Rectilinear Motion
Textbook Reference
| Primary source | OpenStax Calculus Volume 1, Section 4.10: “Antiderivatives” |
| Direct link | https://openstax.org/books/calculus-volume-1/pages/4-10-antiderivatives |
| Textbook used in class | Stewart, Calculus, Section 3.9: “Antiderivatives” (Examples 5-7) |
Opening Scenario
If you are watching a car on a road and you know exactly how fast it is moving at every moment -- but not where it started -- can you determine where it is at any later time?
Not quite: without the starting position, you cannot determine the current position. But with the starting position, the velocity record (together with integration) gives the full position history.
This is the central idea of motion problems using antiderivatives: velocity is the derivative of position, and acceleration is the derivative of velocity. Antidifferentiation reverses these relationships.
Quick Reference
Kinematic relationships (on a straight line):
- $v(t) = s'(t)$: velocity is the derivative of position.
- $a(t) = v'(t) = s''(t)$: acceleration is the derivative of velocity.
Antidifferentiation reverses them:
- $s(t) = \displaystyle\int v(t)\,dt + C_1$ (position from velocity; $C_1$ = initial position).
- $v(t) = \displaystyle\int a(t)\,dt + C_2$ (velocity from acceleration; $C_2$ = initial velocity).
Notation: $s$ = position (displacement from a reference point), $v$ = velocity, $a$ = acceleration. All are functions of time $t$.
Key Concepts
1. Position from Velocity
Example 1. A particle moves along a line with velocity $v(t) = t^2 - 2t$. At time $t = 0$ the particle is at position $s = 3$. Find $s(t)$.
Step 1: Antidifferentiate $v$. $s(t) = \displaystyle\int (t^2 - 2t)\,dt = \frac{t^3}{3} - t^2 + C$.
Step 2: Apply initial condition $s(0) = 3$. $\frac{0}{3} - 0 + C = 3 \Rightarrow C = 3$.
Particular solution: $s(t) = \dfrac{t^3}{3} - t^2 + 3$.
Boxed answer: $s(t) = \dfrac{t^3}{3} - t^2 + 3$.
Interpretation. At $t = 0$: $s(0) = 3$ (starting position). At $t = 3$: $s(3) = 9 - 9 + 3 = 3$ (back to start). The particle left $s = 3$, traveled, and returned.
2. Velocity from Acceleration
Example 2. The acceleration of a particle is $a(t) = 6t - 2$. At $t = 0$, the velocity is $v(0) = 4$ and the position is $s(0) = 1$. Find $s(t)$.
Step 1: Velocity from acceleration. $v(t) = \displaystyle\int (6t - 2)\,dt = 3t^2 - 2t + C_1$.
$v(0) = C_1 = 4$. So $v(t) = 3t^2 - 2t + 4$.
Step 2: Position from velocity. $s(t) = \displaystyle\int (3t^2 - 2t + 4)\,dt = t^3 - t^2 + 4t + C_2$.
$s(0) = C_2 = 1$. So $s(t) = t^3 - t^2 + 4t + 1$.
Boxed answer: $s(t) = t^3 - t^2 + 4t + 1$.
3. Free Fall Under Gravity
Near the surface of the Earth, objects fall with constant downward acceleration $g \approx 9.8$ m/s$^2$ (taking downward as negative: $a = -9.8$ m/s$^2$).
Example 3. An object is dropped (initial velocity $0$) from a height of $500$ m. Find its height $h(t)$ above the ground and the time when it hits the ground. (This is Stewart 3.9, Example 7.)
Setup. $h(t)$ = height above ground. $a(t) = -9.8$ m/s$^2$. Initial conditions: $h(0) = 500$, $v(0) = h'(0) = 0$.
Step 1: Velocity. $v(t) = \displaystyle\int (-9.8)\,dt = -9.8t + C_1$.
$v(0) = C_1 = 0$. So $v(t) = -9.8t$.
Step 2: Position. $h(t) = \displaystyle\int (-9.8t)\,dt = -4.9t^2 + C_2$.
$h(0) = C_2 = 500$. So $h(t) = -4.9t^2 + 500$.
When does it hit the ground? $h(t) = 0$: $$-4.9t^2 + 500 = 0 \Rightarrow t^2 = \frac{500}{4.9} \approx 102.0 \Rightarrow t \approx 10.1 \text{ seconds}.$$
Boxed answer: $h(t) = -4.9t^2 + 500$; the object hits the ground after approximately $10.1$ seconds.
Predict first. The object falls $500$ m. A rough estimate: at free fall, distance $= \frac{1}{2}gt^2 = 4.9t^2$. Setting $4.9t^2 = 500$: $t \approx 10.1$. The calculation confirms the estimate.
4. Understanding Direction from Velocity
The sign of $v(t)$ tells you the direction of motion:
- $v(t) > 0$: moving in the positive direction (to the right, or upward, depending on context).
- $v(t) < 0$: moving in the negative direction.
- $v(t) = 0$: momentarily at rest (might be changing direction).
Example 4. For the particle in Example 1 ($v(t) = t^2 - 2t = t(t-2)$), when is the particle moving in the negative direction?
$v(t) < 0$ when $t(t-2) < 0$, i.e., $0 < t < 2$.
The particle moves in the negative direction for $0 < t < 2$.
“If the position function is decreasing, the particle is moving backward.” That is actually correct -- decreasing $s$ means $s'(t) = v(t) < 0$, so the particle is moving in the negative direction. The issue arises when students confuse position with distance traveled. The particle can return to its starting position (zero displacement) while having traveled a nonzero distance.
Common Errors Summary
| Error | Example | Correction |
|---|---|---|
| Confusing $s$, $v$, and $a$ | Antidifferentiating position to get velocity | Velocity is the derivative of position; antidifferentiate velocity to get position |
| Applying the initial condition to the wrong function | Using $s(0) = 3$ to find $C$ in the velocity equation | Each constant is found from the condition on its own function: $v(0)$ pins $C$ in $v$, $s(0)$ pins $C$ in $s$ |
| Taking $g = 9.8$ without a sign | Writing $a(t) = 9.8$ for a falling object when $h$ is positive upward | If height is measured upward, gravity is negative: $a(t) = -9.8$ m/s$^2$ |
Leveled Practice
Level 1 -- Direct Application
Problem 1. A particle has velocity $v(t) = 3t^2 - 6t$ and $s(0) = 4$. Find $s(t)$.
Show answer
$s(t) = \displaystyle\int (3t^2 - 6t)\,dt = t^3 - 3t^2 + C$.
$s(0) = C = 4$.
Boxed answer: $s(t) = t^3 - 3t^2 + 4$.
Problem 2. A ball is thrown upward with initial velocity $20$ m/s from a height of $2$ m. (Take upward as positive, $a = -9.8$ m/s$^2$.) Find the maximum height reached.
Show answer
$v(t) = -9.8t + 20$ (from $v(0) = 20$).
Maximum height when $v(t) = 0$: $t = 20/9.8 \approx 2.04$ s.
$h(t) = -4.9t^2 + 20t + 2$ (from $h(0) = 2$).
$h(20/9.8) = -4.9(400/96.04) + 20(20/9.8) + 2 \approx -20.41 + 40.82 + 2 = 22.41$ m.
More exactly: $h_{\max} = -4.9(20/9.8)^2 + 20(20/9.8) + 2 = 400/9.8 \cdot (-4.9/9.8 + 1) + 2$.
Let $t^* = 20/9.8$: $h(t^*) = 20t^* - 4.9(t^*)^2 + 2 = 20 \cdot \frac{20}{9.8} - 4.9\cdot \frac{400}{96.04} + 2 = \frac{400}{9.8} - \frac{1960}{96.04} + 2$.
Since $96.04 \approx 9.8^2 = 96.04$: $\frac{1960}{96.04} \approx \frac{400}{19.6} \approx 20.41$.
$h(t^*) \approx 40.82 - 20.41 + 2 = 22.41$ m.
Boxed answer: Maximum height $\approx 22.4$ m (or exactly $2 + \frac{400}{2 \cdot 9.8} = 2 + \frac{200}{9.8} \approx 22.4$ m).
Level 2 -- Multiple Steps
Problem 3. A car decelerates uniformly from $30$ m/s to $0$ in $6$ seconds. How far does it travel during braking?
Show answer
Constant deceleration: $v(t) = 30 + at$. At $t = 6$: $30 + 6a = 0 \Rightarrow a = -5$ m/s$^2$.
$v(t) = 30 - 5t$.
$s(t) = 30t - 2.5t^2 + C$. With $s(0) = 0$ (take start of braking as origin): $C = 0$.
Distance traveled during braking = $s(6) = 180 - 90 = 90$ m.
Boxed answer: The car travels $90$ m during braking.
Mastery Checklist
Mental Model
Motion problems are layered: acceleration is the rate of change of velocity, and velocity is the rate of change of position. Moving down the layers (from $a$ to $v$ to $s$) requires integration. Moving up the layers (from $s$ to $v$ to $a$) requires differentiation.
Each integration introduces one free constant -- one unknown from the information you do not have yet. Each initial condition eliminates one unknown. Two integrations, two conditions: the motion is then completely determined.
The physical interpretation of the constants: the constant from the first integration is the initial velocity; the constant from the second integration is the initial position. Physically, these are the two things you must know to predict future motion from a known force.
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