Overview

Unit 1.4 | Applying Kinematic Equations in 1 Dimension Easily

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Jason Kuma

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Every where you look there is motion. People biking, planes moving, ants crawling. In this lesson we will learn how to analyze this motion mathematically. It’s easier and more fun than you might initially think!

Unit 1 Breakdown

You are on Lesson 4 of 5

  1. Unit 1.1 | Understanding vectors and the Standard Units used in Physics
  2. Unit 1.2 | The Kinematic (motion) variables: Displacement, Velocity, and Acceleration
  3. Unit 1.3 | Graphing motion
  4. Unit 1.4 | Using Kinematic Equations in 1 Dimension [Current Lesson]
  5. Unit 1.5 | Projectile Motion: Using Kinematic Equations in 2 Dimensions

In this lesson you will learn: 

  • The “Big 5” Kinematic equations
  • How to use the equations
  • A simple framework to solve 1-D kinematic questions
  • Applying the framework to solve interesting real-world problems

Introduction

After touching down, how long does it take a plane to stop?

You drop a penny from the two story building. At what speed does it hit the ground?

These are kinematic problems, also known as motion problems. Your job is to find the one of missing kinematic variables: displacement, initial velocity, final velocity, acceleration, or time.

If you like numbers, this is the lesson for you.

And if you hate math, don’t worry… it’s incredibly simple!

Understanding kinematic equations

The “big” 5 kinematic equations, listed below, will be used to solve all kinematic problems.

  1. [katex] v_f = v_0 +at [/katex]
  2. [katex] v_f^2 = v_0^2 + 2a\Delta x [/katex]
  3. [katex]\Delta x = v_0t + \frac{1}{2}at^2 [/katex]
  4. [katex]\Delta x = v_ft – \frac{1}{2}at^2 [/katex]
  5. [katex] \Delta x = \frac{1}{2} t (v_f + v_0) [/katex]

These equations involve the kinematic variables, which was covered in Lesson 1.2 | The kinematic variables.

Variables and their symbols

Let’s quickly recap what each variable means.

  1. Displacement, it’s variable is [katex]\Delta x [/katex]’ (pronounced “delta x”), measured in meters.
    • You might see this written as ‘[katex]s[/katex]’ in some textbooks.
  2. Velocity, it variable is ‘[katex]v[/katex]’, measured in meters/second.

Notice that velocity is split into two variables:

  • initial velocity ‘[katex]v_0[/katex]’ (pronounced “v knot”)
    • NOTE: Sometimes written as ‘[katex]v_i[/katex]’ or ‘[katex]u[/katex]’ in some textbooks
  • final velocity ‘[katex]v_f[/katex]’ (pronounced “v final”)
  1. Acceleration, its variable is ‘[katex]a[/katex]’ and has units of meters/second2.
  2. Time is the last variable. Its symbol is ‘[katex]t[/katex]’, and it’s measured in seconds.

To recap, we have 5 kinematic variables in total listed in the chart below.

Variable nameVariable SymbolUnits
Displacement[katex]\Delta x [/katex][katex]m[/katex]
Initial velocity[katex]v_0 [/katex][katex]\frac{m}{s}[/katex]
Final velocity [katex]v_f[/katex][katex]\frac{m}{s}[/katex]
Acceleration[katex]a[/katex][katex]\frac{m}{s^2}[/katex]
Time [katex]t[/katex][katex]s[/katex]
Five kinematic variables and their symbols.

The “Big 5” equations

Although we have 5 kinematic variables, each equation only uses 4. 

This means, that each equation is doesn’t use exactly 1 variable.

For example, the chart below, shows equation 1 doesn’t use displacement ([katex]\Delta x[/katex]).

EquationVariable not in equation
[katex] v_f = v_0 + at [/katex][katex]\Delta x[/katex]
[katex] (v_f)^2 = (v_0)^2 + 2a\Delta x [/katex][katex]t[/katex]
[katex]\Delta x = v_0t + \frac{1}{2}at^2 [/katex][katex]v_f[/katex]
[katex]\Delta x = v_ft – \frac{1}{2}at^2 [/katex][katex]v_0[/katex]
[katex] \Delta x = \frac{1}{2} t (v_f + v_0) [/katex][katex]a[/katex]
Five kinematic equations and the missing variable in each.

PRO TIP: Memorizing these five equations are a must in helping you solve problems rapidly!

1D Kinematic Problems

One-dimension (1D), refers to an object moving in ONLY in one direction: either horizontally (left and right) OR vertically (up and down). We’ll cover two dimensional (2D) problems in the next lesson.

For now, take a look at this 1D problem: An airplane accelerates down a runway at [katex] 3 \, \frac{m}{s^2}[/katex] for [katex]20[/katex] seconds until it lifts off the ground. Determine the distance traveled before takeoff. 

This might seem hard at first. So let’s use a framework to break down how to solve these questions.

Framework for kinematic problems 

Think of a framework as a template for solving a specific type of problem. This will be the first of many frameworks you will learn in this course.

Here’s my simple framework for solving 1 dimension kinematic problems:

  1. Read the word problem and identify 4 kinematic variables. The problem will ALWAYS give you 3 variables, and ask you to solve for 1 variable
  2. Pick a kinematic equation. To make this easy, look for the variable the problem doesn’t even mention. Find the equation that also doesn’t have it, using the equation table from above.
  3. Plug and chug! Plug in all the given numbers and solve for the unknown variable

This might still sound a bit confusing. So let’s put in into practice.

Once you catch on, revisit this framework, and everything will make much more sense.

Applying the framework

Problem: An airplane accelerates down a runway at [katex] 3 \, \frac{m}{s^2}[/katex] for [katex]20[/katex] seconds until is finally lifts off the ground. Determine the distance traveled before takeoff.

  1. Identify the 4 variables in the problem: [katex]a = 3 \, m/s^2[/katex], [katex]t = 20 \, s[/katex], [katex]v_0 = 0 \, m/s[/katex], [katex]\Delta x = [/katex] what we need to find.
  2. Pick an equation: Notice that this question does not involve the ‘[katex]v_f[/katex]’ variable. Thus, we will pick the equation that does not have [katex]v_f[/katex] : [katex]\Delta x = v_0t + \frac{1}{2}at^2 [/katex]
  3. Plug the numbers into the equation and solve for [katex]\Delta x[/katex]:
    • We find that [katex]\Delta x = 600 \, m[/katex].

PQ – Kinematics in the x and y directions

Using the framework to solve following 1D kinematic problems for both the x and y directions.

Check answers by clicking the full version of each question.

Kinematics in the horizontal (x) direction

Open in UBQ
Question 1
Beginner
Mathematical
An airplane accelerates down a runway at \( 10 \, \text{m/s}^2 \). It reaches a final velocity of \( 200 \, \text{m/s} \) until it finally lifts off the ground. Determine the distance traveled before takeoff.
Open in UBQ
Question 2
Beginner
Mathematical
A car starts from rest and accelerates uniformly over a time of 5 seconds for a distance of 100 m. Determine the acceleration of the car.
Open in UBQ
Question 3
Beginner
Mathematical
A car decelerates from \( 25 \, \text{m/s} \) to \( 5 \, \text{m/s} \) at \( 10 \, \text{m/s}^2 \). How far does the car travel during this deceleration?
Open in UBQ
Question 4
Beginner
Mathematical
A car traveling at \( 20 \, \text{m/s} \) decelerates at a constant rate to a complete stop after traveling \( 40 \, \text{m} \).
Part (a)3 pts
What is the average speed of the car during this process?
Part (b)3 pts

How long does it take for the car to stop?

Kinematics in the vertical (y) direction

Open in UBQ
Question 5
Beginner
Mathematical
A ball is dropped from a window [katex]10 \, [/katex] above the sidewalk. Determine the time it takes for the ball to fall to the sidewalk.
Open in UBQ
Question 6
Beginner
Mathematical
A tennis ball is thrown straight up with an initial speed of \( 22.5 \, \text{m/s} \). It is caught at the same distance above ground.
Part (a)3 pts

How high does the ball rise?

Part (b)3 pts

How long does it take for the ball to reach its highest point?

Part (c)3 pts

How long does the ball remain in the air?

Part (d)3 pts

How fast (speed) was it going just before it is caught?

Part (e)3 pts

What is the velocity and acceleration of the ball at the highest point?

Open in UBQ
Question 7
Beginner
Mathematical
A rock is thrown vertically upward with a velocity of \( 20 \, \text{m/s} \) from the edge of a bridge \( 42 \, \text{m} \) above a river.
Part (a)3 pts

What is the rock’s speed just before it falls into the river?

Part (b)3 pts

How much time does it take from the time the rock is launched to the time when the rock strikes the river water?

Challenge (Bonus) kinematics questions

Open in UBQ
Question 8
Advanced
Mathematical

A car is traveling 20 m/s when the driver sees a child standing on the road. She takes 0.8 s to react then steps on the brakes and slows at 7.0 m/s2. How far does the car go before it stops?

Open in UBQ
Question 9
Advanced
Mathematical

Two students are on a balcony 19.6 m above the street. One student throws a ball vertically downward at 14.7 m/s. At the same instant, the other student throws a ball vertically upward at the same speed. The second ball just misses the balcony on the way down.

Part (a)2 pts

What is the difference in the time the balls spend in the air?

Part (b)2 pts

What is the velocity of each ball as it strikes the ground?

Part (c)3 pts

How far apart are the balls 0.800 s after they are thrown?

PS – 1D Kinematics

In this video we will solve a few problems using my simple 3 step kinematics’ framework. Hopefully, by the end of the video, you’ll see how simple 1D kinematic problems are! 

Helpful problem solving tips 

  • After picking the correct kinematic equation, re-arrange the equation for the variable you are trying to solve for. Then you can plug in the numbers all at once.
  • Label your variables using subscripts. For example if you are working in the horizontal direction, you can label, acceleration as [katex] a_x[/katex] and velocity as [katex] v_x[/katex]. This will make more sense once we start solving 2D problems
  • Get good at finding “hidden variables.” For example, if a problem tells you a car is moving at constant velocity, they are technically also telling you that acceleration = zero.
    • If a problem deals with an object in free fall, you can assume acceleration = [katex]9.81 \, \frac{m}{s^2}[/katex].

Lesson 1.5 Preview

In the next lesson we will apply our problem solving framework to objects moving in two dimensions. We often call these projectiles. Although this may seem harder it’s actually quite simple using the tricks we will show. 

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KinematicsForces
\(\Delta x = v_i t + \frac{1}{2} at^2\)\(F = ma\)
\(v = v_i + at\)\(F_g = \frac{G m_1 m_2}{r^2}\)
\(v^2 = v_i^2 + 2a \Delta x\)\(f = \mu N\)
\(\Delta x = \frac{v_i + v}{2} t\)\(F_s =-kx\)
\(v^2 = v_f^2 \,-\, 2a \Delta x\) 
Circular MotionEnergy
\(F_c = \frac{mv^2}{r}\)\(KE = \frac{1}{2} mv^2\)
\(a_c = \frac{v^2}{r}\)\(PE = mgh\)
\(T = 2\pi \sqrt{\frac{r}{g}}\)\(KE_i + PE_i = KE_f + PE_f\)
 \(W = Fd \cos\theta\)
MomentumTorque and Rotations
\(p = mv\)\(\tau = r \cdot F \cdot \sin(\theta)\)
\(J = \Delta p\)\(I = \sum mr^2\)
\(p_i = p_f\)\(L = I \cdot \omega\)
Simple Harmonic MotionFluids
\(F = -kx\)\(P = \frac{F}{A}\)
\(T = 2\pi \sqrt{\frac{l}{g}}\)\(P_{\text{total}} = P_{\text{atm}} + \rho gh\)
\(T = 2\pi \sqrt{\frac{m}{k}}\)\(Q = Av\)
\(x(t) = A \cos(\omega t + \phi)\)\(F_b = \rho V g\)
\(a = -\omega^2 x\)\(A_1v_1 = A_2v_2\)
ConstantDescription
[katex]g[/katex]Acceleration due to gravity, typically [katex]9.8 , \text{m/s}^2[/katex] on Earth’s surface
[katex]G[/katex]Universal Gravitational Constant, [katex]6.674 \times 10^{-11} , \text{N} \cdot \text{m}^2/\text{kg}^2[/katex]
[katex]\mu_k[/katex] and [katex]\mu_s[/katex]Coefficients of kinetic ([katex]\mu_k[/katex]) and static ([katex]\mu_s[/katex]) friction, dimensionless. Static friction ([katex]\mu_s[/katex]) is usually greater than kinetic friction ([katex]\mu_k[/katex]) as it resists the start of motion.
[katex]k[/katex]Spring constant, in [katex]\text{N/m}[/katex]
[katex] M_E = 5.972 \times 10^{24} , \text{kg} [/katex]Mass of the Earth
[katex] M_M = 7.348 \times 10^{22} , \text{kg} [/katex]Mass of the Moon
[katex] M_M = 1.989 \times 10^{30} , \text{kg} [/katex]Mass of the Sun
VariableSI Unit
[katex]s[/katex] (Displacement)[katex]\text{meters (m)}[/katex]
[katex]v[/katex] (Velocity)[katex]\text{meters per second (m/s)}[/katex]
[katex]a[/katex] (Acceleration)[katex]\text{meters per second squared (m/s}^2\text{)}[/katex]
[katex]t[/katex] (Time)[katex]\text{seconds (s)}[/katex]
[katex]m[/katex] (Mass)[katex]\text{kilograms (kg)}[/katex]
VariableDerived SI Unit
[katex]F[/katex] (Force)[katex]\text{newtons (N)}[/katex]
[katex]E[/katex], [katex]PE[/katex], [katex]KE[/katex] (Energy, Potential Energy, Kinetic Energy)[katex]\text{joules (J)}[/katex]
[katex]P[/katex] (Power)[katex]\text{watts (W)}[/katex]
[katex]p[/katex] (Momentum)[katex]\text{kilogram meters per second (kgm/s)}[/katex]
[katex]\omega[/katex] (Angular Velocity)[katex]\text{radians per second (rad/s)}[/katex]
[katex]\tau[/katex] (Torque)[katex]\text{newton meters (Nm)}[/katex]
[katex]I[/katex] (Moment of Inertia)[katex]\text{kilogram meter squared (kgm}^2\text{)}[/katex]
[katex]f[/katex] (Frequency)[katex]\text{hertz (Hz)}[/katex]

Metric Prefixes

Example of using unit analysis: Convert 5 kilometers to millimeters. 

  1. Start with the given measurement: [katex]\text{5 km}[/katex]

  2. Use the conversion factors for kilometers to meters and meters to millimeters: [katex]\text{5 km} \times \frac{10^3 \, \text{m}}{1 \, \text{km}} \times \frac{10^3 \, \text{mm}}{1 \, \text{m}}[/katex]

  3. Perform the multiplication: [katex]\text{5 km} \times \frac{10^3 \, \text{m}}{1 \, \text{km}} \times \frac{10^3 \, \text{mm}}{1 \, \text{m}} = 5 \times 10^3 \times 10^3 \, \text{mm}[/katex]

  4. Simplify to get the final answer: [katex]\boxed{5 \times 10^6 \, \text{mm}}[/katex]

Prefix

Symbol

Power of Ten

Equivalent

Pico-

p

[katex]10^{-12}[/katex]

Nano-

n

[katex]10^{-9}[/katex]

Micro-

µ

[katex]10^{-6}[/katex]

Milli-

m

[katex]10^{-3}[/katex]

Centi-

c

[katex]10^{-2}[/katex]

Deci-

d

[katex]10^{-1}[/katex]

(Base unit)

[katex]10^{0}[/katex]

Deca- or Deka-

da

[katex]10^{1}[/katex]

Hecto-

h

[katex]10^{2}[/katex]

Kilo-

k

[katex]10^{3}[/katex]

Mega-

M

[katex]10^{6}[/katex]

Giga-

G

[katex]10^{9}[/katex]

Tera-

T

[katex]10^{12}[/katex]

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