Topic 4.2 Notes – Straight-Line Motion: Connecting Position, Velocity, and Acceleration

Position, Velocity, and Acceleration

We’re modeling motion along a line, so everything depends on time $t$.

• Position: $x(t)$
  Where the particle is at time $t$.
• Velocity: $$v(t) = x'(t)$$
  Instantaneous rate of change of position.
• Acceleration: $$a(t) = v'(t) = x''(t)$$
  Instantaneous rate of change of velocity.

Each derivative tells you how the previous quantity is changing.

Units matter

If:

• $x$ is in meters
• $t$ is in seconds

Then:

• $v$ is meters/second
• $a$ is meters/second$^2$

On FRQs, including units in your final answer can earn a point. Don’t skip them.

Direction, Speed, and What the Signs Mean

Velocity tells you direction

Velocity is signed.

• $v(t) > 0$ → moving right (positive direction)
• $v(t) < 0$ → moving left (negative direction)
• $v(t) = 0$ → at rest (could be turning around)

If a question asks when the particle is moving right, you’re solving $v(t) > 0$.

Speed is different from velocity

Speed is the magnitude of velocity:

$$\text{speed} = |v(t)|$$

Speed is never negative.

If a problem says “how fast is the particle moving,” they want a positive number, even if velocity is negative.

Acceleration and speeding up vs slowing down

Acceleration tells you how velocity is changing, not just whether it’s positive or negative.

Here’s the key idea:

If velocity and acceleration have the same sign, the particle is speeding up.
If they have opposite signs, it’s slowing down.

This shows up constantly on tests. Students often say “acceleration is positive so it’s speeding up.” That’s only true if velocity is also positive.

How to Solve Rectilinear Motion Problems

There are three common setups.

1. You’re given position $x(t)$

Example:
$x(t) = t^3 - 6t^2 + 9t$

• Velocity: $$v(t) = x'(t) = 3t^2 - 12t + 9$$
• Acceleration: $$a(t) = x''(t) = 6t - 12$$

If asked for velocity at $t = 2$, differentiate first, then plug in.

A common mistake is plugging in before differentiating.

2. You’re given velocity $v(t)$

• Acceleration → differentiate
• Speed → take absolute value

If they ask when the particle is at rest, solve $v(t) = 0$.

3. Analyzing motion on an interval

Suppose you need to know when a particle is speeding up.

1. Find $v(t)$.
2. Find $a(t)$.
3. Determine where each is positive or negative.
4. Compare signs on each interval.

A quick sign chart usually keeps this organized, especially on FRQs where clarity matters.

Graph Connections You Should Recognize

Sometimes you won’t get formulas. You’ll get graphs. The three panels below show the same motion represented as position, velocity, and acceleration over time.

Position, velocity, and acceleration for the same particle

From a position graph $x(t)$

• Slope of tangent line = velocity
• Increasing position → $v(t) > 0$
• Concave up → $a(t) > 0$
• Concave down → $a(t) < 0$

Horizontal tangent points often mean $v(t) = 0$.

From a velocity graph $v(t)$

• Slope = acceleration
• Above the axis → moving right
• Below the axis → moving left
• Increasing velocity → positive acceleration

On calculator multiple choice, they love giving a velocity graph and asking about speeding up. You must look at both the sign of $v$ and whether the graph is rising or falling.