Position, Velocity &
Acceleration

Master the calculus of straight-line motion: how derivatives and integrals connect where an object is, how fast it moves, and how its speed changes.

AP Calculus AB

What Is a Position Function?

A position function $ s(t) $ tells you where a particle is on a number line at any time $ t $. The input is time; the output is location.

Velocity as the Derivative

Velocity measures the rate of change of position with respect to time. It is the first derivative of the position function:

$ v(t) > 0 $

Particle moves in the positive direction (right)

$ v(t) < 0 $

Particle moves in the negative direction (left)

$ v(t) = 0 $

Particle is momentarily at rest (possible direction change)

Acceleration as the Second Derivative

Acceleration measures the rate of change of velocity. It is the second derivative of position, or the first derivative of velocity:

Acceleration tells you whether the particle is speeding up or slowing down: when $ v(t) $ and $ a(t) $ share the same sign, speed increases; when they have opposite signs, speed decreases.

The Derivative Chain

Going Backwards with Integrals

Integration reverses differentiation. If you know acceleration, you can recover velocity; if you know velocity, you can recover position:

Speed vs. Velocity

Students often confuse these two. Velocity is a signed quantity (direction matters). Speed is always non-negative:

Velocity

$ v(t) = s'(t) $
Can be positive, negative, or zero. Tells direction.

Speed

$ |v(t)| = |s'(t)| $
Always $ \ge 0 $. Tells magnitude only.

When Does a Particle Speed Up?

Scenario	$ v(t) $	$ a(t) $	Speed
Moving right, accelerating right	+	+	Increasing
Moving left, accelerating left	−	−	Increasing
Moving right, accelerating left	+	−	Decreasing
Moving left, accelerating right	−	+	Decreasing

Displacement vs. Total Distance

Displacement

$$ \Delta s = \int_a^b v(t)\,dt $$ Net change in position. Can be negative.

Total Distance

$$ D = \int_a^b |v(t)|\,dt $$ Sum of all movement. Always $ \ge 0 $.

One cart, three meanings: position, displacement, and distance

A blue dynamics cart rests on a straight laboratory rail with a photogate and end buffers; an observer stands behind the setup.

A cart can reverse direction while total distance continues to increase. Signed position changes give displacement; their absolute values give distance.

Position can return while distance grows

A cart can reverse direction while total distance continues to increase. Signed position changes give displacement; their absolute values give distance.

How to Solve Motion Problems

1

Identify what you're given. Is it $ s(t) $, $ v(t) $, or $ a(t) $?

2

Differentiate or integrate to reach the quantity the problem asks for.

3

Use initial conditions to solve for $ +C $ after integration.

4

Interpret the sign. Positive/negative velocity = direction. Compare signs for speeding up/slowing down.

5

For total distance, find where $ v(t) = 0 $, split the integral, and use absolute values.

Example 1: Finding Velocity

A particle moves along a line with position $ s(t) = t^3 - 6t^2 + 9t + 2 $. Find the velocity at $ t = 1 $ and determine the direction of motion.

Example 2: Graphing the Motion

Using the same $ s(t) = t^3 - 6t^2 + 9t + 2 $, here are $ s(t) $, $ v(t) $, and $ a(t) $:

Example 3: From Acceleration to Position

A particle has acceleration $ a(t) = 6t - 4 $, with $ v(0) = 5 $ and $ s(0) = 1 $. Find $ s(t) $.

Example 4: Total Distance Traveled

Find the total distance traveled by the particle with $ v(t) = t^2 - 4 $ on $ [0, 3] $.

Common Mistakes

✕ Confusing Speed and Velocity

Speed is $ |v(t)| $, always non-negative. Don't say "speed is negative."

✕ Forgetting $+C$

When integrating, always add the constant and use initial conditions to find it.

✕ Using Displacement for Distance

$ \int_a^b v(t)\,dt $ gives displacement, not total distance. Use $ \int_a^b |v(t)|\,dt $ for distance.

Summary of Relationships

Quantity	Symbol	Derivative	Integral
Position	$ s(t) $	$ s'(t) = v(t) $	---
Velocity	$ v(t) $	$ v'(t) = a(t) $	$ \int v\,dt = s(t) + C $
Acceleration	$ a(t) $	---	$ \int a\,dt = v(t) + C $
Displacement	$ \Delta s $	$ \int_a^b v(t)\,dt = s(b) - s(a) $
Total Distance	$ D $	$ \int_a^b \|v(t)\|\,dt $

Question 1 of 5

A particle has position $ s(t) = 2t^3 - 9t^2 + 12t $. What is $ v(t) $?

Velocity is the first derivative of position: $ v(t) = s'(t) = 6t^2 - 18t + 12 $. Option (c) is the second derivative (acceleration), and (d) is the antiderivative.

Question 2 of 5

At time $ t = 2 $, a particle has $ v(2) = -3 $ and $ a(2) = 5 $. Is the particle speeding up or slowing down?

The particle is slowing down. Velocity is negative and acceleration is positive: they have opposite signs. The magnitudes of $ v $ and $ a $ do not determine speeding up/slowing down: only their signs do.

Question 3 of 5

Given $ a(t) = 10 $ and $ v(0) = -3 $, find $ v(t) $.

Integrate acceleration: $ v(t) = \int 10\,dt = 10t + C $. Use $ v(0) = -3 $: $ C = -3 $. So $ v(t) = 10t - 3 $. Option (d) mistakenly integrates twice.

Question 4 of 5

A particle has velocity $ v(t) = t - 2 $ on $ [0, 4] $. What is the total distance traveled?

$ v(t) = 0 $ at $ t = 2 $. On $ [0,2] $, $ v < 0 $: distance = $ \int_0^2 (2-t)\,dt = 2 $. On $ [2,4] $, $ v > 0 $: distance = $ \int_2^4 (t-2)\,dt = 2 $. Total = $ 2 + 2 = 4 $. The displacement $ \int_0^4 (t-2)\,dt = 0 $, which is choice (a): the displacement trap.

Question 5 of 5

If $ s(t) = \sin(t) $, what is the acceleration $ a(t) $?

First derivative: $ v(t) = \cos(t) $. Second derivative: $ a(t) = -\sin(t) $. Option (a) is velocity, not acceleration. Remember: acceleration is the second derivative of position.

Quantity	Symbol	Derivative	Integral
Position	\( s(t) \)	\( s'(t) = v(t) \)	---
Velocity	\( v(t) \)	\( v'(t) = a(t) \)	\( \int v\,dt = s(t) + C \)
Acceleration	\( a(t) \)	---	\( \int a\,dt = v(t) + C \)
Displacement	\( \Delta s \)	\( \int_a^b v(t)\,dt = s(b) - s(a) \)
Total Distance	\( D \)	\( \int_a^b \|v(t)\|\,dt \)

Position, Velocity &Acceleration

What Is a Position Function?

Velocity as the Derivative

Acceleration as the Second Derivative

The Derivative Chain

Going Backwards with Integrals

Speed vs. Velocity

When Does a Particle Speed Up?

Displacement vs. Total Distance

One cart, three meanings: position, displacement, and distance

Position can return while distance grows

How to Solve Motion Problems

Example 1: Finding Velocity

Example 2: Graphing the Motion

Example 3: From Acceleration to Position

Example 4: Total Distance Traveled

Common Mistakes

Summary of Relationships

Question 1 of 5

Question 2 of 5

Question 3 of 5

Question 4 of 5

Question 5 of 5

Position, Velocity &
Acceleration