{"id":11292,"date":"2026-07-09T14:28:03","date_gmt":"2026-07-09T14:28:03","guid":{"rendered":"https:\/\/www.myengineeringbuddy.com\/blog\/motion-in-two-dimensions-2d-in-physics\/"},"modified":"2026-07-12T04:24:27","modified_gmt":"2026-07-12T04:24:27","slug":"motion-in-two-dimensions-2d-in-physics","status":"publish","type":"post","link":"https:\/\/www.myengineeringbuddy.com\/blog\/motion-in-two-dimensions-2d-in-physics\/","title":{"rendered":"Motion in Two Dimensions (2D) in Physics: A Complete Guide"},"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>2D motion occurs in a plane and can be split into independent x and y components.<\/li>\n<li>Projectile and circular motion are the most common examples of 2D motion.<\/li>\n<li>Kinematics equations apply separately to each axis when acceleration is constant.<\/li>\n<li>The displacement vector in 2D is \u0394r = \u0394x \u00ee + \u0394y \u0135.<\/li>\n<li>In circular motion, force and momentum vectors are always perpendicular (90\u00b0).<\/li>\n<\/ul><\/div>\n\n<p>Before we begin our discussion on motion in two dimensions (2D), let us answer these 3 questions. Students who want deeper support with these concepts can work with an <a href=\"https:\/\/www.myengineeringbuddy.com\/subject\/physics\/\">online physics tutor<\/a> for personalised guidance.<\/p>\n\n<h2>What is motion in 2 dimensions (2D)?<\/h2>\n\n<p>The motion of a particle in a plane is called 2-dimensional motion.<\/p>\n\n<p>Example: An insect crawling on your laptop or mobile screen but not flying above it.<\/p>\n\n<h2>Can there be motion in 2 dimensions (2D)?<\/h2>\n\n<p>Yes! Motion in two dimensions is possible. It is pretty common to have 2-dimensional motion.<\/p>\n\n<p>Some examples of 2D motion are Projectile motion, Circular motion, motion in a plane (planar motion), etc. Those studying for standardised exams can find targeted support through an <a href=\"https:\/\/www.myengineeringbuddy.com\/subject\/ap-physics\/\">AP Physics tutor<\/a>.<\/p>\n\n<h2>How do we tackle the motion in 2D?<\/h2>\n\n<p>Break the motion into two separate components, x, and y, and treat each component as a separate motion. But since it is a motion of a single object so, at one time, the body can only be in one place. Using this constraint, we can correlate the two separate equations and solve them for unknowns.<\/p>\n\n<h3>Why does this strategy work?<\/h3>\n\n<p>The concept of motion, when a particle moves along a straight line, can be used for motion in a plane. When the motion of a particle is in-plane, we generally consider the plane of motion as the x-y plane. We choose the origin at the place from where the motion starts, and then we consider motion along any two convenient mutually perpendicular directions as one-dimensional motion. Motion in two perpendicular directions is chosen as the x and y-axes.<\/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\/1-removebg-preview-2-300x209.png\" alt=\"Diagram showing motion in x-y plane with perpendicular axes\">\n\n<h3>Position<\/h3>\n\n<p>Suppose a particle moves in a plane along the curve as shown in figure. At time ${t_1}$ the particle is at point P and at some later time ${t_2}$, the particle is at Q. As the particle moves from P to Q in the time interval $Delta t = {t_2} &#8211; {t_1}$, the position vector changes from ${bf{r}}_1$ to ${bf{r}}_2$.<\/p>\n\n<p>From the triangle rule of vector addition, we can write<\/p>\n\n<p>${bf{OP}} + {bf{PQ}} = {bf{OQ}}quad Rightarrow quad {bf{PQ}} = {bf{OQ}} &#8211; {bf{OP}}quad Rightarrow quad {bf{PQ}} = {{bf{r}}_2} &#8211; {{bf{r}}_1}$<\/p>\n\n<p>${bf{Delta r}} = {{bf{r}}_2} &#8211; {{bf{r}}_1}$<\/p>\n\n<p>Here, ${{bf{r}}_1} = {x_1}hat i + {y_1}hat j$ and ${{bf{r}}_2} = {x_2}hat i + {y_2}hat j$<\/p>\n\n<p>${{{bf{r}}_2} &#8211; {{bf{r}}_1} = ({x_2} &#8211; {x_1})hat i + ({y_2} &#8211; {y_1})hat j = Delta xhat i + Delta yhat j}$<\/p>\n\n<p>$Delta {bf{r}} = Delta xhat i + Delta yhat j$; where $Delta x = ({x_2} &#8211; {x_1})$ and $Delta y = ({y_2} &#8211; {y_1})$<\/p>\n\n<p>So we see here that motion in an x-y plane can be simplified into two separate motions; one in the x-direction and one in the y-direction. Hence the strategy that we discussed to tackle the 2D motion works.<\/p>\n\n<h3>Velocity<\/h3>\n\n<p>We define the particle&#8217;s average velocity during the time interval $Delta t$ as the ratios of the displacement of that time interval.<\/p>\n\n<p>${{{bf{v}}_{av}} = frac{{{bf{Delta r}}}}{{Delta t}} = frac{{Delta x,hat i + Delta y,hat j}}{{Delta t}}}$<\/p>\n\n<p>${{{bf{v}}_{av}} = frac{{{bf{Delta r}}}}{{Delta t}} = frac{{Delta x,hat i}}{{Delta t}} + frac{{Delta y,hat j}}{{Delta t}}}$<\/p>\n\n<p>${mathop {lim }limits_{Delta t to 0} frac{{{bf{Delta r}}}}{{Delta t}} = mathop {lim }limits_{Delta t to 0} left( {frac{{Delta x}}{{Delta t}}} right)hat i + mathop {lim }limits_{Delta t to 0} left( {frac{{Delta y}}{{Delta t}}} right)hat j}$<\/p>\n\n<p>${{{bf{v}}_{inst.}} = {v_x}hat i + {v_y}hat j}$<\/p>\n\n<p>${{{bf{v}}_{inst.}}}$ is instantaneous velocity.<\/p>\n\n<h3>Acceleration<\/h3>\n\n<p>The average acceleration of a particle as it moves from P to Q is defined as the ratio of the change in the instantaneous velocity vector.<\/p>\n\n<p>${{{bf{a}}_{av}} = frac{{{bf{Delta v}}}}{{Delta t}}}$<\/p>\n\n<p>and ${{{bf{a}}_{{rm{inst}}{rm{.}}}} = mathop {lim }limits_{Delta t to 0} frac{{{bf{Delta v}}}}{{Delta t}} = frac{{d{bf{v}}}}{{dt}}}$<\/p>\n\n<p>${{{bf{a}}_{{rm{inst}}{rm{.}}}} = frac{d}{{dt}}left( {{v_x}hat i + {v_y}hat j} right) = frac{{d{v_x}}}{{dt}}hat i + frac{{d{v_y}}}{{dt}}hat j}$<\/p>\n\n<p>${{{bf{a}}_{{rm{inst}}{rm{.}}}} = {bf{a}} = {a_x}hat i + {a_y}hat j}$<\/p>\n\n<p>Because ${bf{a}}$ is normally assumed as constant, its component ${{{bf{a}}_x}}$ and ${{{bf{a}}_y}}$ are also constants. Therefore we can apply the kinematics equations to the x and y components of the velocity vector and displacement vector as follows.<\/p>\n\n<p>${bf{V}} = {{{bf{V}}_0}} + {bf{a}}tleft{ {begin{array}{*{20}{c}}{{v_x} = {{({v_x})}_0} + {a_x}t}\\{{v_y} = {{({v_y})}_0} + {a_y}t}end{array}} right.$<\/p>\n\n<p>${bf{r}} = {{{bf{V}}_0}} + frac{1}{2}{bf{a}}{t^2}left{ {begin{array}{*{20}{c}}{x = {{({v_x})}_0}t + frac{1}{2}{a_x}{t^2}}\\{y = {{({v_y})}_0}t + frac{1}{2}{a_y}{t^2}}end{array}} right.$<\/p>\n\n<p>${v^2} = u_0^2 + 2asleft{ {begin{array}{*{20}{c}}{v_x^2 = ({v_x})_0^2 + 2{a_x}{s_x}}\\{v_y^2 = ({v_y})_0^2 + 2{a_y}{s_y}}end{array}} right.$<\/p>\n\n<p>In other words, two-dimensional motion with constant acceleration is equivalent to two independent motions in the x and y directions having constant accelerations ${a_x}$ and ${a_y}$.<\/p>\n\n<h2>Examples of two-dimensional motion<\/h2>\n\n<h3>Projectile motion<\/h3>\n\n<p>When an object is thrown at some angle to the horizontal, the projectile&#8217;s motion will be in the horizontal and vertical directions. Suppose the motion of the projectile is in the x-y plane of the coordinate system. The object&#8217;s motion will be along two mutually perpendicular directions together, say the x and y axes of the coordinate system.<\/p>\n\n<p>Consider motion along the horizontal direction (x-axis) and motion along the vertical direction (y-axis). Students preparing for IB exams can get additional support from an <a href=\"https:\/\/www.myengineeringbuddy.com\/subject\/ib-physics-hl-sl\/\">IB Physics HL\/SL tutor<\/a>.<\/p>\n\n<p>$begin{array}{*{20}{c}}{{v_x} = ucos theta }\\{x = x}\\{{a_x} = 0}end{array}$<\/p>\n\n<p>$begin{array}{*{20}{c}}{{v_y} = usin theta }\\{y = y}\\{{a_y} = &#8211; g}end{array}$<\/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\/2-removebg-preview-1-2-300x263.png\" alt=\"Diagram illustrating projectile motion in x-y plane with horizontal and vertical components\">\n\n<p>So, the equation of kinematics for one-dimensional motion can be used.<\/p>\n\n<p>Watch this <a href=\"https:\/\/www.youtube.com\/watch?v=BZwizmCI_g0\" target=\"_blank\" rel=\"noopener\">video on 2D motion<\/a>, precisely projectile motion, to better understand the concept.<\/p>\n\n<p><div class=\"video-shortcode\"><iframe title=\"Kinematic Equations 2D\" width=\"1200\" height=\"675\" src=\"https:\/\/www.youtube.com\/embed\/BZwizmCI_g0?feature=oembed\" frameborder=\"0\" allow=\"accelerometer; autoplay; clipboard-write; encrypted-media; gyroscope; picture-in-picture; web-share\" referrerpolicy=\"strict-origin-when-cross-origin\" allowfullscreen><\/iframe><\/div><\/p>\n\n<p>For a broader look at how these mechanics topics appear in exams, see this guide to <a href=\"https:\/\/www.myengineeringbuddy.com\/blog\/ap-physics-1-exam-prep-2026-advanced-mastering-mechanics-circuits\/\">AP Physics 1 exam prep: mastering mechanics and circuits<\/a>.<\/p>\n\n<h3>Circular motion<\/h3>\n\n<p>Circular motion is a type of motion in a plane in which a particle moves around a fixed point such that its distance from a fixed point is constant.<\/p>\n\n<p>Suppose a particle moves on a circular path uniformly, as shown in the figure.<\/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\/3-removebg-preview-2-300x283.png\" alt=\"Diagram of a particle moving on a circular path with radius R and angle theta\">\n\n<p>Here, R is the distance from the fixed point, the center of this circle, to the particle.<\/p>\n\n<p>${bf{R}} = Rcos theta hat i + Rsin theta hat j$<\/p>\n\n<p>$x = Rcos theta ,;,,,y = Rsin theta$<\/p>\n\n<p>${bf{v}} = &#8211; vsin theta hat i + vcos theta hat j$<\/p>\n\n<p>${v_x} = ( &#8211; vsin theta ),;,,{v_y} = (vcos theta )$<\/p>\n\n<p>${bf{a}} = acos theta ( &#8211; hat i) + asin theta ( &#8211; hat j)$<\/p>\n\n<p>${a_x} = &#8211; acos theta$<\/p>\n\n<p>${a_y} = &#8211; asin theta$<\/p>\n\n<p>Click the link for an <a href=\"https:\/\/www.myengineeringbuddy.com\/circular-motion\/\">in-depth guide on circular motion<\/a>. Those curious about related celestial mechanics may also find value in exploring topics with an <a href=\"https:\/\/www.myengineeringbuddy.com\/subject\/astronomy\/\">astronomy tutor<\/a>.<\/p>\n\n<h2>Solved problems on two-dimensional motion<\/h2>\n\n<h3>Example 1<\/h3>\n\n<p>A helicopter on a flood relief mission flying horizontally with speed u at an altitude H has to drop a food packet for a victim standing on the ground. At what distance from the victim should the packet be dropped? The victim stands in the vertical plane of the helicopter&#8217;s motion.<\/p>\n\n<p><strong>Solution<\/strong><\/p>\n\n<p>We can solve this problem by using the concept involved in motion in a plane.<\/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\/4-removebg-preview-2-300x122.png\" alt=\"Diagram showing helicopter dropping food packet at horizontal distance D from victim at altitude H\">\n\n<p>Suppose the velocity of the food packet at the time of release is v and is horizontal. The vertical velocity at the time of release is zero.<\/p>\n\n<p>The motion of the food packet along the y-axis.<\/p>\n\n<p>$y = {u_y} + frac{1}{2}{a_y}{t^2}$<\/p>\n\n<p>Here, $y = H,;,,{u_y} = 0,;,,{a_y} = g$<\/p>\n\n<p>$H = 0 + frac{1}{2}g{t^2}quad Rightarrow quad t = sqrt {frac{{2H}}{g}}$<\/p>\n\n<p>Again, the motion of the food packet along the x-axis,<\/p>\n\n<p>$x = {u_x}t + frac{1}{2}{a_x}{t^2}$<\/p>\n\n<p>Here, let $x = D,;,,{u_x} = u,;,,{a_x} = 0$<\/p>\n\n<p>$Rightarrow quad D = ut + 0quad Rightarrow quad D = ut$<\/p>\n\n<p>$D = usqrt {frac{{2H}}{g}}$<\/p>\n\n<p>Therefore, the required distance equal to $D = usqrt {frac{{2H}}{g}}$.<\/p>\n\n<h3>Example 2<\/h3>\n\n<p>A projectile is given an initial velocity of $(hat i + 2hat j)$ m\/s where $hat i$ is along the ground and $hat j$ is along the vertical of $g = 10,{rm{m\/}}{{rm{s}}^{rm{2}}}$ then find equation of trajectory.<\/p>\n\n<p><strong>Solution<\/strong><\/p>\n\n<p>This problem is based on the concept of motion in a plane.<\/p>\n\n<p>We can study the two-dimensional motion as one-dimensional along the x-axis and y-axis since its superposition gives the resultant motion.<\/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\/5-removebg-preview-2-300x241.png\" alt=\"Diagram showing projectile trajectory with initial velocity components along x and y axes\">\n\n<p>${({v_x})_i} = 1,{rm{m\/s}},{rm{;}},,{{rm{a}}_x} = 0$<\/p>\n\n<p>${({v_y})_i} = 2,{rm{m\/s}},{rm{;}},,{{rm{a}}_y} = &#8211; 10,{rm{m\/}}{{rm{s}}^2}$<\/p>\n\n<p>Using motion of the projectile along x-axis,<\/p>\n\n<p>$x = {u_x}t + frac{1}{2}{a_x}{t^2}$<\/p>\n\n<p>So, $x = 1 times t + frac{1}{2} times (0){t^2}$<\/p>\n\n<p>$x = t$ &nbsp;&nbsp;&nbsp;\u2026(i)<\/p>\n\n<p>Again, using motion of the projectile along y-axis,<\/p>\n\n<p>$y = {u_y}t + frac{1}{2}{a_y}{t^2}$<\/p>\n\n<p>$y = 2 times t + frac{1}{2} times ( &#8211; 10){t^2}$<\/p>\n\n<p>So, $y = 2t &#8211; 5{t^2}$<\/p>\n\n<p>Putting the value of t = x in the above equation,<\/p>\n\n<p>$y = 2x &#8211; 5{x^2}$<\/p>\n\n<p>This is the required equation of trajectory.<\/p>\n\n<h3>Example 3<\/h3>\n\n<p>The particle has initial velocity $(3hat i + 4hat j)$ and an acceleration of $(0.4hat i + 0.3hat j)$. Find speed of particle after 10 sec.<\/p>\n\n<p><strong>Solution<\/strong><\/p>\n\n<p>Here, in this question, it is given that<\/p>\n\n<p>$({u_x}) = 3,unit,;,,{a_x} = 0.4,unit$<\/p>\n\n<p>$({u_y}) = 4,unit,;,,{a_y} = 0.3,unit$<\/p>\n\n<p>Since the acceleration of the particle is constant so, we can use<\/p>\n\n<p>${({v_x})_f} = {({v_x})_i} + {a_x}t$ and ${({v_y})_f} = {({v_y})_i} + {a_y}t$<\/p>\n\n<p>${({v_x})_f} = 3 + (0.4 times 10)$ and ${({v_y})_f} = 4 + (0.3 times 10)$<\/p>\n\n<p>So, ${({v_x})_f} = 3 + 4$ and ${({v_y})_f} = 4 + 3$<\/p>\n\n<p>${({v_x})_f} = 7,unit$ and ${({v_y})_f} = 7,unit$<\/p>\n\n<p>Now, we can write<\/p>\n\n<p>${{{bf{v}}_f} = {({v_x})_f}hat i + {({v_y})_f}hat j}$<\/p>\n\n<p>${{{bf{v}}_f} = (7,unit)hat i + (7,unit)hat j}$<\/p>\n\n<p>Speed of particle $ = left| {{{bf{v}}_f}} right|$<\/p>\n\n<p>$ = sqrt {{{(7,unit)}^2} + {{(7,unit)}^2}}$<\/p>\n\n<p>$ = (7,unit)sqrt 2 = 7sqrt 2 ,,unit$<\/p>\n\n<h3>Example 4<\/h3>\n\n<p>A particle moves in x-y plane under the action of a force ${bf{F}}$ such that the value of its linear momentum ${bf{P}}$ at any time t is ${P_x} = 2cos t,;,,{P_y} = 2sin t$. Find the angle $theta$ between ${bf{F}}$ and ${bf{P}}$ at a given time t.<\/p>\n\n<p><strong>Solution<\/strong><\/p>\n\n<p>Here momentum vector of the particle can be written as<\/p>\n\n<p>${bf{P}} = {P_x}hat i + {P_y}hat j$<\/p>\n\n<p>${bf{P}} = 2cos t,hat i + 2sin t,hat j$<\/p>\n\n<p>Now, from Newton&#8217;s 2<sup>nd<\/sup> law of motion, we can obtain force as<\/p>\n\n<p>${bf{F}} = frac{{d{bf{P}}}}{{dt}}quad Rightarrow quad {bf{F}} = frac{d}{{dt}}left( {2cos t,hat i + 2sin t,hat j} right)$<\/p>\n\n<p>Now, using scalar product of two vectors for calculating angle between ${bf{F}}$ &amp; ${bf{P}}$,<\/p>\n\n<p>${bf{P}} cdot {bf{F}} = left| {bf{P}} right|left| {bf{F}} right|,cos theta$<\/p>\n\n<p>$cos theta = frac{{{bf{P}} cdot {bf{F}}}}{{left| {bf{P}} right|left| {bf{F}} right|}} = frac{{(2cos t,hat i + 2sin t,hat j) cdot ( &#8211; 2sin t,hat i + 2cos t,hat j)}}{{left| {bf{P}} right|left| {bf{F}} right|}}$<\/p>\n\n<p>$cos theta = frac{{ -,4sin t cdot cos t + 4sin t cdot cos t}}{{sqrt {{{(2cos t)}^2} + {{(2sin t)}^2}} sqrt {{{( &#8211; 2sin t)}^2} + {{(2cos t)}^2}} }}$<\/p>\n\n<p>$cos theta = 0quad Rightarrow quad theta = 90\u00b0$<\/p>\n\n<p>So, the angle between the force vector and momentum vector is 90\u00ba.<\/p>\n\n<p>For more context on how 2D motion concepts appear in A-Level exams, see this <a href=\"https:\/\/www.myengineeringbuddy.com\/blog\/a-level-physics-a-blueprint-2026-exam-traps-fixed\/\">A-Level Physics blueprint and 2026 exam traps guide<\/a>. Students looking for broader exam strategies may also find this guide on <a href=\"https:\/\/www.myengineeringbuddy.com\/blog\/ap-physics-c-score-3-to-5-gap-guide\/\">closing the AP Physics C score gap from 3 to 5<\/a> useful. If you are considering working with a tutor, this <a href=\"https:\/\/www.myengineeringbuddy.com\/blog\/find-a-great-online-physics-tutor-a-freshmans-guide-to-getting-real-help\/\">freshman&#8217;s guide to finding a great online physics tutor<\/a> is a helpful starting point.<\/p>\n\n<p>Related: <a href=\"https:\/\/www.myengineeringbuddy.com\/kinematics-equations\/\">Kinematics equations<\/a> | <a href=\"https:\/\/www.myengineeringbuddy.com\/vectors-in-physics\/\">Vectors in physics<\/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<h2>Related Reading<\/h2>\n<ul>\n<li><a href=\"https:\/\/www.myengineeringbuddy.com\/blog\/mastering-physics-homework-help\/\">Mastering Physics Homework Help<\/a><\/li>\n<li><a href=\"https:\/\/www.myengineeringbuddy.com\/blog\/how-to-learn-physics\/\">How to Learn Physics<\/a><\/li>\n<li><a href=\"https:\/\/www.myengineeringbuddy.com\/blog\/flying-spiders-physics-behind-it\/\">Flying Spiders: The Physics Behind It<\/a><\/li>\n<li><a href=\"https:\/\/www.myengineeringbuddy.com\/blog\/how-to-solve-a-physics-problem\/\">How to Solve a Physics Problem<\/a><\/li>\n<\/ul>\n","protected":false},"excerpt":{"rendered":"<p>Key Takeaways 2D motion occurs in a plane and can  [&#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-11292","post","type-post","status-publish","format-standard","hentry","category-uncategorized"],"_links":{"self":[{"href":"https:\/\/www.myengineeringbuddy.com\/blog\/wp-json\/wp\/v2\/posts\/11292","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=11292"}],"version-history":[{"count":1,"href":"https:\/\/www.myengineeringbuddy.com\/blog\/wp-json\/wp\/v2\/posts\/11292\/revisions"}],"predecessor-version":[{"id":12168,"href":"https:\/\/www.myengineeringbuddy.com\/blog\/wp-json\/wp\/v2\/posts\/11292\/revisions\/12168"}],"wp:attachment":[{"href":"https:\/\/www.myengineeringbuddy.com\/blog\/wp-json\/wp\/v2\/media?parent=11292"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.myengineeringbuddy.com\/blog\/wp-json\/wp\/v2\/categories?post=11292"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.myengineeringbuddy.com\/blog\/wp-json\/wp\/v2\/tags?post=11292"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}