to the basic definition of velocity.

Velocity is the change in displacement divided by time taken, or “v = ds/dt.”

So, differentiating the displacement function gives us velocity. Similarly, if we multiply “dt” on both sides, we get the integral of velocity with time, which gives us the change in displacement or displacement itself.

Chapter 3

In conclusion, differentiating velocity gives us acceleration, while integrating velocity gives us displacement. And if we integrate acceleration, we get back velocity. Likewise, differentiating displacement will also give us back velocity.

Now, we must know that we can only follow these formulas if we are given acceleration, velocity, and displacement as functions of time.

However, in some questions, we are given acceleration as a function of displacement.

In all such situations, we can use the formula: “integral of v dv = integral of a ds.”

Chapter 4

Now, lets Practice a few Questions:

**Solved Example : We are given a free-falling object with variable acceleration. The object starts from rest and its velocity changes with time according to the equation v = 9.8t – 4.9t ^{2}. **

**Our task is to find the displacement of the object after 3 seconds.**

**Solution: **The object

starts from rest, and its velocity changes with time according to the equation

v = 9.8t – 4.9t^{2}. We need to find the displacement of the object

after 3 seconds.

*Step 1: Integration of Velocity Equation*

To find the displacement (s), we need to integrate the velocity equation with respect to time (t).

The velocity equation is **v = 9.8t – 4.9t ^{2}**

The integral of v dt will yield the displacement function s(t).

**Step 2: Integrating the Equation**

Integrate the equation **v = 9.8t – 4.9t ^{2}** with respect to time (t).

The result will be the displacement function s(t) in terms of t and the constant of integration, C.

**Step 3: Finding the Constant of Integration**

After integrating, we get

**s = 9.8t ^{2}/2 – 4.9t^{3}/3 + C,**

where C is the constant of integration.

**Step 4: Calculating Displacement at t = 3 seconds**

- Plug in the initial limit of time (0 seconds) and the final limit (3 seconds) into the displacement function s(t).
- Evaluate s at t = 3 seconds to find the displacement of the object at that time.

**Step 5: Solving for the Displacement**

Evaluate the expression

**s = (4.9(3) ^{2} – 4.9(3)^{3}/3 + C) – (4.9(0)^{2} – 4.9(0)^{3}/3 + C).**

Answers: we get s = 44.1 meters as the displacement of the object after 3 seconds.

**Solved Example: Suppose a car starts from rest and accelerates with an acceleration given by a = 2t, where a is the acceleration in m/s ^{2} and t is the time in seconds. **

**Find the distance traveled by the car after 5 seconds. **

**Solution**

*Step 1 & 2 : Sketch a Diagram & Identify Knowns and Unknowns*

**Step 3: Select the Appropriate Equation(s)**

We need an equation that relates the knowns (t, a) to the unknown (s). In this case, we can use the equations:

**dv = ∫ a dt**

**ds = ∫ v dt**

**Step 4: Solve the Equation(s)**

Plug in the known values into the equation:

**dv = ∫ (2t) dt**

Integrating the equation, we get:

**v = t ^{2} + C **

Now since the initial velocity is Zero, so at t = 0, v = 0

**0 = 0****2 + C**

**we get C = 0**

**v = t2**

**ds = ∫ v dt**

**ds = ∫ ****t2**** dt**

**s = ****t3 ****/ 3 + C**

Again here, since the initial velocity is Zero, so at t = 0, s = 0

0 = 0**3 + C**

**again, we get C = 0**

These formulas, involving differentiation and integration, provide the necessary mathematical tools to handle variable acceleration problems and analyze the complex motion of objects experiencing changing velocities due to external forces and factors.

Let’s try to use these formulas in the coming chapters.

But if you want a calculator to simply answer all your questions you can simply use this link: Kinematics Questions Calculator