Introduction

Welcome! This topic focuses on using derivatives to analyze motion along a straight line. The essential knowledge tells us that the derivative can be used to solve rectilinear motion problems involving position, speed, velocity, and acceleration. Understanding these relationships allows us to analyze the motion of objects moving in one dimension.

Position, Velocity, and Acceleration Relationships

The Fundamental Connection

For an object moving along a straight line, if position is given by $s(t)$:

Position: $s(t)$ (measured from a reference point)

Velocity: $v(t)=s^{\prime }(t)=\frac{ds}{dt}$ (rate of change of position)

Acceleration: $a(t)=v^{\prime }(t)=s^{\prime \prime }(t)=\frac{d^{2}s}{d{t}^2}$ (rate of change of velocity)

Example 1:

If $s(t)=2t^3-9t^2+12t-3$ represents position in meters after $t$ seconds:

$v(t)=s^{\prime }(t)=6t^2-18t+12$ meters per second

$a(t)=v^{\prime }(t)=12t-18$ meters per second squared

Understanding the Derivative Chain

Position → Velocity → Acceleration

Each step involves taking the derivative with respect to time:

$s(t)\stackrel{\text{derivative}}{\to }v(t)\stackrel{\text{derivative}}{\to }a(t)$

Example 2:

For $s(t)=t^4-4t^3+6t^2$:

$v(t)=4t^3-12t^2+12t$

$a(t)=12t^2-24t+12$

At $t=2$ seconds:

Velocity vs. Speed

Key Distinction

Velocity: $v(t)=s^{\prime }(t)$ (can be positive, negative, or zero)

Speed: $\text{speed}(t)=|v(t)|=|s^{\prime }(t)|$ (always non-negative)

Example 3:

If $v(t)=3t^2-12t+9$:

At $t=1$: $v(1)=3-12+9=0$ (velocity is zero, speed is 0)

At $t=2$: $v(2)=12-24+9=-3$ (velocity is -3, speed is 3)

At $t=4$: $v(4)=48-48+9=9$ (velocity is 9, speed is 9)

Analyzing Motion Characteristics

Determining Direction of Motion

Moving Right/Forward: $v(t)>0$

Moving Left/Backward: $v(t)<0$

At Rest: $v(t)=0$

Example 4:

For $s(t)=t^3-6t^2+9t$:

$v(t)=3t^2-12t+9=3(t^2-4t+3)=3(t-1)(t-3)$

$v(t)=0$ when $t=1$ or $t=3$

For $0<t<1$: $v(t)>0$ (moving right)

For $1<t<3$: $v(t)<0$ (moving left)

For $t>3$: $v(t)>0$ (moving right)

Analyzing Acceleration

Speeding Up: velocity and acceleration have the same sign

Slowing Down: velocity and acceleration have opposite signs

Example 5:

Using $v(t)=3t^2-12t+9$ and $a(t)=6t-12$:

At $t=0.5$: $v(0.5)=3.75>0$ and $a(0.5)=-9<0$ (slowing down)

At $t=2$: $v(2)=-3<0$ and $a(2)=0$ (neither speeding up nor slowing down)

At $t=4$: $v(4)=9>0$ and $a(4)=12>0$ (speeding up)

Finding Critical Points in Motion

When Objects Are at Rest

Set $v(t)=s^{\prime }(t)=0$ and solve for $t$.

These are times when the object momentarily stops moving.

Example 6:

For $s(t)=2t^3-15t^2+36t-20$:

$v(t)=6t^2-30t+36=6(t^2-5t+6)=6(t-2)(t-3)$

The object is at rest at $t=2$ seconds and $t=3$ seconds.

Maximum and Minimum Positions

When $v(t)=0$, the object might be at a maximum or minimum position.

Use the second derivative or analyze the sign of $v(t)$ around these points.

Example 7:

From Example 6, at critical points:

At $t=2$: $a(2)=6(2)-30=-18<0$ (maximum position)

At $t=3$: $a(3)=6(3)-30=-12<0$ (also maximum position)

Check positions: $s(2)=16-60+72-20=8$ and $s(3)=54-135+108-20=7$

So the object reaches its highest position at $t=2$ seconds.

Displacement vs. Distance Traveled

Key Concepts

Displacement: $s(t_2)-s(t_1)$ (net change in position)

Distance Traveled: Total length of path traveled

Example 8:

If $s(t)=t^3-6t^2+9t$ from $t=0$ to $t=4$:

Displacement: $s(4)-s(0)=(64-96+36)-0=4$ meters

Distance Traveled: Object changes direction at $t=1$ and $t=3$

$s(0)=0$, $s(1)=4$, $s(3)=0$, $s(4)=4$

Distance = $|s(1)-s(0)|+|s(3)-s(1)|+|s(4)-s(3)|$

Distance = $|4-0|+|0-4|+|4-0|=4+4+4=12$ meters

Problem-Solving Strategies

Systematic Approach

Common Question Types

"When is the object at rest?" Solve $v(t)=0$

"When is the object moving right/left?" Determine where $v(t)>0$ or $v(t)<0$

"When is the object speeding up/slowing down?" Compare signs of $v(t)$ and $a(t)$

"What is the maximum speed?" Find critical points of $|v(t)|$

Practice Section