Kinematic Equations (College Board AP® Physics 1: Algebra-Based)

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Leander Oates

Written by: Leander Oates

Reviewed by: Caroline Carroll

Kinematic equations

  • In physics, motion can be represented in a number of different ways:

    • Motion diagrams

    • Figures and data

    • Graphs

    • Equations

    • Narrative descriptions

  • For objects in a state of constant or uniform acceleration, there are three kinematic equations that can be used to describe their instantaneous linear motion in one dimension

    • Each of the three kinematics equations is listed on the equation sheet and will be provided in the exams

  • For all of these equations, the following conditions apply:

    • acceleration is uniform

      • therefore, average and instantaneous acceleration are equal

    • motion is along a straight line

      • motion is presented in the x direction but can also be applied to the y direction

  • For all these equations:

    • Time interval, increment t space equals space t

      • The assumption is that the timer is started from zero, t subscript 0 space equals space 0

    • Change in velocity, increment v subscript x space equals space v subscript x space minus space v subscript x 0 end subscript

    • Displacement, increment x space equals space x space minus space x subscript 0

  • In exam questions, some of the information required for the calculation may be assumed

  • Phrases and situations to look out for:

    • The object begins at rest

      • The object is initially stationary; therefore, the initial velocity is zero

      • This leads to any expression containing initial velocity canceling to zero

    • Falling objects or objects in free fall

      • The acceleration is equal to the acceleration of free fall at Earth's surface, g space equals space 10 space straight m divided by straight s squared

Kinematic equation 1

  • This equation is used when position and displacement are not required

v subscript x space equals space v subscript x 0 end subscript space plus space a subscript x t

  • Where:

    • v subscript x = final velocity in the x direction, measured in straight m divided by straight s

    • v subscript x 0 end subscript = initial velocity in the x direction, measured in straight m divided by straight s

    • a subscript x = constant acceleration in the x direction, measured in straight m divided by straight s squared

    • t = time interval, measured in straight s

Kinematic equation 2

  • This equation is used when final velocity is not required

x space equals space x subscript 0 space plus space v subscript x 0 end subscript t space plus space 1 half a subscript x t squared

  • Where:

    • x = final position in the x direction, measured in straight m

    • x subscript 0 = initial position in the x direction, measured in straight m

    • v subscript x 0 end subscript = initial velocity in the x direction, measured in straight m divided by straight s

    • t = time interval, measured in straight s

    • a subscript x = constant acceleration in the x direction, measured in straight m divided by straight s squared

Kinematic equation 3

  • This equation is used when time is not required

v subscript x superscript 2 space equals space v subscript x 0 end subscript superscript 2 space plus space 2 a subscript x open parentheses x space minus space x subscript 0 close parentheses

  • Where:

    • v subscript x = final velocity in the x direction, measured in straight m divided by straight s

    • v subscript x space 0 end subscript = initial velocity in the x direction, measured in straight m divided by straight s

    • a subscript x = constant acceleration in the x direction, measured in straight m divided by straight s squared

    • x space minus space x subscript 0 = displacement, measured in straight m

Other helpful equations in kinematics

  • Displacement can be calculated using velocity and time when acceleration is not required

increment x space equals space 1 half open parentheses v subscript x 0 end subscript space plus space v subscript x close parentheses t

  • Where:

    • increment x = displacement, measured in straight m

    • v subscript x 0 end subscript = initial velocity in the x direction, measured in straight m divided by straight s

    • v subscript x = final velocity in the x direction, measured in straight m divided by straight s

    • t = time interval, measured in straight s

  • Final position can be calculated when initial velocity is not required using the following equation:

x space equals space x subscript 0 space plus space v subscript x t space minus space 1 half a subscript x t squared

  • Where:

    • x = final position in the x direction, measured in straight m

    • x subscript 0 = initial position in the x direction, measured in straight m

    • v subscript x = final velocity in the x direction, measured in straight m divided by straight s

    • t = time interval, measured in straight s

    • a subscript x = constant acceleration in the x direction, measured in straight m divided by straight s squared

Table of kinematics equations

Equation

Quantity not required

v subscript x space equals space v subscript x 0 end subscript space plus space a subscript x t

open parentheses x space minus space x subscript 0 close parentheses

x space equals space x subscript 0 space plus space v subscript x 0 end subscript t space plus space 1 half a subscript x t squared

v subscript x

v subscript x superscript 2 space equals space v subscript x 0 end subscript superscript 2 space plus space 2 a subscript x open parentheses x space minus space x subscript 0 close parentheses

t

increment x space equals space 1 half open parentheses v subscript x 0 end subscript space plus space v subscript x close parentheses t

a subscript x

x space equals space x subscript 0 space plus space v subscript x t space minus space 1 half a subscript x t squared

v subscript x 0 end subscript

Examiner Tips and Tricks

It can be quicker to choose a kinematic equation based on the quantity that is not required for the calculation. The table above is a handy way to locate the relevant equation based on the quantity not required.

Worked Example

A rock is dropped from a bridge and strikes the water with an impact velocity of 18 space straight m divided by straight s.

Calculate the height of the bridge.

Answer:

Step 1: Check for any implied quantities

  • The rock is dropped; this implies that the initial velocity is zero

  • The rock is in free fall; this implies that the acceleration is equal to the acceleration due to gravity at Earth's surface

Step 2: List the known quantities

  • Taking the positive direction to be downward

  • Initial velocity, v subscript y 0 end subscript space equals space 0 space straight m divided by straight s

  • Final velocity, v subscript y space equals space 18 space straight m divided by straight s

  • Uniform acceleration, a subscript y space equals space 10 space straight m divided by straight s squared

Step 3: Choose the relevant kinematic equation

  • The question asks for the height of the bridge, which is equal to the displacement of the rock

  • The quantity not required in this calculation is time

v subscript y superscript 2 space equals space v subscript y 0 end subscript superscript 2 space plus space 2 a subscript y open parentheses y space minus space y subscript 0 close parentheses

Step 4: Check if any of the quantities cancel to zero

  • Initial velocity is zero, therefore:

v squared subscript y space equals space 0 space plus space 2 a subscript y open parentheses y space minus space y subscript 0 close parentheses

v squared subscript y space equals space 2 a subscript y open parentheses y space minus space y subscript 0 close parentheses

Step 5: Rearrange the equation

  • The question asks for displacement, so make open parentheses y space minus space y subscript 0 close parentheses the subject

open parentheses y space minus space y subscript 0 close parentheses space equals space fraction numerator v subscript y superscript 2 over denominator 2 a subscript y end fraction

Step 6: Substitute in the known values to calculate

increment x space equals space fraction numerator 18 squared over denominator 2 space times space 10 end fraction

increment x space equals space 16.2 space straight m

Examiner Tips and Tricks

There is only one numerical value given in the question above; two of the values are implied. At first glance, it may seem like you don't have enough information to solve the problem. This should be seen as a clue that there is extra information hiding in the wording of the question, or that you may have already calculated a value in a previous question part.

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Leander Oates

Author: Leander Oates

Expertise: Physics

Leander graduated with First-class honours in Science and Education from Sheffield Hallam University. She won the prestigious Lord Robert Winston Solomon Lipson Prize in recognition of her dedication to science and teaching excellence. After teaching and tutoring both science and maths students, Leander now brings this passion for helping young people reach their potential to her work at SME.

Caroline Carroll

Author: Caroline Carroll

Expertise: Physics Subject Lead

Caroline graduated from the University of Nottingham with a degree in Chemistry and Molecular Physics. She spent several years working as an Industrial Chemist in the automotive industry before retraining to teach. Caroline has over 12 years of experience teaching GCSE and A-level chemistry and physics. She is passionate about creating high-quality resources to help students achieve their full potential.