🎯⭐ INTERACTIVE LESSON

Motion with Variable Acceleration

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Motion with Variable Acceleration - Complete Interactive Lesson

Part 1: Acceleration as a Function of Time

Variable Acceleration — a(t) Functions

Part 1 of 7

In AP Physics C, acceleration is often not constant. When $a = a(t)$ is a function of time, we must use calculus — the kinematic equations for constant $a$ no longer apply.

When Acceleration Depends on Time

Examples of time-dependent acceleration:

Physical Situation	$a(t)$
Rocket with linearly increasing thrust	$a_0 + bt$
Oscillating force	$A \cos (ω t)$

The Calculus Approach

Since $a(t) = dv/dt$ , we integrate to find velocity:

$v(t) = v_0 + \int_0^t a(t')\,dt'$

Then integrate velocity to find position:

$x(t) = x_0 + \int_0^t v(t')\,dt'$

Polynomial Acceleration

The most common case on AP exams is polynomial $a(t)$ .

Example: Quadratic Acceleration

$a(t) = 2 - 3t + t^2$ , with , .

Sinusoidal Acceleration

When $a(t) = a_0\sin(\omega t)$ (or cosine), the motion is oscillatory.

Worked Example

$a(t) = 10\sin(2t)$ m/s², , .

Exponential Acceleration

Exponentially decaying acceleration models many real systems (e.g., drag-limited motion, RC circuits).

Example: Decaying Thrust

A rocket has $a(t) = 20e^{-t/5}$ m/s², $v(0) = 0$ .

Part 2: v(t) from Integration

Variable Acceleration — v(t) from Integration

Part 2 of 7

This section focuses on the technique of integrating $a(t)$ to find $v(t)$ , with careful attention to initial conditions and physical interpretation.

The Fundamental Formula

$v(t) = v(t_0) + \int_{t_0}^t a(t')\,dt'$

Part 3: x(t) from Double Integration

Variable Acceleration — x(t) from Double Integration

Part 3 of 7

When given $a(t)$ , finding $x(t)$ requires two successive integrations, each with its own constant determined by initial conditions.

The Two-Step Process

$a (t) \overset{\int}{\to} v (t) = v_{0} + \int_{0}^{t} a (t^{'}) d t^{'} \overset{}{}$

Part 4: a(v) & Separation of Variables

Variable Acceleration — a(v) and Separation of Variables

Part 4 of 7

When acceleration depends on velocity — $a = a(v)$ — we can no longer simply integrate with respect to time. Instead, we use separation of variables.

Common Physical Examples

Situation	$a(v)$
Linear drag (low speed)	$a$

Part 5: a(x) & Energy Methods

Variable Acceleration — a(x) and the Energy Method

Part 5 of 7

When acceleration depends on position — $a = a(x)$ — we use the chain rule identity:

$a = \frac{dv}{dt} = \frac{dv}{dx} \cdot \frac{dx}{dt} = v\frac{dv}{dx}$

Part 6: Problem-Solving Workshop

Variable Acceleration — Problem-Solving Workshop

Part 6 of 7

This workshop brings together all the variable-acceleration techniques. The key is recognizing which variable the acceleration depends on to choose the right method.

Method Selection Guide

$a$ depends on...	Method	Key Equation
$t$ only	Direct integration	$v = v_0 + \int a(t)\,dt$

Part 7: Review & Applications

Variable Acceleration — Review & Applications

Part 7 of 7 — Comprehensive Review

Master Formula Sheet

Scenario	Method	Key Formula
$a = a(t)$	Integrate w.r.t. $t$	$v = v_0 + \int a\,dt$

Step 1: Find $v(t)$ :

$v(t) = 4 + \int_0^t (2 - 3t' + t'^2)\,dt' = 4 + 2t - \frac{3t^2}{2} + \frac{t^3}{3}$

Step 2: Find $x(t)$ :

$x(t) = \int_0^t v(t')\,dt' = \int_0^t \left(4 + 2t' - \frac{3t'^2}{2} + \frac{t'^3}{3}\right)dt'$

$= 4t + t^2 - \frac{t^3}{2} + \frac{t^4}{12}$

If $a(t)$ is a polynomial of degree $n$ :

$v(t)$ is a polynomial of degree $n + 1$
$x(t)$ is a polynomial of degree $n + 2$

Each integration raises the degree by one.

$v(t) = \int_0^t 10\sin(2t')\,dt' = 10\left[-\frac{\cos(2t')}{2}\right]_0^t = -5\cos(2t) + 5 = 5[1 - \cos(2t)]$

Note: $v(t) \ge 0$ always! The particle never moves backward.

$x(t) = \int_0^t 5[1-\cos(2t')]\,dt' = 5\left[t' - \frac{\sin(2t')}{2}\right]_0^t = 5t - \frac{5}{2}\sin(2t)$

$v$ oscillates between $0$ and $10$ m/s.
$x$ has a linear trend ( $5t$ ) with sinusoidal oscillations superimposed.
The average velocity is $\bar{v} = 5$ m/s.

$v(t) = \int_0^t 20e^{-t'/5}\,dt' = 20(-5)\left[e^{-t'/5}\right]_0^t = -100(e^{-t/5} - 1) = 100(1 - e^{-t/5})$

As $t \to \infty$ : $v \to 100$ m/s. The acceleration approaches zero, and the velocity approaches a constant — this is the terminal velocity.

$x(t) = \int_0^t 100(1 - e^{-t'/5})\,dt' = 100\left[t + 5e^{-t/5} - 5\right] = 100t + 500e^{-t/5} - 500$

At large $t$ : $x \approx 100t - 500$ → linear growth (constant velocity).

Choosing the Right Form

Given	Method
$a(t)$ explicit	Direct integration
$a$ from a graph	Compute area under $a$ - $t$ curve
$a$ from a table	Numerical integration (trapezoidal rule)

Role of Initial Conditions

The constant of integration is determined by the initial condition $v(t_0) = v_0$ .

Why It Matters

Consider $a(t) = 3t^2$ . The indefinite integral gives:

$v(t) = t^3 + C$

Without knowing $v_0$ , we can't determine $C$ . Different initial conditions give different motions:

Initial Condition	$v(t)$	Behavior
$v(0) = 0$	$t^3$

Worked Example

$a(t) = \frac{6}{(1+t)^2}$ with $v (0$ m/s.

$v(t) = 3 + \int_0^t \frac{6}{(1+t')^2}\,dt' = 3 + \left[-\frac{6}{1+t'}\right]_0^t = 3 - \frac{6}{1+t} + 6 = 9 - \frac{6}{1+t}$

Check: $v(0) = 9 - 6 = 3$ ✓

As $t \to \infty$ : $v \to 9$ m/s (terminal velocity).

Graphical Integration

When $a(t)$ is given as a graph, the change in velocity is the area under the curve.

$\Delta v = \int_{t_1}^{t_2} a\,dt = \text{area under } a\text{-}t \text{ graph}$

Example: Piecewise Linear Acceleration

Suppose the $a$ - $t$ graph shows:

$a = 4$ m/s² for $0 \le t \le 3$
$a = -2$ m/s² for $3 < t \le 9$

With $v(0) = 0$ :

$v(3) = 0 + 4 \times 3 = 12 \text{ m/s}$

$v(9) = 12 + (-2)(6) = 12 - 12 = 0 \text{ m/s}$

The particle speeds up, then slows down and stops at $t = 9$ s.

Trapezoidal Rule for Data Tables

Given data points $(t_i, a_i)$ :

$\Delta v \approx \sum_{i} \frac{a_i + a_{i+1}}{2} \Delta t_i$

Piecewise Acceleration Functions

Many real situations involve acceleration that changes form at specific times.

Example: Two-Phase Acceleration

$a(t) = \begin{cases} 6t & 0 \le t \le 2 \\ 12 - 3(t-2) & t > 2 \end{cases}$

with $v(0) = 0$ .

Phase 1 ( $0 \le t \le 2$ ):

$v(t) = \int_0^t 6t'\,dt' = 3t^2$

$v(2) = 12$ m/s

Phase 2 ( $t > 2$ ):

$v(t) = 12 + \int_2^t [12 - 3(t'-2)]\,dt' = 12 + \left[12t' - \frac{3(t'-2)^2}{2}\right]_2^t$

$= 12 + 12(t-2) - \frac{3(t-2)^2}{2}$

Key Rule

Always match the value at the boundary: $v$ must be continuous (no sudden jumps in velocity).

a (t) \int v (t) = v_{0} + \int_{0}^{t} a (t^{'}) d t^{'} \int x (t) = x_{0} + \int_{0}^{t} v (t^{'}) d t^{'}

Initial Conditions Required

You need two initial conditions:

$v(t_0) = v_0$ — determines constant from first integration
$x(t_0) = x_0$ — determines constant from second integration

Worked Example: Cubic Position from Linear Acceleration

Problem: $a(t) = 4 - 2t$ m/s², $v(0) = 0$ , $x(0) = 5$ m. Find the position when the particle first stops.

Solution

Step 1: Velocity

$v(t) = \int_0^t (4 - 2t')\,dt' = 4t - t^2$

Step 2: Find when $v = 0$

$4t - t^2 = t(4 - t) = 0 \implies t = 0 \text{ or } t = 4$

The particle stops again at $t = 4$ s.

Step 3: Position

$x(t) = 5 + \int_0^t (4t' - t'^2)\,dt' = 5 + 2t^2 - \frac{t^3}{3}$

$x(4) = 5 + 32 - \frac{64}{3} = 5 + \frac{96 - 64}{3} = 5 + \frac{32}{3} = \frac{47}{3} \approx 15.67 \text{ m}$

Verification

Check units: $[a] =$ m/s², after two integrations: $[x] =$ m ✓

Check: at $t = 0$ , $x = 5$ ✓, $v = 0$ ✓

Double Integration: Sinusoidal Case

Problem: $a(t) = -\omega^2 A\sin(\omega t)$ , $v(0) = \omega A$ , $x(0) = 0$ .

Velocity:

$v(t) = \omega A + \int_0^t [-\omega^2 A\sin(\omega t')]\,dt'$

$= \omega A + \omega A[\cos(\omega t') ]_0^t = \omega A + \omega A(\cos\omega t - 1) = \omega A\cos(\omega t)$

Position:

$x(t) = \int_0^t \omega A\cos(\omega t')\,dt' = A\sin(\omega t)$

This is simple harmonic motion: $x = A\sin(\omega t)$ .

Verification

$\frac{d^2x}{dt^2} = -\omega^2 A\sin(\omega t) = -\omega^2 x$

Indeed, $a = -\omega^2 x$ — the hallmark equation of SHM.

General Strategy for Double Integration

Step-by-Step Process

Integrate $a(t)$ to get $v(t)$ : Include $+C_1$ .
Apply first initial condition: $v(t_0) = v_0$ → solve for $C_1$ .
Integrate $v(t)$ to get $x(t)$ : Include $+C_2$ .
Apply second initial condition: $x(t_0) = x_0$ → solve for $C_2$ .

Common Mistake

Forgetting to apply initial conditions results in the wrong $C_1$ and $C_2$ . Always check that your answer satisfies both initial conditions.

Quick Check Table

After Integration	Degree Increase	New Constant
$a \to v$	$+1$	$C_1 = v_0$

For $a(t) = \text{polynomial of degree } n$ :

$v(t)$ has degree $n+1$
$x(t)$ has degree $n+2$

Starting from $a = dv/dt$ :

$\frac{dv}{dt} = a(v)$

$\frac{dv}{a(v)} = dt$

Integrate both sides:

$\int_{v_0}^{v} \frac{dv'}{a(v')} = \int_0^t dt' = t$

Linear Drag: $a = -bv$

This models drag force proportional to velocity (e.g., motion through a viscous fluid at low speeds).

$\frac{dv}{dt} = -bv \implies \frac{dv}{v} = -b\,dt$

$\ln\frac{v}{v_0} = -bt \implies v(t) = v_0 e^{-bt}$

Finding Position

$x(t) = \int_0^t v_0 e^{-bt'}\,dt' = \frac{v_0}{b}(1 - e^{-bt})$

Key Features

Quantity	Value
Time constant	$\tau = 1/b$
$v$ at $t = \tau$	$v_{0} / e \approx 0.37$

Physical Insight

The velocity decays exponentially. The object covers a finite total distance $v_0/b$ even though it never truly stops (it asymptotically approaches zero velocity).

Falling with Quadratic Drag: $a = g - cv^2$

For an object falling under gravity with quadratic air resistance:

$m\frac{dv}{dt} = mg - bv^2 \implies \frac{dv}{dt} = g - \frac{b}{m}v^2 = g\left(1 - \frac{v^2}{v_T^2}\right)$

where the terminal velocity is $v_T = \sqrt{mg/b}$ .

Separation of Variables

$\frac{dv}{1 - v^2/v_T^2} = g\,dt$

Using partial fractions:

$\frac{v_T}{2}\left[\frac{1}{v_T+v} + \frac{1}{v_T-v}\right]dv = g\,dt$

$\frac{v_T}{2}\ln\frac{v_T+v}{v_T-v} = gt$

Solution (starting from rest)

$v(t) = v_T\tanh\left(\frac{gt}{v_T}\right)$

Behavior

At small $t$ : $v \approx gt$ (free fall, drag negligible)
As $t \to \infty$ : $v \to v_T$ (terminal velocity)

Finding Position When $a = a(v)$

There are two approaches:

Approach 1: Find $v(t)$ first, then integrate

$x(t) = \int_0^t v(t')\,dt'$

Approach 2: Use the chain rule directly

Since $a = v\frac{dv}{dx}$ :

$a(v) = v\frac{dv}{dx}$

Separate:

$a(v)\,dx = v\,dv \implies dx = \frac{v\,dv}{a(v)}$

$x - x_0 = \int_{v_0}^{v} \frac{v'\,dv'}{a(v')}$

Example: Linear Drag

$a = -bv$ :

$x = \int_{v_0}^{v} \frac{v'\,dv'}{-bv'} = -\frac{1}{b}\int_{v_0}^{v} dv' = \frac{v_0 - v}{b}$

As $v \to 0$ : $x \to v_0/b$ ✓ (matches our earlier result).

The Energy Method Identity

$a(x) = v\frac{dv}{dx}$

Separating variables:

$\int_{x_0}^{x} a(x')\,dx' = \int_{v_0}^{v} v'\,dv' = \frac{v^2 - v_0^2}{2}$

$\boxed{v^2 = v_0^2 + 2\int_{x_0}^{x} a(x')\,dx'}$

This is the generalization of $v^2 = v_0^2 + 2a\Delta x$ for variable acceleration!

Application: Spring-Mass System

A mass on a spring has restoring force $F = -kx$ , giving:

$a(x) = -\frac{k}{m}x = -\omega^2 x$

where $\omega = \sqrt{k/m}$ .

Using the Energy Method

Starting from rest at $x = A$ (amplitude):

$v^2 = 0 + 2\int_A^x (-\omega^2 x')\,dx' = -\omega^2(x^2 - A^2) = \omega^2(A^2 - x^2)$

$v = \omega\sqrt{A^2 - x^2}$

Key Results

Position	Speed
$x = 0$ (equilibrium)	$v_{\max} = \omega A$
(endpoints)

Connection to Energy

$\frac{1}{2}mv^2 + \frac{1}{2}kx^2 = \frac{1}{2}kA^2$

The energy method is literally the work-energy theorem!

Application: Gravity with Varying Distance

Newton's law of gravitation: $a(r) = -\frac{GM}{r^2}$ (toward center, so negative for outward $r$ ).

Escape Velocity

Starting at the surface ( $r = R$ ) with speed $v_0$ , how fast must we go to escape ( $r \to \infty$ , $v \to 0$ )?

$v^2 = v_0^2 + 2\int_R^{\infty} \left(-\frac{GM}{r^2}\right)dr = v_0^2 + 2\left[\frac{GM}{r}\right]_R^{\infty} = v_0^2 - \frac{2GM}{R}$

For escape: set $v \to 0$ as $r \to \infty$ :

$0 = v_0^2 - \frac{2GM}{R} \implies v_{\text{esc}} = \sqrt{\frac{2GM}{R}}$

For Earth: $v_{\text{esc}} = \sqrt{2gR} \approx 11.2$ km/s.

The Power of the Energy Method

This problem would be extremely difficult to solve by integrating $a(t)$ directly, since $a$ depends on $x$ , which depends on $t$ in a complicated way. The $v\,dv = a\,dx$ approach bypasses time entirely!

Turning Points and Bounded Motion

From $v^2 = v_0^2 + 2\int_{x_0}^x a(x')\,dx'$ , a turning point occurs where $v = 0$ :

$v_0^2 + 2\int_{x_0}^{x_{\text{turn}}} a(x')\,dx' = 0$

Example: Potential Well

$a(x) = -8x + 2x^3$ , with $v = 0$ at $x = 0$ .

$v^2 = 2\int_0^x (-8x' + 2x'^3)\,dx' = 2\left[-4x^2 + \frac{x^4}{2}\right] = -8x^2 + x^4$

Turning points: $v = 0$ when $x^4 - 8x^2 = 0 \implies x^2(x^2-8) = 0$

So $x = 0$ (start) or $x = \pm 2\sqrt{2}$ .

But we need $v^2 \ge 0$ : $x^4 - 8x^2 \ge 0 \implies x^2 \ge 8$ .

This means the particle stays at $x = 0$ (it's a turning point where motion reverses... but actually $v^2 < 0$ for small displacements from $x = 0$ ). This indicates is an .

Worked Problem 1: Rocket with Drag

A rocket in space has thrust $F = F_0$ and experiences drag $F_d = -bv$ . Its mass $m$ is constant. Find $v(t)$ .

$ma = F_0 - bv \implies \frac{dv}{dt} = \frac{F_0}{m} - \frac{b}{m}v$

Let $\alpha = F_0/m$ and $\beta = b/m$ :

$\frac{dv}{dt} = \alpha - \beta v$

Separate: $\frac{dv}{\alpha - \beta v} = dt$

$-\frac{1}{\beta}\ln|\alpha - \beta v| = t + C$

With $v(0) = 0$ : $C = -\frac{1}{\beta}\ln\alpha$ .

$-\frac{1}{\beta}\ln\frac{\alpha - \beta v}{\alpha} = t$

$v(t) = \frac{\alpha}{\beta}(1 - e^{-\beta t}) = \frac{F_0}{b}(1 - e^{-bt/m})$

Terminal velocity: $v_T = F_0/b$ (when thrust balances drag).

Time constant: $\tau = m/b$ (time to reach $\approx 63\%$ of $v_T$ ).

Worked Problem 2: Position-Dependent Force

A bead slides along a wire with acceleration $a(x) = 6 - 2x$ . The bead starts from rest at $x = 0$ . Find where it reaches maximum speed.

Solution Using the Energy Method

Max speed occurs where $a = 0$ (acceleration changes from positive to negative):

$6 - 2x = 0 \implies x = 3$

Verify: for $x < 3$ , $a > 0$ (speeding up); for $x > 3$ , $a < 0$ (slowing down).

Finding Maximum Speed

$v^2 = 2\int_0^3 (6 - 2x)\,dx = 2[6x - x^2]_0^3 = 2(18 - 9) = 18$

$v_{\max} = \sqrt{18} = 3\sqrt{2} \approx 4.24 \text{ m/s}$

Finding Turning Point

The bead stops when $v = 0$ again:

$0 = 2\int_0^x (6-2x')\,dx' = 2(6x - x^2) \implies x(6-x) = 0$

So $x = 6$ (the other root is the starting point $x = 0$ ).

Dimensional Analysis Check

Variable acceleration problems are prone to algebra errors. Use dimensional analysis as a sanity check.

Rules

$[a] =$ m/s²
$[v] =$ m/s
$[x] =$ m
$[t] =$ s

Example Checks

For $a = -bv$ (linear drag):

$[b] = [a/v] = (\text{m/s}^2)/(\text{m/s}) = 1/\text{s}$ ✓
: dimensionless ✓

For $a = -cv^2$ (quadratic drag):

$[c] = [a/v^2] = 1/\text{m}$
$v = v_0/(1 + cv_0 t)$ : dimensionless ✓

Common Dimensional Errors

Exponent not dimensionless → wrong formula
Answer units don't match what's asked → algebra error
Time constant has wrong dimensions → check coefficients

Force Model	$v(t)$
$a = -bv$ (linear drag)	$v_0 e^{-bt}$
$a = -cv^2$ (quadratic drag)	$\frac{v_0}{1 + cv_0 t}$
$a = g - cv^2$ (gravity + drag)	$v_T\tanh(gt/v_T)$
$a = -\omega^2 x$ (spring)	$v = \omega\sqrt{A^2 - x^2}$

AP-Style Free Response Problem

Problem: A particle of mass $m$ moves along the $x$ -axis. At $t = 0$ , it is at $x = 0$ with velocity $v_0 > 0$ . It experiences a retarding force $F = -\alpha v^{1/2}$ where $\alpha > 0$ is a constant.

(a) Show that the velocity as a function of time is:

$v(t) = \left(\sqrt{v_0} - \frac{\alpha t}{2m}\right)^2$

Solution: $m\frac{dv}{dt} = -\alpha v^{1/2}$

$v^{-1/2}dv = -\frac{\alpha}{m}dt$

$2v^{1/2} = -\frac{\alpha}{m}t + C$

At $t = 0$ : $C = 2\sqrt{v_0}$ . So , giving . ✓

(b) Find the time $T$ when the particle stops.

$\sqrt{v_0} - \frac{\alpha T}{2m} = 0 \implies T = \frac{2m\sqrt{v_0}}{\alpha}$

(c) Find the total distance traveled.

$x = \int_0^T v\,dt = \int_0^T \left(\sqrt{v_0} - \frac{\alpha t}{2m}\right)^2 dt$

Let $u = \sqrt{v_0} - \frac{\alpha t}{2m}$ , :

$x = -\frac{2m}{\alpha}\int_{\sqrt{v_0}}^{0} u^2\,du = \frac{2m}{\alpha} \cdot \frac{v_0^{3/2}}{3} = \frac{2mv_0^{3/2}}{3\alpha}$

Connections to Other Topics

Variable acceleration is not just kinematics — it connects to nearly every topic in AP Physics C.

Newton's Second Law

$a = F_{\text{net}}/m$

When forces depend on position (springs, gravity), velocity (drag), or time (varying thrust), acceleration is variable.

Work-Energy Theorem

$\int F\,dx = \Delta KE$

This is just $m \cdot v\,dv = m \cdot a\,dx$ integrated — the energy method!

Differential Equations Preview

Many AP Physics C problems reduce to first-order ODEs:

Physics	ODE	Solution Type
RC circuit	$\frac{dq}{dt} = -q/(RC)$	Exponential decay

The mathematical techniques are the same in all cases!

Topic Complete!

You've mastered Variable Acceleration for AP Physics C:

Part	Topic	Status
1	$a(t)$ functions	✅
2	$v(t)$ from integration	✅
3	$x(t)$ from double integration	✅
4	$a(v)$ and separation of variables	✅
5	$a(x)$ and the energy method	✅
6	Problem-solving workshop	✅
7	Review & applications	✅

AP Exam Tip: On free-response problems involving variable acceleration, first identify what $a$ depends on. If $a = a(t)$ , integrate directly. If $a = a(v)$ , separate variables. If $a = a (x)$ , use . State your method clearly and show every step of the separation and integration — setup points are often worth more than the final answer.

[12t′−23(t′−2)2]2t

v_{0} / e \approx 0.37 v_{0}

The characteristic time scale is

v_T/g

2∫R∞(−r2GM)dr=

unstable equilibrium

[bt] = (1/\text{s})(\text{s}) =

[cv_0 t] = (1/\text{m})(\text{m/s})(\text{s}) =

\sqrt{v} = \sqrt{v_0} - \frac{\alpha t}{2m}

v = (\sqrt{v_0} - \frac{\alpha t}{2m})^2

du = -\frac{\alpha}{2m}dt

Motion with Variable Acceleration

Motion with Variable Acceleration - Complete Interactive Lesson

Part 1: Acceleration as a Function of Time

Variable Acceleration — a(t) Functions

When Acceleration Depends on Time

The Calculus Approach

Polynomial Acceleration

Example: Quadratic Acceleration

Sinusoidal Acceleration

Worked Example

Exponential Acceleration

Example: Decaying Thrust

Part 2: v(t) from Integration

Variable Acceleration — v(t) from Integration

The Fundamental Formula

Part 3: x(t) from Double Integration

Variable Acceleration — x(t) from Double Integration

The Two-Step Process

Part 4: a(v) & Separation of Variables

Variable Acceleration — a(v) and Separation of Variables

Common Physical Examples

Part 5: a(x) & Energy Methods

Variable Acceleration — a(x) and the Energy Method

Part 6: Problem-Solving Workshop

Variable Acceleration — Problem-Solving Workshop

Method Selection Guide

Part 7: Review & Applications

Variable Acceleration — Review & Applications

Master Formula Sheet

Pattern

Analysis

Terminal velocity

Position

Choosing the Right Form

Role of Initial Conditions

Why It Matters

Worked Example

Graphical Integration

Example: Piecewise Linear Acceleration

Trapezoidal Rule for Data Tables

Piecewise Acceleration Functions

Example: Two-Phase Acceleration

Key Rule

Initial Conditions Required

Worked Example: Cubic Position from Linear Acceleration

Solution

Verification

Double Integration: Sinusoidal Case

Verification

General Strategy for Double Integration

Step-by-Step Process

Common Mistake

Quick Check Table

The Technique

Linear Drag: a=−bva = -bva=−bv

Finding Position

Key Features

Physical Insight

Falling with Quadratic Drag: a=g−cv2a = g - cv^2a=g−cv2

Separation of Variables

Solution (starting from rest)

Behavior

Finding Position When a=a(v)a = a(v)a=a(v)

Approach 1: Find v(t)v(t)v(t) first, then integrate

Approach 2: Use the chain rule directly

Example: Linear Drag

The Energy Method Identity

Application: Spring-Mass System

Using the Energy Method

Key Results

Connection to Energy

Application: Gravity with Varying Distance

Escape Velocity

The Power of the Energy Method

Turning Points and Bounded Motion

Example: Potential Well

Worked Problem 1: Rocket with Drag

Worked Problem 2: Position-Dependent Force

Solution Using the Energy Method

Finding Maximum Speed

Linear Drag: $a = -bv$

Falling with Quadratic Drag: $a = g - cv^2$

Finding Position When $a = a(v)$

Approach 1: Find $v(t)$ first, then integrate