{"id":11293,"date":"2026-07-09T14:28:03","date_gmt":"2026-07-09T14:28:03","guid":{"rendered":"https:\/\/www.myengineeringbuddy.com\/blog\/kinematics-equations\/"},"modified":"2026-07-12T04:24:27","modified_gmt":"2026-07-12T04:24:27","slug":"kinematics-equations","status":"publish","type":"post","link":"https:\/\/www.myengineeringbuddy.com\/blog\/kinematics-equations\/","title":{"rendered":"Kinematics Equations Explained: Formulas, Derivations and Examples"},"content":{"rendered":"\n<div style=\"background-color:#f8f8f8; border-left:4px solid #d0d0d0; padding:12px 16px; margin-bottom:20px;\"><strong>Key Takeaways<\/strong>\n<ul>\n<li>Kinematics studies motion in terms of space and time, without considering forces.<\/li>\n<li>Four core kinematic equations describe motion under constant acceleration.<\/li>\n<li>The same equation framework extends to projectile motion, free fall, and rotation.<\/li>\n<li>Calculus-based derivations confirm all three standard kinematic equations.<\/li>\n<li>Nonuniform acceleration requires differential equations rather than the standard four formulas.<\/li>\n<\/ul><\/div>\n\n<h2>What Is Kinematics<\/h2>\n\n<p>When we observe nature, we see many objects in motion. From experience, we recognize that motion represents the continuous change in the position of an object with time. We study dynamics in two parts: kinematics and kinetics.<\/p>\n\n<p>In kinematics, we study motion in terms of space and time without knowing the cause of motion (the forces that caused the motion). In kinetics, we are also concerned about the cause of motion (forces). Students working through these concepts in broader engineering contexts may find it useful to connect with an <a href=\"https:\/\/www.myengineeringbuddy.com\/subject\/electrical-engineering\/\">electrical engineering tutor<\/a> for related coursework in circuits and electromagnetism.<\/p>\n\n<h2>Kinematic Equations List<\/h2>\n\n<p>${v_x} = {v_{ox}} + {a_x}t$ \u2026(1)<\/p>\n\n<p>$x = {x_0} + {v_{ox}}t + frac{1}{2}{a_x}{t^2}$ \u2026(2)<\/p>\n\n<p>$v_x^2 = v_{ox}^2 + 2{a_x}(x &#8211; {x_0})$ \u2026(3)<\/p>\n\n<p>$(x &#8211; {x_0}) = frac{1}{2}left( {{v_{ox}} + {v_x}} right)t$ \u2026(4)<\/p>\n\n<p><em>Equations (1), (2) &amp; (3) (and sometimes (4)) are the equations of motion with constant acceleration. These are also known as equations of kinematics or kinematic equations.<\/em><\/p>\n\n<p>${v_x}:$ velocity of particle along x-axis at time t<\/p>\n\n<p>${v_{ox}}:$ velocity of particle along x-axis at t = 0<\/p>\n\n<p>${a_x}:$ constant acceleration of particle along x-axis<\/p>\n\n<p>x: position of the particle at time t on the x-axis<\/p>\n\n<p>${x_o}:$ position of particle at time t = 0 on the x-axis<\/p>\n\n<p>t: it represents general time.<\/p>\n\n<h2>How to Use Kinematic Equations<\/h2>\n\n<p>We have the first equation of kinematics ${v_x} = {v_{ox}} + {a_x}t$. This expression enables us to determine the velocity at any time t if the initial velocity, the acceleration, and the elapsed time are given.<\/p>\n\n<p>We have the second equation of kinematics $x = {x_o} + {v_{ox}}t + frac{1}{2}{a_x}{t^2}$. This expression enables us to determine the distance traveled during any time t if the initial velocity, acceleration, and time elapsed are known.<\/p>\n\n<p>We have a third equation that lets us determine the final velocity if the initial velocity, acceleration, and displacement are known. Those studying related applied mechanics will find these principles also underpin <a href=\"https:\/\/www.myengineeringbuddy.com\/subject\/structural-analysis\/\">structural analysis<\/a> coursework.<\/p>\n\n<h2>Solved Problems on Each Kinematic Equation<\/h2>\n\n<h3>Example 1<\/h3>\n\n<p><strong>An athlete takes 20 sec to reach his maximum speed of 18.0 km\/h. What is the magnitude of average acceleration?<\/strong><\/p>\n\n<p>Ans. The athlete is running along the x-axis with constant acceleration; we can use<\/p>\n\n<p>${v_x} = {v_{ox}} + {a_x}t$<\/p>\n\n<p>$18 times frac{5}{18},text{m\/s} = 0 + {a_x} times 2,text{sec}$<\/p>\n\n<p>$5,text{m\/s} = {a_x} times 2,text{sec}$<\/p>\n\n<p>${a_x} = 2.5,text{m\/s}^2$<\/p>\n\n<p>Therefore, the magnitude of the average acceleration is 2.5 m\/s<sup>2<\/sup>.<\/p>\n\n<h3>Example 2<\/h3>\n\n<p><strong>An object having a velocity of 4.0 m\/s is accelerated at the rate of 1.2 m\/s<sup>2<\/sup> for 5.0 sec. Find the distance traveled during the period of acceleration.<\/strong><\/p>\n\n<p>Ans. Suppose the object is moving along the +ve x-axis. Assuming that the initial location of the particle is at the origin.<\/p>\n\n<p>Here, $x &#8211; {x_o} = x &#8211; 0 = x$<\/p>\n\n<p>${v_{ox}} = 4.0,text{m\/s}$, ${a_x} = 1.2,text{m\/s}^2$, $t = 5,text{sec}$.<\/p>\n\n<p>Using kinematics equation,<\/p>\n\n<p>$x = {x_o} + {v_{ox}}t + frac{1}{2}{a_x}{t^2}$<\/p>\n\n<p>$Rightarrow x &#8211; {x_o} = {v_{ox}}t + frac{1}{2}{a_x}{t^2}$<\/p>\n\n<p>$Rightarrow x = left[ {4 times 5 + frac{1}{2} times 1.2 times {(5)}^2} right]text{meter}$<\/p>\n\n<p>$Rightarrow x = left[ {20 + frac{1.2}{2} times 25} right]text{meter}$<\/p>\n\n<p>$Rightarrow x = 35,text{meter}$<\/p>\n\n<h3>Example 3<\/h3>\n\n<p><strong>A person traveling at 43.2 km\/h applies the brake giving a deceleration of 6.0 m\/s<sup>2<\/sup> to his scooter. How far will it travel before stopping?<\/strong><\/p>\n\n<p>Ans. Suppose the person is traveling along the positive x-axis and initially that person was at the origin (x = 0). He is moving along a straight line with constant negative acceleration.<\/p>\n\n<p>Using kinematics equation,<\/p>\n\n<p>$v_x^2 = v_{ox}^2 + 2{a_x}(x &#8211; {x_o})$<\/p>\n\n<p>Here, ${v_x} = 0$, ${v_{ox}} = 12,text{m\/s}$, ${a_x} = -6.0,text{m\/s}^2$, $(x &#8211; {x_o}) = ?$<\/p>\n\n<p>$Rightarrow quad {0^2} = {(12,text{m\/s})^2} + 2(-6,text{m\/s}^2)(x)$ [because $x &#8211; {x_o} = x &#8211; 0 = x$]<\/p>\n\n<p>$Rightarrow quad 12x = 144$<\/p>\n\n<p>$Rightarrow quad x = 12,text{m}$<\/p>\n\n<p>So, the person will travel a distance of 12 m before stopping.<\/p>\n\n<h2>Kinematic Equations for Projectile Motion<\/h2>\n\n<p>The projectile&#8217;s motion can be discussed separately for the horizontal and vertical parts. We take origin $({x_o} = 0,,{y_o} = 0)$ at the point of projection. The instant when the particle was projected is taken as t = 0.<\/p>\n\n<p>The plane of motion is taken as the x-y plane, the horizontal line OX as the x-axis and the vertical line OY as the y-axis. Vertically upward direction is taken as the positive direction of the y-axis.<\/p>\n\n<img class=\"lazyload\" decoding=\"async\" src=\"data:image\/gif;base64,R0lGODlhAQABAAAAACH5BAEKAAEALAAAAAABAAEAAAICTAEAOw==\" data-orig-src=\"https:\/\/myengineeringbuddy.com\/wp-content\/uploads\/2022\/06\/Fig-1-300x217.png\" alt=\"Projectile motion diagram showing horizontal and vertical components of motion\">\n\n<p>We have, ${v_{ox}} = {v_o}cos{alpha_o}$, ${a_x} = 0$<\/p>\n\n<p>${v_{oy}} = {v_o}sin{alpha_o}$, ${a_y} = -g$<\/p>\n\n<p><strong>Horizontal motion<\/strong><\/p>\n\n<p>${a_x} = 0$, we have<\/p>\n\n<p>${v_x} = {v_{ox}} + {a_x}t$<\/p>\n\n<p>$Rightarrow quad {v_x} = {v_{ox}}$<\/p>\n\n<p>Again, $x &#8211; {x_0} = {v_{ox}}t + frac{1}{2}{a_x}{t^2}$ [Here, ${a_x} = 0$, ${v_{ox}} = {v_o}sin{alpha_o}$, ${x_o} = 0$]<\/p>\n\n<p>$Rightarrow quad x &#8211; {x_0} = {v_{ox}}t$<\/p>\n\n<p>$Rightarrow quad x = {x_0} + {v_{ox}}t$<\/p>\n\n<p><strong>Vertical Motion<\/strong><\/p>\n\n<p>The acceleration of the particle is g in the downward direction. Thus ${a_y} = -g$ [Here, ${a_y} = -g$, ${v_{oy}} = {v_o}sin{alpha_o}$, ${y_o} = 0$]<\/p>\n\n<p>We can write, ${v_y} = {v_{oy}} + {a_y}t$<\/p>\n\n<p>$Rightarrow {v_y} = {v_{oy}} &#8211; gt$<\/p>\n\n<p>and $y &#8211; {y_o} = {v_{oy}}t + frac{1}{2}{a_y}{t^2}$<\/p>\n\n<p>$Rightarrow y = {y_o} + {v_{oy}}t + frac{1}{2}{a_y}{t^2}$<\/p>\n\n<p>Again, $v_y^2 = v_{oy}^2 + 2{a_y}(y &#8211; {y_o})$<\/p>\n\n<h2>Kinematic Equations for Free Fall<\/h2>\n\n<p>A typical example of motion in a straight line with constant acceleration is the free fall of a body near the earth&#8217;s surface.<\/p>\n\n<img class=\"lazyload\" decoding=\"async\" src=\"data:image\/gif;base64,R0lGODlhAQABAAAAACH5BAEKAAEALAAAAAABAAEAAAICTAEAOw==\" data-orig-src=\"https:\/\/myengineeringbuddy.com\/wp-content\/uploads\/2022\/06\/Fig-5-300x244.png\" alt=\"Diagram illustrating free-fall motion with upward positive direction and downward acceleration due to gravity\">\n\n<p>Here, we have taken the origin O at the starting point and the upward direction as positive. Assuming the initial position coordinate ${y_o}$ and initial velocity is ${v_{oy}}$.<\/p>\n\n<p>The y-acceleration is downward (i.e., along the \u2013ve y-direction) therefore, ${a_y} = -g$.<\/p>\n\n<p>After time t, its position coordinate can be written as<\/p>\n\n<p>$y &#8211; {y_o} = {v_{oy}}t + frac{1}{2}{a_y}{t^2}$<\/p>\n\n<p>$Rightarrow y = {y_o} + {v_{oy}}t + frac{1}{2}(-g){t^2}$<\/p>\n\n<p>Here in this case, ${y_o} = 0$, ${v_{oy}} = 0$<\/p>\n\n<p>After time t, its velocity coordinate ${v_y} = {v_{oy}} + {a_y}t$<\/p>\n\n<p>$Rightarrow {v_y} = {v_{oy}} &#8211; gt$<\/p>\n\n<p>Here ${v_{oy}} = 0$ and ${a_y} = -g$<\/p>\n\n<p>We can also write, $v_y^2 = v_{oy}^2 + 2{a_y}(y &#8211; {y_o})$<\/p>\n\n<p>$Rightarrow v_y^2 = v_{oy}^2 + 2(-g)(y &#8211; {y_o})$<\/p>\n\n<p>Here, ${a_y} = -g$ and ${y_o} = 0$. Students in life-sciences programs who encounter free-fall and motion concepts in biomechanics may benefit from working with a <a href=\"https:\/\/www.myengineeringbuddy.com\/subject\/biomedical-engineering\/\">biomedical engineering tutor<\/a> for deeper applied context.<\/p>\n\n<h2>Kinematic Equations for Rotation<\/h2>\n\n<p>When a body revolves about a stationary axis (usually z-axis), the position of the body is described by an angular position coordinate $theta$. If the angular acceleration is constant, then $theta$, ${omega_2}$ and ${alpha_2}$ are related by simple kinematic equations.<\/p>\n\n<img class=\"lazyload\" decoding=\"async\" src=\"data:image\/gif;base64,R0lGODlhAQABAAAAACH5BAEKAAEALAAAAAABAAEAAAICTAEAOw==\" data-orig-src=\"https:\/\/myengineeringbuddy.com\/wp-content\/uploads\/2022\/06\/Fig-4-295x300.png\" alt=\"Diagram of a rotating body about a stationary z-axis showing angular position and acceleration\">\n\n<p><strong>Kinematic equations in the case of rotation are as follows:<\/strong><\/p>\n\n<p>$theta &#8211; {theta_o} = {omega_{oz}} + frac{1}{2}{alpha_z}{t^2}$<\/p>\n\n<p>$Rightarrow {omega_z} = {omega_{oz}} + {alpha_z}t$<\/p>\n\n<p>$Rightarrow omega_z^2 = omega_{oz}^2 + 2{alpha_z}(theta &#8211; {theta_o})$<\/p>\n\n<p>Here, ${theta_o}:$ Angular position of the body at time t = 0<\/p>\n\n<p>$theta:$ Angular position of the body at time t = t<\/p>\n\n<p>${alpha_z}:$ Constant angular acceleration of the body<\/p>\n\n<p>${omega_{oz}}:$ Angular velocity of the body at time t = 0<\/p>\n\n<p>${omega_z}:$ Angular velocity of the body at time t = t<\/p>\n\n<p>Those preparing for standardized engineering exams that include rotational dynamics can find structured support through <a href=\"https:\/\/www.myengineeringbuddy.com\/subject\/test-preparation\/\">test preparation tutoring<\/a>.<\/p>\n\n<h2>Kinematic Equation Derivation Using Calculus<\/h2>\n\n<p>Let a particle be moving along the x-axis. Initially, the particle was at $x = {x_o}$ with initial velocity ${v_{ox}}$.<\/p>\n\n<img class=\"lazyload\" decoding=\"async\" src=\"data:image\/gif;base64,R0lGODlhAQABAAAAACH5BAEKAAEALAAAAAABAAEAAAICTAEAOw==\" data-orig-src=\"https:\/\/myengineeringbuddy.com\/wp-content\/uploads\/2022\/06\/Fig-3-300x125.png\" alt=\"Diagram of a particle moving along the x-axis used for kinematic equation derivation\">\n\n<p>The particle has a constant acceleration along the x-axis. From the definition of acceleration,<\/p>\n\n<p>${a_x} = frac{d{v_x}}{dt} quad Rightarrow quad d{v_x} = {a_x},dt$<\/p>\n\n<p>Integrating both sides with proper limits,<\/p>\n\n<p>$intlimits_{{v_{ox}}}^{{v_x}} {d{v_x}} = intlimits_0^t {{a_x}},dt$<\/p>\n\n<p>$Rightarrow {v_x} &#8211; {v_{ox}} = {a_x}(t &#8211; 0)$<\/p>\n\n<p>$Rightarrow quad {v_x} = {v_{ox}} + {a_x}t$<\/p>\n\n<p>Again from the definition of velocity,<\/p>\n\n<p>${v_x} = frac{dx}{dt} Rightarrow frac{dx}{dt} = {v_{ox}} + {a_x}t$<\/p>\n\n<p>$Rightarrow dx = ({v_{ox}} + {a_x}t),dt$<\/p>\n\n<p>Now, integrating both sides with proper limits,<\/p>\n\n<p>$intlimits_{{x_o}}^x {dx} = intlimits_0^t {({v_{ox}} + {a_x}t),dt}$<\/p>\n\n<p>$Rightarrow x &#8211; {x_o} = {v_{ox}}(t &#8211; 0) + {a_x}left[ frac{t^2}{2} right]_0^t$<\/p>\n\n<p>$Rightarrow quad x &#8211; {x_o} = {v_{ox}}t + frac{1}{2}{a_x}{t^2}$<\/p>\n\n<p>Again starting from, ${a_x} = frac{d{v_x}}{dt}$<\/p>\n\n<p>$frac{d{v_x}}{dx} cdot frac{dx}{dt} = {a_x} Rightarrow {v_x} cdot frac{d{v_x}}{dt} = {a_x}$<\/p>\n\n<p>$Rightarrow {v_x} cdot d{v_x} = {a_x}(dx)$<\/p>\n\n<p>Now, integrating both sides,<\/p>\n\n<p>$intlimits_{{v_{ox}}}^{{v_x}} {{v_x},dx} = intlimits_{{x_o}}^x {{a_x}(dx)} Rightarrow left[ frac{v_x^2}{2} right]_{{v_{ox}}}^{{v_x}} = {a_x}(x &#8211; {x_o})$<\/p>\n\n<p>$Rightarrow quad v_x^2 = v_{ox}^2 + 2a_x(x &#8211; {x_o})$<\/p>\n\n<p>Students who find calculus-based derivations challenging alongside their chemical engineering coursework can get support from a <a href=\"https:\/\/www.myengineeringbuddy.com\/subject\/chemical-engineering\/\">chemical engineering tutor<\/a> for the broader mathematical methods involved.<\/p>\n\n<h3>Example<\/h3>\n\n<p><strong>A particle starts with an initial velocity of 2.5 m\/s along the positive x-direction and accelerates uniformly at the rate of 0.5 m\/s<sup>2<\/sup>. Find the distance traveled by it in the first two seconds.<\/strong><\/p>\n\n<p>Ans. Let the particle start its motion from the origin with an initial velocity of ${v_{ox}} = 2.5,text{m\/s}$, ${a_x} = 0.5,text{m\/s}^2$<\/p>\n\n<p>Using, ${a_x} = frac{d{v_x}}{dt} Rightarrow d{v_x} = {a_x},dt$<\/p>\n\n<img class=\"lazyload\" decoding=\"async\" src=\"data:image\/gif;base64,R0lGODlhAQABAAAAACH5BAEKAAEALAAAAAABAAEAAAICTAEAOw==\" data-orig-src=\"https:\/\/myengineeringbuddy.com\/wp-content\/uploads\/2022\/06\/Fig-2-300x130.png\" alt=\"Graph showing velocity versus time for a uniformly accelerating particle\">\n\n<p>Now, integrating both sides,<\/p>\n\n<p>$intlimits_{{v_{ox}}}^{{v_x}} {d{v_x}} = {a_x}intlimits_0^t {dt}$<\/p>\n\n<p>$Rightarrow {v_x} &#8211; 2.5 = frac{1}{2}(t &#8211; 0) Rightarrow {v_x} = 2.5 + frac{t}{2}$<\/p>\n\n<p>Again using ${v_x} = frac{dx}{dt}$<\/p>\n\n<p>$Rightarrow frac{dx}{dt} = 2.5 + frac{t}{2} Rightarrow dx = left( 2.5 + frac{t}{2} right)dt$<\/p>\n\n<p>Now, integrating both sides,<\/p>\n\n<p>$intlimits_{{x_o}}^x {dx} = intlimits_0^2 {left( 2.5 + frac{t}{2} right)dt}$<\/p>\n\n<p>$Rightarrow x &#8211; {x_o} = 2.5(2 &#8211; 0) + frac{1}{2}frac{(2)^2}{2}$<\/p>\n\n<p>Since the particle does not turn back, the distance traveled equals 6 m.<\/p>\n\n<h2>Kinematics Equations for Nonuniform Acceleration<\/h2>\n\n<p>If the acceleration of a particle during its motion does not remain constant, then both motion and acceleration are nonuniform.<\/p>\n\n<p>As by definition, ${v_x} = frac{dx}{dt}$, ${a_x} = frac{d^2x}{dt^2}$<\/p>\n\n<p>If x is a function of time, the second derivative of displacement with respect to time represents acceleration. If velocity is a function of position, then by chain rule of differentiation,<\/p>\n\n<p>${a_x} = frac{d{v_x}}{dt} = frac{d{v_x}}{dx} cdot frac{dx}{dt}$<\/p>\n\n<p>$Rightarrow {a_x} = vfrac{d{v_x}}{dx}$<\/p>\n\n<p>For students exploring how these differential methods connect to real-world language and communication modeling, our overview of <a href=\"https:\/\/www.myengineeringbuddy.com\/blog\/communication-media-studies-tutoring-empowering-tomorrows-media-savvy-students\/\">communication and media studies tutoring<\/a> shows how analytical thinking transfers across disciplines.<\/p>\n\n<h3>Example<\/h3>\n\n<p><strong>The acceleration ${a_x}$ of a particle moving in a straight line varies with its displacement $x = 2{a_x}$. The velocity of the particle is zero at zero displacement. Find the corresponding velocity displacement equation.<\/strong><\/p>\n\n<p>Ans. We have, ${a_x} = 2x$<\/p>\n\n<p>$Rightarrow frac{d{v_x}}{dx} cdot frac{dx}{dt} = 2x$<\/p>\n\n<p>$Rightarrow {v_x}(d{v_x}) = 2x(dx)$<\/p>\n\n<p>Now, integrating both sides with proper limits,<\/p>\n\n<p>$intlimits_0^{{v_x}} {{v_x}(d{v_x})} = intlimits_0^x {2x(dx)}$<\/p>\n\n<p>$Rightarrow frac{v_x^2}{2} = 2 cdot frac{x^2}{2} Rightarrow v_x^2 = 2x^2$<\/p>\n\n<p>$Rightarrow quad {v_x} = pmsqrt{2},x$<\/p>\n\n<h2>Resources<\/h2>\n\n<h3>Kinematics Equations Calculator<\/h3>\n\n<p>While there are many online calculators on kinematics equations, we suggest you use <a href=\"https:\/\/physicscatalyst.com\/calculators\/physics\/kinematics-calculator.php\" target=\"_blank\" rel=\"noopener\">Physics Catalyst&#8217;s calculator<\/a> on the same.<\/p>\n\n<h3>Online Tutoring in Kinematics Equations and Related Physics Topics<\/h3>\n\n<p>My Engineering Buddy provides online physics tutoring that covers topics like kinematics equations and much more.<\/p>\n\n<p>Related: <a href=\"https:\/\/www.myengineeringbuddy.com\/motion-in-two-dimensions-2d-in-physics\/\">Motion in two dimensions 2d in physics<\/a> | <a href=\"https:\/\/www.myengineeringbuddy.com\/significant-figures\/\">Significant figures<\/a> | <a href=\"https:\/\/www.myengineeringbuddy.com\/homework-help\/physics-homework-help\/\">Physics Homework Help<\/a> | <a href=\"https:\/\/www.myengineeringbuddy.com\/online-tutoring\/online-physics-tutor\/\">Physics Tutor Online<\/a><\/p>\n\n<p>Learners who want to see how structured online support works across different subjects can read about <a href=\"https:\/\/www.myengineeringbuddy.com\/blog\/language-tutoring-personalized-support-for-students\/\">language tutoring and personalized student support<\/a>, or explore how technical platforms are taught in our guide to <a href=\"https:\/\/www.myengineeringbuddy.com\/blog\/apache-tutoring-online-master-apache-technologies\/\">mastering Apache technologies online<\/a>. For a look at how AI writing tools compare to human support, see our <a href=\"https:\/\/www.myengineeringbuddy.com\/blog\/paraphrasing-tool-ai-reviews-alternatives-pricing-offerings\/\">review of paraphrasing tools, AI alternatives, and pricing<\/a>.<\/p>\n\n<h2>Related Reading<\/h2>\n<ul>\n<li><a href=\"https:\/\/www.myengineeringbuddy.com\/blog\/front-end-development-tutoring-empowering-students-for-success-in-a-digital-world\/\">Front-End Development Tutoring: Empowering Students for Success in a Digital World<\/a><\/li>\n<li><a href=\"https:\/\/www.myengineeringbuddy.com\/blog\/forensic-science-tutoring-guide\/\">Forensic Science Tutoring Guide<\/a><\/li>\n<li><a href=\"https:\/\/www.myengineeringbuddy.com\/blog\/engineering-presentation-templates\/\">Engineering Presentation Templates<\/a><\/li>\n<li><a href=\"https:\/\/www.myengineeringbuddy.com\/blog\/psychology-tutoring-a-complete-guide-for-students-and-parents\/\">Psychology Tutoring: A Complete Guide for Students and Parents<\/a><\/li>\n<\/ul>\n","protected":false},"excerpt":{"rendered":"<p>Key Takeaways Kinematics studies motion in terms of space and  [&#8230;]<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[1],"tags":[],"class_list":["post-11293","post","type-post","status-publish","format-standard","hentry","category-uncategorized"],"_links":{"self":[{"href":"https:\/\/www.myengineeringbuddy.com\/blog\/wp-json\/wp\/v2\/posts\/11293","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/www.myengineeringbuddy.com\/blog\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/www.myengineeringbuddy.com\/blog\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/www.myengineeringbuddy.com\/blog\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/www.myengineeringbuddy.com\/blog\/wp-json\/wp\/v2\/comments?post=11293"}],"version-history":[{"count":1,"href":"https:\/\/www.myengineeringbuddy.com\/blog\/wp-json\/wp\/v2\/posts\/11293\/revisions"}],"predecessor-version":[{"id":12169,"href":"https:\/\/www.myengineeringbuddy.com\/blog\/wp-json\/wp\/v2\/posts\/11293\/revisions\/12169"}],"wp:attachment":[{"href":"https:\/\/www.myengineeringbuddy.com\/blog\/wp-json\/wp\/v2\/media?parent=11293"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.myengineeringbuddy.com\/blog\/wp-json\/wp\/v2\/categories?post=11293"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.myengineeringbuddy.com\/blog\/wp-json\/wp\/v2\/tags?post=11293"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}