Connecting Position, Velocity, and Acceleration of Functions Using Integrals  AP Calc Study Guide 2024
Introduction
Welcome to the fascinating world of motion and calculus! 🚀 In this guide, we'll explore how integrals are used to connect the dots between position, velocity, and acceleration. Imagine understanding the secrets behind how fast your dream car can go, or why you feel like you're being sucked into your seat when it accelerates! We're making math come alive with some integral insights. 😊📈
The Trio of Motion: Position, Velocity, and Acceleration
Before we dive into the calculus, let's recall some fundamental concepts of motion. If you need a refresher on rectilinear motion (a fancy term for things moving in a straight line), check out our guide from Unit 4 or our AP Physics 1 material. 🚗
Understanding Position, Velocity, and Acceleration
Position (s(t))
In mathematics, position (s(t)) describes the location of an object at a particular time. Think of it as the GPS coordinates of your spacecraft traveling across the galaxy. For example, if you’re parked at the beginning of a race track, that’s your starting position, s(t).
Now, let's visualise this with a caterpillar:
 From t=0 to t=0.5, the caterpillar moves in a positive direction (straight outta its comfort zone!), indicated by the positive slope of the graph.
 At t=0.5, it pauses to catch its breath, so the graph shows a horizontal line.
 From t=1 to t=2, our little friend changes its mind and heads back home, shown by the graph's negative slope.
Understanding and interpreting these graphs is essential to get a good grasp on the relationship among position, velocity, and acceleration. 🚜
Velocity (v(t))
Velocity is the rate of change of position over time, which tells us how fast and which way the object is moving. Mathematically, velocity is the derivative of the position function:
[ v(t) = \frac{d{s(t)}}{dt} = s'(t) ]
It's like the speedometer in your dream car; not only does it tell you how fast you’re going, but also whether you’re cruising forward or reversing back.
Here's a fun note: On a position vs. time graph, the velocity is the slope. Let's imagine a velocity vs. time graph:
 A line with a positive slope means you're accelerating smoothly, like entering a highway.
 A horizontal line indicates constant velocity; you're on cruise control.
 A negative slope? Time to slam the brakes, pal!
Acceleration (a(t))
Acceleration is the change in velocity over time. It’s what you feel when a rollercoaster takes off or hits the brakes. Mathematically, it's the derivative of velocity:
[ a(t) = \frac{d{v(t)}}{dt} = v'(t) ]
And since velocity is the derivative of position, acceleration can also be found as the second derivative of position:
[ a(t) = \frac{d^2s(t)}{dt^2} = s''(t) ]
Visualise this using a velocity vs. time graph. When the segments are straight:
 Positive slope > Positive acceleration (speeding up).
 Horizontal > Zero acceleration (steady speed).
 Negative slope > Negative acceleration (slowing down).
Using Integration to Find Position, Velocity, and Acceleration
Integrals aren't just something you dread for your exams; they help you figure out realworld problems! Let's see how.
Integrating Velocity to Find Position
To locate an object at any moment, integrate its velocity function and add the initial position:
[ s(t) = \int v(t) dt + C ]
Where C is the constant of integration (the initial position). Integrating the velocity gives you the change in position over that interval, a.k.a. displacement.
[ \text{Displacement} = s_f  s_i = \Delta s = \int v(t) dt ]
If you need the total distance travelled (including all zigzags), integrate the absolute value of velocity:
[ \text{Distance travelled} = \int v(t) dt ]
Integrating Acceleration to Find Velocity
To find the velocity, integrate acceleration and add the initial velocity:
[ v(t) = \int a(t) dt + C ]
Where C is the initial velocity.
Practice Problems: Mastering Motion 🎓
Example 1
Suppose an object has a velocity function ( v(t) = 2t + 5 ) and starts at position ( s(0) = 10 ). Find the position function ( s(t) ):

Integrate velocity: [ s(t) = \int (2t + 5) dt + 10 ]

Solution: [ s(t) = \frac{1}{2} t^2 + 5t + 10 ]
Example 2
Given ( v(t) = 3t  2 ) on ([1,4]) with ( s(1) = 5 ), find displacement and distance travelled.

Displacement: [ \text{Displacement} = \int_{1}^{4} (3t  2) dt ] [ = \left[ \frac{3}{2} t^2  2t \right]_{1}^{4} ] [ = \frac{9}{2} ]

Distance Travelled: [ \text{Distance travelled} = \int_{1}^{4} 3t  2 dt ]
Since velocity is always positive here: [ = \int_{1}^{4} (3t  2) dt ] [ = \frac{9}{2} ]
Surprisingly, in this case, the object's path didn't involve extra zigzagging, resulting in equal displacement and distance travelled.
Conclusion
Congratulations! We've navigated through the dynamic world of calculus, connecting position, velocity, and acceleration using integrals. Remember:
 Velocity is the derivate of position.
 Acceleration is the derivative of velocity.
 Displacement is the net change in position.
 Distance travelled includes every twist and turn.
These concepts aren't just for passing calculus—they're useful tools for understanding the physical world around us. So, keep practicing, and soon you’ll be a master of motion. 🚀
Key Terms to Review:
Acceleration: Rate at which velocity changes. Definite Integral: Calculates exact area under a curve over an interval. Derivative: Measures rate of change at a point. Dv/dt: Derivative of velocity (rate of change). Integral: Calculates areas, volumes, and accumulation. Position: Location relative to a reference point. Velocity: Rate of change of position.
Keep practicing, and may your integrals always converge! 🍀