Unit 4 · Topic 4.2 Beta

Straight-Line Motion: Position, Velocity, Acceleration

For a particle on a line, v(t) = x′(t) and a(t) = v′(t) = x″(t) .

Practice 1: Implementing Mathematical ProcessesPractice 3: Justification

Part 1 · Hook

Why this matters

A runner jogs east, slows, turns around and sprints back. Her position, velocity and acceleration are a function and its first two derivatives. The exam asks about this kind of motion every year: which way is the particle going, is it speeding up, when does it turn around. Each answer comes from a sign.

Part 2 · Before you start

What this builds on

Part 3 · Prerequisite check

Quick check before you start

1. If $f(x) = x^3 - 2x$ , what is $f''(x)$ ?

$6x$
$3x^2 - 2$
$6$
$6x - 2$

Show the answer

$f'(x) = 3x^2 - 2$ , then $f''(x) = 6x$ .

Correct: $6x$ :
$3x^2 - 2$ :
$6$ :
$6x - 2$ :

2. $D'(4) = -3$ for a depth $D$ in meters and $t$ in hours means

at 4 hours the depth is decreasing at 3 meters per hour
the depth is 3 meters at 4 hours
the depth fell 3 meters in 4 hours
the depth is increasing at 3 meters per hour

Show the answer

A negative derivative is a decreasing quantity, at that rate.

Correct: at 4 hours the depth is decreasing at 3 meters per hour:
the depth is 3 meters at 4 hours:
the depth fell 3 meters in 4 hours:
the depth is increasing at 3 meters per hour:

Part 4 · See it

See it first

For

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

: velocity and acceleration, with shading where they share a sign (speeding up), and the path the particle traces on the line. LevlPrep original diagram.

Part 5 · Step by step

How it works, step by step

Position $x(t)$ changes with timeits rate of change is the velocity, $v(t) = x'(t)$
Velocity changes with timeits rate of change is the acceleration, $a(t) = v'(t) = x''(t)$
The sign of $v$ gives the direction: right (or up) when $v > 0$ , left when $v < 0$ ; a direction change needs $v$ to change sign
Compare the signs of $v$ and $a$ same sign: speeding up; opposite signs: slowing down

Part 6 · Key ideas

Key ideas

Velocity has a sign; speed is $|v(t)|$ and is never negative.
The particle is at rest when $v(t) = 0$ , and changes direction only where $v$ changes sign.
Speeding up when $v$ and $a$ have the same sign. State both signs in a justification.
Units: position in meters gives velocity in meters per second and acceleration in meters per second per second.

Part 7 · Misconception

A common mistake

The wrong idea: Negative acceleration means the particle is slowing down.

What actually happens: It depends on the velocity too. With $v < 0$ and $a < 0$ , the particle moves left faster and faster: it is speeding up. Slowing down needs $v$ and $a$ with opposite signs.

Part 8 · Check yourself

Check yourself

Exam-style questions. Anything you miss goes into your review queue.

Graph

Position of a particle: $x(t) = t^3 - 6t^2 + 9t$

1. A particle moves along the $x$ -axis with position $x(t) = t^3 - 6t^2 + 9t$ meters at time $t$ seconds, as graphed. Find its velocity at $t = 2$ , with units.

No calculator. Type the exact value (a fraction, π, e or a root is fine), then its unit.

Show the answer

$v(t)$ $= x'(t)$ $= 3t^2 - 12t + 9$ , so $v(2) = 12 - 24 + 9 = -3$ meters per second.

Answer: -3 meters per second

2. Find the acceleration of the particle at $t = 3$ , with units.

No calculator. Type the exact value (a fraction, π, e or a root is fine), then its unit.

Show the answer

$a(t) = v'(t) = 6t - 12$ , so $a(3) = 6$ meters per second per second.

Answer: 6 meters per second per second

3. At $t = 2.5$ seconds, is the particle speeding up or slowing down?

Slowing down, because $v(2.5) < 0$ and $a(2.5) > 0$
Speeding up, because $a(2.5) > 0$
Speeding up, because $v(2.5)$ and $a(2.5)$ are both negative
Slowing down, because $v(2.5) < 0$

Show the answer

$v(2.5)$ $= 3(2.5)^2 - 12(2.5)$ $+ 9$ $= -2.25$ and $a(2.5) = 3$ . They have opposite signs, so the speed is decreasing: slowing down.

Correct: Slowing down, because $v(2.5) < 0$ and $a(2.5) > 0$ : $v(2.5) = -2.25$ and $a(2.5) = 3$ : opposite signs.
Speeding up, because $a(2.5) > 0$ : A positive acceleration speeds the particle up only if it is moving in the positive direction.
Speeding up, because $v(2.5)$ and $a(2.5)$ are both negative: $a(2.5) = 6(2.5) - 12 = 3$ is positive.
Slowing down, because $v(2.5) < 0$ : The sign of $v$ alone gives the direction of motion, not whether speed grows.

4. On which interval is the particle moving to the left?

$2 < t < 4$
$1 < t < 3$
$0 < t < 2$
$3 < t < 4$

Show the answer

Moving left means $v(t) < 0$ . Since $v(t) = 3(t - 1)(t - 3)$ , that is $1 < t < 3$ , where the graph of $x$ falls.

$2 < t < 4$ : $v$ is positive for $t > 3$ ; the acceleration is positive on $2 < t < 4$ .
Correct: $1 < t < 3$ : $v(t)$ $= 3(t - 1)(t - 3) < 0$ there, and the graph of $x$ falls.
$0 < t < 2$ : The particle moves right until $t = 1$ .
$3 < t < 4$ : $v > 0$ there: the position increases.

5. At which times does the particle change direction, and why?

$t = 2$ , because $a(2) = 0$
$t = 0$ and $t = 3$ , because $x = 0$ there
$t = 1$ and $t = 3$ , because $v$ changes sign there
$t = 1$ and $t = 3$ , because $x(t) = 4$ or $x(t) = 0$ there

Show the answer

The direction changes where $v$ changes sign. $v(t) = 3(t - 1)(t - 3)$ changes sign at $t = 1$ (positive to negative) and $t = 3$ (negative to positive).

$t = 2$ , because $a(2) = 0$ : A zero of $a$ is where the velocity stops falling; the particle still moves left.
$t = 0$ and $t = 3$ , because $x = 0$ there: Being at the origin says nothing about the direction of motion.
Correct: $t = 1$ and $t = 3$ , because $v$ changes sign there: $v$ goes from positive to negative at 1 and from negative to positive at 3.
$t = 1$ and $t = 3$ , because $x(t) = 4$ or $x(t) = 0$ there: The right times, but the reason must be about $v$ : it changes sign.

6. At $t = 3$ , a particle has $v(3) = -4$ and $a(3) = -2$ . The particle is

slowing down, because $a(3)$ is negative
slowing down, because $v(3)$ is negative
at rest, because $v(3)$ and $a(3)$ are both negative
speeding up, because $v(3)$ and $a(3)$ share a sign

Show the answer

Speeding up when $v$ and $a$ share a sign; slowing down when they differ. Here both are negative: speeding up while moving left.

slowing down, because $a(3)$ is negative: A negative acceleration speeds up a particle moving in the negative direction.
slowing down, because $v(3)$ is negative: The sign of $v$ alone gives direction only.
at rest, because $v(3)$ and $a(3)$ are both negative: At rest means $v = 0$ ; here the particle moves left.
Correct: speeding up, because $v(3)$ and $a(3)$ share a sign: Both negative: the acceleration points the way the particle moves.

Part 9 · Summary

Summary

For a particle on a line,

v(t) = x'(t)

and

a(t) = v'(t) = x''(t)

. The sign of

v

gives the direction, and the direction changes only where

v

changes sign. Speed is

|v|

; it increases when

v

and

a

have the same sign and decreases when their signs differ.

Part 10 · Up next

What comes next

Part 11 · Connections

Why this matters

What this builds on

Quick check before you start

See it first

How it works, step by step

Key ideas

A common mistake

Check yourself

Position of a particle: x(t)=t3−6t2+9tx(t) = t^3 - 6t^2 + 9t

Summary

What comes next

Connections

Position of a particle: $x(t) = t^3 - 6t^2 + 9t$