Physics: Foundations of Physical Science

Part II — Motion and Forces · Chapter 8

Kinematic Equations

8.1 Why Kinematic Equations Are Useful

In Chapter 7 we defined velocity as Δx/Δt and acceleration as Δv/Δt. These definitions are precise and correct, but they leave open a practical problem: if an object moves with constant acceleration, and we know some combination of its initial velocity, final velocity, acceleration, time, and displacement, how do we find the others?

The kinematic equations are the answer. They are algebraic relationships among five kinematic quantities — displacement (Δx), initial velocity (v_i), final velocity (v_f), acceleration (a), and time (t) — derived specifically for the case of constant acceleration. Each equation links four of the five variables. By identifying which three are known and which one is unknown, you can select the correct equation and solve directly, without iteration.

These equations are not independent laws of nature in the way that Newton’s laws are. They are mathematical consequences of the definitions of velocity and acceleration under the specific condition of constant acceleration. Their power lies in systematic application, not in memorization by brute force.

8.2 Conditions for Using the Kinematic Equations

The kinematic equations apply only when all of the following conditions are satisfied:

The acceleration is constant: same magnitude and direction throughout the time interval. If acceleration changes during the motion, the equations give wrong answers.
The motion is in one dimension: along a straight line. The equations can be applied separately to each dimension in 2D or 3D problems (Chapter 9), but each application is one-dimensional.
The time interval is clearly defined: you must specify both the start and end of the interval for which the equations apply.

Example — Valid situation

A car accelerates at a constant 2.5 m/s² from rest for 10 s. Constant acceleration ✓. One dimension along the road ✓. Kinematic equations apply.

Example — Invalid situation

A ball thrown upward, where air resistance is significant. The acceleration changes with speed. Kinematic equations do not apply directly — you would need to neglect air resistance first.

8.3 Position, Displacement, Velocity, Acceleration, and Time

Before writing any kinematic equation, identify which of the five kinematic variables are involved in the problem. A clear symbol list and sign convention prevents most errors.

Table 8.1 Kinematic variables used in constant-acceleration problems.
Symbol	Name	Unit
Δx (or d)	Displacement	m
v_i (or v₀)	Initial velocity	m/s
v_f (or v)	Final velocity	m/s
a	Acceleration	m/s²
t (or Δt)	Time interval	s

8.4 Initial Velocity vs. Final Velocity

Initial velocity (v_i or v₀) is the velocity at the beginning of the time interval you are analyzing. Final velocity (v_f or v) is the velocity at the end of that interval. These labels are defined by the time interval chosen, not by the overall problem.

In multi-step problems, the final velocity of one phase becomes the initial velocity of the next. Carrying this value correctly between phases is essential.

Example — Identifying v_i and v_f

A car traveling at 20 m/s brakes to a stop. The analysis interval begins at the moment braking starts: v_i = +20 m/s, v_f = 0. Now suppose you analyze only the second half of the braking interval. The v_i for that sub-interval is whatever velocity the car had at the midpoint.

8.5 The Four Main Kinematic Equations

The four kinematic equations are derived from the definitions of average velocity and constant acceleration. You do not need to re-derive them each time, but knowing that they come from definitions — not experimental laws — helps you remember their structure and limits.

Equation 1: $v_{f} = v_{i} + a t$ ← missing Δx
Equation 2: $Δx = v_{i} t + ½ a t^{2}$ ← missing v_f
Equation 3: $v_{f}^{2} = v_{i}^{2} + 2 a Δx$ ← missing t
Equation 4: $Δx = ½ (v_{i} + v_{f}) t$ ← missing a

Table 8.2 The four main kinematic equations and the variable each one omits.
Equation	Contains	Missing Variable
v_f = v_i + at	v_i, v_f, a, t	Displacement Δx
Δx = v_it + ½at²	Δx, v_i, a, t	Final velocity v_f
v_f² = v_i² + 2aΔx	Δx, v_i, v_f, a	Time t
Δx = ½(v_i + v_f)t	Δx, v_i, v_f, t	Acceleration a

8.6 Choosing the Right Equation

The most reliable method for choosing the correct kinematic equation is to identify the missing variable — the one not mentioned in the problem — and select the equation that does not contain it.

The Selection Process

List all five variables: Δx, v_i, v_f, a, t.
Label each as Given, Find, or Missing (not mentioned).
Identify the one variable not mentioned in the problem (not given, not asked for).
Select the equation that is missing that variable.
Solve algebraically before substituting numbers.

Example — Equation selection example

Known: v_i = 0, a = +3.0 m/s², t = 5.0 s. Find: Δx. Missing: v_f. → Use Δx = v_it + ½at² (Equation 2, which does not contain v_f).

Example — Another selection

Known: v_i = +20 m/s, v_f = 0, Δx = +80 m. Find: a. Missing: t. → Use v_f² = v_i² + 2aΔx (Equation 3, which does not contain t).

8.7 Solving for Displacement

Displacement is found when the acceleration, time, and initial (and possibly final) velocity are known.

Worked Example 8.1 — Car Accelerating from Rest

Problem: A car starts from rest and accelerates at +2.0 m/s² for 8.0 s. How far does it travel?

List variables: v_i = 0 (starts from rest); a = +2.0 m/s²; t = 8.0 s; Δx = ?; v_f = missing.
Select equation: Missing v_f → Equation 2: Δx = v_it + ½at².
Substitute: Δx = (0)(8.0) + ½(2.0)(8.0)² = 0 + ½(2.0)(64) = 64 m.
Check: Units: (m/s²)(s²) = m ✔. Reasonable: 64 m in 8 s for a car ✔.

Answer: Δx = +64 m

Worked Example 8.2 — Given Both Velocities and Time

Problem: A cyclist slows from 12.0 m/s to 4.0 m/s over 5.0 s. How far does she travel?

List variables: v_i = +12.0 m/s; v_f = +4.0 m/s; t = 5.0 s; Δx = ?; a = missing.
Select equation: Missing a → Equation 4: Δx = ½(v_i + v_f)t.
Substitute: Δx = ½(12.0 + 4.0)(5.0) = ½(16.0)(5.0) = 40.0 m.
Check: Units: (m/s)(s) = m ✔.

Answer: Δx = +40.0 m

8.8 Solving for Velocity

Worked Example 8.3 — Final Velocity After Acceleration

Problem: A rocket starts from rest and accelerates upward at +15.0 m/s². What is its velocity after 6.0 s?

List variables: v_i = 0; a = +15.0 m/s²; t = 6.0 s; v_f = ?; Δx = missing.
Select equation: Missing Δx → Equation 1: v_f = v_i + at.
Substitute: v_f = 0 + (15.0)(6.0) = +90.0 m/s.
Check: Units: (m/s²)(s) = m/s ✔.

Answer: v_f = +90.0 m/s (upward)

Worked Example 8.4 — Final Velocity Without Time

Problem: A car traveling at 25.0 m/s brakes with a = −4.0 m/s². What is its velocity after traveling 50.0 m?

List variables: v_i = +25.0 m/s; a = −4.0 m/s²; Δx = +50.0 m; v_f = ?; t = missing.
Select equation: Missing t → Equation 3: v_f² = v_i² + 2aΔx.
Substitute: v_f² = (25.0)² + 2(−4.0)(50.0) = 625 − 400 = 225.
Solve: v_f = √225 = +15.0 m/s (positive: still moving in positive direction).
Check: Units: m²/s² under root → m/s ✔. Speed decreased from 25 to 15 m/s with negative a ✔.

Answer: v_f = +15.0 m/s

8.9 Solving for Acceleration

Worked Example 8.5 — Finding Acceleration from Velocities and Displacement

Problem: A car traveling at 20.0 m/s comes to a stop in 80.0 m. Find the acceleration.

List variables: v_i = +20.0 m/s; v_f = 0; Δx = +80.0 m; a = ?; t = missing.
Select equation: Missing t → Equation 3: v_f² = v_i² + 2aΔx.
Rearrange: a = (v_f² − v_i²) / (2Δx) = (0 − 400) / (2 × 80.0) = −400 / 160.
Calculate: a = −2.5 m/s².
Check: Negative acceleration with positive velocity → slowing down ✔. Units: m²/s² / m = m/s² ✔.

Answer: a = −2.5 m/s²

Worked Example 8.6 — Finding Acceleration from Velocities and Time

Problem: A ball’s velocity changes from +5.0 m/s to +17.0 m/s in 3.0 s. Find the acceleration.

List variables: v_i = +5.0 m/s; v_f = +17.0 m/s; t = 3.0 s; a = ?; Δx = missing.
Select equation: Missing Δx → Equation 1: v_f = v_i + at → a = (v_f − v_i) / t.
Calculate: a = (17.0 − 5.0) / 3.0 = 12.0 / 3.0 = +4.0 m/s².

Answer: a = +4.0 m/s²

8.10 Solving for Time

Worked Example 8.7 — Time to Reach a Given Velocity

Problem: A train starts from rest and accelerates at +0.50 m/s². How long until it reaches 30.0 m/s?

List variables: v_i = 0; a = +0.50 m/s²; v_f = +30.0 m/s; t = ?; Δx = missing.
Select equation: Missing Δx → Equation 1: v_f = v_i + at → t = (v_f − v_i) / a.
Calculate: t = (30.0 − 0) / 0.50 = 60.0 s.

Answer: t = 60.0 s

Worked Example 8.8 — Time from Displacement and Velocity

Problem: An object with v_i = +3.0 m/s and a = +2.0 m/s² travels Δx = +20.0 m. How long does this take?

List variables: v_i = +3.0 m/s; a = +2.0 m/s²; Δx = +20.0 m; t = ?; v_f = missing.
Select equation: Missing v_f → Equation 2: Δx = v_it + ½at².
Rearrange: 20.0 = 3.0t + ½(2.0)t² → t² + 3t − 20 = 0.
Quadratic formula: t = [−3 ± √(9 + 80)] / 2 = [−3 ± √89] / 2 = [−3 ± 9.43] / 2.
Select positive root: t = (−3 + 9.43) / 2 = 6.43 / 2 = 3.2 s. Discard the negative root.

Answer: t ≈ 3.2 s

8.11 Free Fall as Constant Acceleration

Free fall is motion under the influence of gravity alone, with no air resistance. Near Earth’s surface, free fall produces a constant downward acceleration whose magnitude is:

g = 9.81 m/s² (downward)

All four kinematic equations apply directly to free fall, with a = −9.81 m/s² if upward is positive, or a = +9.81 m/s² if downward is positive. The choice of sign is yours — state it clearly and be consistent.

Worked Example 8.9 — Ball Dropped from Rest

Problem: A ball is dropped from a height of 45.0 m. (a) How long does it take to reach the ground? (b) What is its velocity just before hitting?

Setup: upward positive; origin at ground; initial position = +45.0 m; final position = 0; so Δx = −45.0 m. v_i = 0 (dropped). a = −9.81 m/s².
Find t: Missing v_f → Equation 2: Δx = v_it + ½at² → −45.0 = 0 + ½(−9.81)t² → t² = 9.174 → t = 3.03 s.
Find v_f: Missing Δx → Equation 1: v_f = 0 + (−9.81)(3.03) = −29.7 m/s.
Check: Negative velocity confirms downward motion at impact ✔. Speed = 29.7 m/s ≈ 107 km/h, reasonable for a 45 m fall ✔.

Worked Example 8.10 — Ball Thrown Upward

Problem: A ball is thrown straight up with v_i = +18.0 m/s. (a) How high does it go? (b) How long to reach the peak? (c) What is the velocity at t = 1.5 s?

Setup: upward positive; a = −9.81 m/s²; at peak: v_f = 0.
(a) Max height: Equation 3: 0 = (18.0)² + 2(−9.81)Δx → Δx = 324 / 19.62 = 16.5 m.
(b) Time to peak: Equation 1: 0 = 18.0 + (−9.81)t → t = 18.0 / 9.81 = 1.84 s.
(c) v at t = 1.5 s: Equation 1: v_f = 18.0 + (−9.81)(1.5) = 18.0 − 14.7 = +3.3 m/s (still rising).

8.12 Sign Conventions in Kinematics

Sign errors are the single most common source of mistakes in kinematic problems. A systematic approach prevents them.

The Sign Convention Protocol

Choose positive direction before writing any equations. State it explicitly.
Assign signs to v_i, v_f, a, and Δx based on that choice. Quantities in the positive direction are positive; opposite direction is negative.
Carry signs through all algebra. Do not strip the sign until the very end.
Interpret the sign of the answer in terms of direction, not magnitude.

Table 8.3 Free-fall signs under different positive-direction choices.
Situation	With Upward Positive	With Downward Positive
Object thrown upward, initial velocity	v_i = +20 m/s	v_i = −20 m/s
Gravitational acceleration	a = −9.81 m/s²	a = +9.81 m/s²
Displacement downward 30 m	Δx = −30 m	Δx = +30 m
Velocity of falling object	v < 0 (downward)	v > 0 (downward)

8.13 Multi-Step Kinematics Problems

Many realistic problems involve two or more phases of motion, each with a different constant acceleration. The key is to analyze each phase separately, carrying the final state of one phase as the initial state of the next.

Worked Example 8.11 — Accelerate then Brake

Problem: A car starts from rest, accelerates at +3.0 m/s² for 10.0 s, then brakes at −5.0 m/s² until it stops. Find the total distance traveled.

Phase 1 — Acceleration: v_i = 0; a = +3.0 m/s²; t = 10.0 s.
Phase 1 — Find v_f: v_f = 0 + 3.0(10.0) = +30.0 m/s.
Phase 1 — Find Δx₁: Δx₁ = 0(10) + ½(3.0)(100) = +150 m.
Phase 2 — Braking: v_i = +30.0 m/s; v_f = 0; a = −5.0 m/s².
Phase 2 — Find Δx₂: Equation 3: 0 = (30.0)² + 2(−5.0)Δx₂ → Δx₂ = 900 / 10 = +90 m.
Total distance: 150 + 90 = 240 m.

Answer: The car travels 240 m total.

8.14 Checking Answers with Units and Reasonableness

After every calculation, apply two checks before accepting the answer.

Check 1: Units

Substitute the units of each variable into the equation and confirm the units of the answer are correct. If you solved for displacement, the units must be meters. If you solved for velocity, the units must be m/s. A unit mismatch means either the wrong equation was used or an algebra error was made.

Example — Unit check

From Δx = v_it + ½at²: units = (m/s)(s) + (m/s²)(s²) = m + m = m ✔. Both terms have units of meters, confirming dimensional consistency.

Check 2: Reasonableness

Compare the answer to your physical intuition. A car braking from 100 km/h: stopping distance of 50–100 m is reasonable; 0.05 m is not. A ball thrown upward: peak height of 10–20 m for a strong throw is reasonable; 200 m is not. If the answer is off by orders of magnitude, check for a factor-of-ten error — the most common arithmetic mistake.

Compare to known physical scales from Table 7.1.
Check whether the direction (sign) of the answer makes sense.
Check whether the final speed is less than the initial speed when decelerating, and greater when accelerating.
For free fall, confirm the answer gives g ≈ 9.81 m/s² when you back-calculate.

8.15 Common Mistakes with Kinematic Equations

Mistake 1: Using the wrong equation

Each equation is designed for a specific missing variable. Using an equation that contains an unknown variable forces you to solve a simultaneous system unnecessarily. Always identify the missing variable first and select accordingly.

Mistake 2: Using wrong sign for acceleration

For free fall with upward positive: a = −9.81 m/s². Students often write +9.81, obtaining velocities and displacements with wrong signs. State the sign convention, then assign a = −9.81 m/s².

Mistake 3: Confusing instantaneous and average velocity

In Equation 2, v_i is the instantaneous initial velocity — not the average velocity over the interval. Do not substitute Δx/Δt for v_i.

Mistake 4: Applying the equations to non-constant acceleration

If a problem has friction that depends on speed, or gravity at high altitude, or any scenario where acceleration changes, the four equations cannot be applied directly. State explicitly: “assume constant acceleration” if you are making that approximation.

Mistake 5: Forgetting initial velocity is not zero

Many students default to v_i = 0 unless the problem explicitly says “starts from rest.” Read the problem. If the object was already moving, v_i is that given velocity, not zero.

Mistake 6: Discarding the negative square root without checking

When solving v_f² = ... for v_f, you take a square root. This gives two solutions: positive and negative. Both may be physically valid. Check which makes sense given the direction of motion. Do not automatically take the positive root.

Mistake 7: Not carrying units through algebra

Solve the equation symbolically (with variables, not numbers) before substituting. Then substitute numbers with units. This makes unit checking automatic and error-finding easy.

Chapter Summary

The four kinematic equations solve constant-acceleration, one-dimensional motion problems by relating the five variables: displacement (Δx), initial velocity (v_i), final velocity (v_f), acceleration (a), and time (t).

The equations apply only when acceleration is constant throughout the interval.
Eq. 1: v_f = v_i + at ← missing Δx.
Eq. 2: Δx = v_it + ½at² ← missing v_f.
Eq. 3: v_f² = v_i² + 2aΔx ← missing t.
Eq. 4: Δx = ½(v_i + v_f)t ← missing a.
Selection strategy: identify the variable not mentioned in the problem; use the equation missing that variable.
Free fall: a = −9.81 m/s² when upward is positive. At the peak: v = 0, but a = −9.81 m/s², not zero.
Multi-step problems: solve each phase separately; v_f of phase n becomes v_i of phase n+1.
Always check units and physical reasonableness after calculating.
Solve equations symbolically before substituting numbers; carry signs through all algebra.

Key Terms

kinematic equations: Four algebraic equations relating the five kinematic variables (Δx, v_i, v_f, a, t) for constant-acceleration motion.
initial velocity (v_i): The velocity of an object at the beginning of the time interval being analyzed.
final velocity (v_f): The velocity of an object at the end of the time interval being analyzed.
constant acceleration: Acceleration that does not change in magnitude or direction during the time interval; the required condition for the kinematic equations.
free fall: Motion under gravity alone with no air resistance; produces constant acceleration of g = 9.81 m/s² downward near Earth’s surface.
g: Symbol for the magnitude of gravitational acceleration near Earth’s surface: g = 9.81 m/s².
missing variable: The kinematic variable that is neither given nor asked for in a problem; used to select the appropriate kinematic equation.
multi-step kinematics: A problem involving two or more phases of motion, each with a different constant acceleration; solved by analyzing each phase separately.
quadratic formula: The algebraic formula for solving ax² + bx + c = 0; arises when solving Equation 2 for time: x = [−b ± √(b² − 4ac)] / 2a.

Concept Review Questions

Answer in complete sentences.

What condition must be satisfied for the kinematic equations to be valid? Give an example of a situation where this condition is met and one where it is not.
Why is there no single equation that contains all five kinematic variables? How does this structure help you choose the right equation?
A ball is at the peak of its trajectory after being thrown upward. (a) What is its velocity? (b) What is its acceleration? (c) Is the acceleration constant throughout the throw?
In free fall with upward positive, what sign should you assign to a? What sign should you assign to the displacement if the object falls?
Explain in your own words the strategy of identifying the “missing variable” to select the correct kinematic equation.
What is the difference between applying the kinematic equations to a single-phase versus a multi-step problem? What connects the two phases?
A student gets a displacement of 350 m for a car braking from 20 m/s in 2 seconds. Without calculating yourself, explain two checks the student could apply to realize the answer is wrong.
Why does solving Equation 2 for time sometimes give two answers? How do you decide which to use?

Equation Selection Practice

For each problem, list all five variables as given, find, or missing. State which equation you would use and why. Do not solve.

Known: v_i = 0, a = +5.0 m/s², t = 4.0 s. Find: Δx.
Known: v_i = +12 m/s, v_f = +30 m/s, a = +3.0 m/s². Find: Δx.
Known: v_i = +20 m/s, Δx = +100 m, t = 8.0 s. Find: a.
Known: Δx = +64 m, v_i = 0, a = +2.0 m/s². Find: v_f.
Known: v_i = +15 m/s, v_f = 0, a = −3.0 m/s². Find: t.
Known: v_i = +8.0 m/s, a = +4.0 m/s², Δx = +90 m. Find: t.

One-Dimensional Motion Problems

Solve completely. Show variable list, equation selection, algebraic solution, substitution with units, and units check.

A car starts from rest and accelerates at 2.5 m/s² for 12.0 s. Find: (a) final speed; (b) distance traveled.
A truck traveling at 22.0 m/s brakes with a = −3.5 m/s². Find: (a) velocity after 4.0 s; (b) stopping distance.
A bicycle accelerates from 4.0 m/s to 12.0 m/s over 80.0 m. Find the acceleration.
A ball rolls with initial velocity +6.0 m/s and constant acceleration −1.5 m/s². (a) When does it stop? (b) How far does it roll before stopping? (c) What is its velocity at t = 5.0 s?
Two cars start from the same point. Car A has constant velocity +20 m/s. Car B starts from rest and accelerates at +4.0 m/s². (a) After how many seconds does Car B match Car A’s speed? (b) How far has each car traveled at that moment?
A sprinter accelerates from rest at 4.0 m/s² for 2.5 s, then runs at constant velocity. Find: (a) maximum velocity; (b) distance covered in the first 2.5 s; (c) total distance in 10.0 s.

Free-Fall Problems

Use g = 9.81 m/s². Unless stated otherwise, use upward as positive so a = −9.81 m/s².

A rock is dropped from a bridge 60.0 m above the water. (a) How long to reach the water? (b) What is the impact speed?
A ball is thrown straight up at 20.0 m/s. (a) Maximum height? (b) Time to reach the peak? (c) Time to return to the starting point?
A coin is thrown downward from a building with v_i = −5.0 m/s and falls 30.0 m to the ground. (a) How long does it fall? (b) What is the speed at impact?
A stone is thrown upward at 15.0 m/s from the top of a 40.0 m cliff. (a) What is the maximum height above the cliff top? (b) How long does the stone take to hit the base of the cliff?
Two balls are dropped simultaneously from heights of 20.0 m and 45.0 m. (a) Which hits first? (b) What is the time difference between impacts?
A ball is thrown upward at 12.0 m/s. What is its velocity 1.0 s, 2.0 s, and 2.5 s after release? What is its height at each of those times?

Mixed Kinematics Practice

A car traveling at 30.0 m/s sees a red light 80.0 m ahead and brakes at −6.0 m/s². Does it stop in time? Show your work.
A rocket starts from rest and accelerates at +25.0 m/s². (a) Speed after 200 m? (b) Time to reach that speed? (c) Sketch the v-t graph for this motion.
A ball is thrown horizontally from a 50.0 m cliff at 15.0 m/s. Consider only the vertical motion: (a) How long to fall 50.0 m? (b) What is the vertical velocity at impact? Use downward positive for this problem only.
A car accelerates from 0 to 28.0 m/s in 6.0 s. It then travels at constant velocity for 20.0 s. It then brakes to a stop in 4.0 s. Find: (a) acceleration in phase 1; (b) deceleration in phase 3; (c) total distance traveled.
Challenge: A ball is thrown upward at 18.0 m/s from a rooftop that is 25.0 m above the ground. Find the time and velocity when the ball hits the ground. Hint: define upward positive, Δx = −25.0 m.

Chapter 8 Checkpoint

Skills to confirm before moving on from Chapter 8.
I can…	Section
State the condition required for the kinematic equations to apply	8.2
List all five kinematic variables with symbols, meanings, and units	8.3
Write all four kinematic equations from memory	8.5
Identify the missing variable and select the correct equation	8.6
Solve for any of the five variables given three known values	8.7–8.10
Apply the kinematic equations to free fall (g = 9.81 m/s², with correct sign)	8.11
Choose and apply a consistent sign convention throughout a problem	8.12
Solve two-phase kinematics problems by treating each phase separately	8.13
Check answers using unit analysis and physical reasonableness	8.14
Identify and avoid the seven most common kinematic equation mistakes	8.15