{"id":222,"date":"2026-06-13T16:31:28","date_gmt":"2026-06-13T16:31:28","guid":{"rendered":"https:\/\/physicsfundamentalsinfo.com\/blog\/?p=222"},"modified":"2026-06-13T16:31:29","modified_gmt":"2026-06-13T16:31:29","slug":"suvat-equations","status":"publish","type":"post","link":"https:\/\/physicsfundamentalsinfo.com\/blog\/kinematics\/suvat-equations\/","title":{"rendered":"What Are the SUVAT Equations?"},"content":{"rendered":"\n<div class=\"pf-citation\"><div class=\"eyebrow\">Definition<\/div><p>\n\nThe SUVAT equations are five equations of motion that describe an object moving with constant (uniform) acceleration, linking displacement (s), initial velocity (u), final velocity (v), acceleration (a) and time (t). Each equation leaves out one of these variables, so you choose the one that matches the quantities you know and the one you want.\n\n<\/p><\/div>\n<p>Press the accelerator and your car doesn&#8217;t jump straight to 60 \u2014 it climbs there, second by second. Hit the brakes and it doesn&#8217;t stop dead; it slides to a halt over a distance you&#8217;d rather not underestimate.<\/p>\n<p>Every one of those moments \u2014 speeding up, slowing down, a ball arcing through the air \u2014 obeys the same five compact formulas. Physicists call them the SUVAT equations, and once you know them, motion stops being a mystery and becomes something you can predict with a pencil.<\/p>\n<h2>What Are the SUVAT Equations?<\/h2>\n<p>SUVAT is a memory aid, not a Latin word. Each letter stands for one of the five quantities that describe an object moving in a straight line with steady acceleration:<\/p>\n<ul>\n<li><strong>s<\/strong> \u2014 displacement: how far it moves, and in which direction (metres).<\/li>\n<li><strong>u<\/strong> \u2014 initial velocity: the speed it starts with (metres per second).<\/li>\n<li><strong>v<\/strong> \u2014 final velocity: the speed it ends with (metres per second).<\/li>\n<li><strong>a<\/strong> \u2014 acceleration: how quickly the velocity changes (metres per second squared).<\/li>\n<li><strong>t<\/strong> \u2014 time: how long the motion lasts (seconds).<\/li>\n<\/ul>\n<p>Here is the precise version. The SUVAT equations are a set of five formulas that connect these quantities for any object whose acceleration stays constant throughout the motion. Hand the equations any three of the five values, and they give you the other two.<\/p>\n<p>That last sentence hides the whole trick. A typical question quietly tells you three of the quantities and asks for a fourth \u2014 so the real skill is spotting which three you already have.<\/p>\n<svg viewBox=\"0 0 640 300\" role=\"img\" aria-label=\"An object accelerating along a straight line: it starts with a small initial velocity u and ends with a larger final velocity v after acceleration a acts over a displacement s during time t.\" xmlns=\"http:\/\/www.w3.org\/2000\/svg\" style=\"width:100%;height:auto;max-width:640px;display:block;margin:24px auto;background:#F5F2EA;border:1px solid #D9CFB8;border-radius:6px;\">\n<line x1=\"250\" y1=\"95\" x2=\"378\" y2=\"95\" stroke=\"#7A1F2B\" stroke-width=\"3\"><\/line>\n<polygon points=\"390,95 376,89 376,101\" fill=\"#7A1F2B\"><\/polygon>\n<text x=\"314\" y=\"80\" font-family=\"Manrope, Arial, sans-serif\" font-size=\"15\" fill=\"#7A1F2B\" text-anchor=\"middle\" font-weight=\"700\">a (acceleration)<\/text>\n<line x1=\"40\" y1=\"210\" x2=\"600\" y2=\"210\" stroke=\"#0A1628\" stroke-width=\"2\"><\/line>\n<rect x=\"60\" y=\"170\" width=\"46\" height=\"40\" rx=\"4\" fill=\"#142139\"><\/rect>\n<rect x=\"470\" y=\"170\" width=\"46\" height=\"40\" rx=\"4\" fill=\"#142139\" fill-opacity=\"0.5\"><\/rect>\n<line x1=\"110\" y1=\"190\" x2=\"150\" y2=\"190\" stroke=\"#C8932A\" stroke-width=\"3\"><\/line>\n<polygon points=\"158,190 148,185 148,195\" fill=\"#C8932A\"><\/polygon>\n<text x=\"128\" y=\"162\" font-family=\"Georgia, serif\" font-size=\"18\" font-style=\"italic\" fill=\"#0A1628\" text-anchor=\"middle\">u<\/text>\n<line x1=\"520\" y1=\"190\" x2=\"588\" y2=\"190\" stroke=\"#C8932A\" stroke-width=\"3\"><\/line>\n<polygon points=\"596,190 586,185 586,195\" fill=\"#C8932A\"><\/polygon>\n<text x=\"556\" y=\"162\" font-family=\"Georgia, serif\" font-size=\"18\" font-style=\"italic\" fill=\"#0A1628\" text-anchor=\"middle\">v<\/text>\n<line x1=\"83\" y1=\"252\" x2=\"493\" y2=\"252\" stroke=\"#0A1628\" stroke-width=\"1.5\"><\/line>\n<line x1=\"83\" y1=\"244\" x2=\"83\" y2=\"260\" stroke=\"#0A1628\" stroke-width=\"1.5\"><\/line>\n<line x1=\"493\" y1=\"244\" x2=\"493\" y2=\"260\" stroke=\"#0A1628\" stroke-width=\"1.5\"><\/line>\n<text x=\"288\" y=\"278\" font-family=\"Manrope, Arial, sans-serif\" font-size=\"15\" fill=\"#0A1628\" text-anchor=\"middle\">s (displacement) \u2014 covered in time t<\/text>\n<\/svg>\n<p style=\"text-align:center;font-size:13px;color:#1F2E47;font-style:italic;margin-top:-8px;\">The five SUVAT quantities in one picture: an object starts at velocity u, accelerates at a, and reaches velocity v after travelling a displacement s in time t.<\/p>\n<p>In practice, students lose more marks to muddled units than to muddled algebra. Convert everything to metres, seconds, and metres per second <em>before<\/em> you substitute, and most errors vanish.<\/p>\n<h2>The Five SUVAT Equations<\/h2>\n<p>There are five SUVAT equations, and each one leaves out a different variable. That is precisely why there are five: whichever quantity you don&#8217;t know, one equation doesn&#8217;t need it.<\/p>\n<p>Linking final velocity, initial velocity, acceleration and time (no displacement):<\/p>\n<div class=\"pf-formula\">v = u + at<\/div>\n<p>Linking displacement to initial velocity, acceleration and time (no final velocity):<\/p>\n<div class=\"pf-formula\">s = ut + \u00bdat\u00b2<\/div>\n<p>Linking displacement to both velocities and time (no acceleration):<\/p>\n<div class=\"pf-formula\">s = \u00bd(u + v)t<\/div>\n<p>Linking the two velocities to acceleration and displacement (no time):<\/p>\n<div class=\"pf-formula\">v\u00b2 = u\u00b2 + 2as<\/div>\n<p>And linking displacement to final velocity, acceleration and time (no initial velocity):<\/p>\n<div class=\"pf-formula\">s = vt \u2212 \u00bdat\u00b2<\/div>\n<h3>What each symbol means<\/h3>\n<ul>\n<li><strong>s<\/strong> = displacement \u2014 metres (m)<\/li>\n<li><strong>u<\/strong> = initial velocity \u2014 metres per second (m\/s)<\/li>\n<li><strong>v<\/strong> = final velocity \u2014 metres per second (m\/s)<\/li>\n<li><strong>a<\/strong> = acceleration \u2014 metres per second squared (m\/s\u00b2)<\/li>\n<li><strong>t<\/strong> = time \u2014 seconds (s)<\/li>\n<\/ul>\n<h3>Which SUVAT equation should you use?<\/h3>\n<p>The fastest method: write down what you know, mark what you want, then pick the equation that contains those quantities \u2014 and skips the one you neither have nor need.<\/p>\n<div class=\"pf-table-scroll\" style=\"display:block;width:100%;max-width:100%;overflow-x:auto;-webkit-overflow-scrolling:touch;margin:1.5em 0;\">\n<table style=\"width:100%;border-collapse:collapse;word-break:break-word;\">\n<thead>\n<tr style=\"background:#0A1628;color:#FAF6EE;\">\n<th style=\"padding:10px;border:1px solid #D9CFB8;text-align:left;\">Equation<\/th>\n<th style=\"padding:10px;border:1px solid #D9CFB8;text-align:left;\">Links these<\/th>\n<th style=\"padding:10px;border:1px solid #D9CFB8;text-align:left;\">Leaves out<\/th>\n<th style=\"padding:10px;border:1px solid #D9CFB8;text-align:left;\">Use it when you don&#8217;t know\u2026<\/th>\n<\/tr>\n<\/thead>\n<tbody>\n<tr><td style=\"padding:10px;border:1px solid #D9CFB8;\"><strong>v = u + at<\/strong><\/td><td style=\"padding:10px;border:1px solid #D9CFB8;\">u, v, a, t<\/td><td style=\"padding:10px;border:1px solid #D9CFB8;\">s<\/td><td style=\"padding:10px;border:1px solid #D9CFB8;\">displacement<\/td><\/tr>\n<tr style=\"background:#F5F2EA;\"><td style=\"padding:10px;border:1px solid #D9CFB8;\"><strong>s = ut + \u00bdat\u00b2<\/strong><\/td><td style=\"padding:10px;border:1px solid #D9CFB8;\">s, u, a, t<\/td><td style=\"padding:10px;border:1px solid #D9CFB8;\">v<\/td><td style=\"padding:10px;border:1px solid #D9CFB8;\">final velocity<\/td><\/tr>\n<tr><td style=\"padding:10px;border:1px solid #D9CFB8;\"><strong>s = \u00bd(u + v)t<\/strong><\/td><td style=\"padding:10px;border:1px solid #D9CFB8;\">s, u, v, t<\/td><td style=\"padding:10px;border:1px solid #D9CFB8;\">a<\/td><td style=\"padding:10px;border:1px solid #D9CFB8;\">acceleration<\/td><\/tr>\n<tr style=\"background:#F5F2EA;\"><td style=\"padding:10px;border:1px solid #D9CFB8;\"><strong>v\u00b2 = u\u00b2 + 2as<\/strong><\/td><td style=\"padding:10px;border:1px solid #D9CFB8;\">s, u, v, a<\/td><td style=\"padding:10px;border:1px solid #D9CFB8;\">t<\/td><td style=\"padding:10px;border:1px solid #D9CFB8;\">time<\/td><\/tr>\n<tr><td style=\"padding:10px;border:1px solid #D9CFB8;\"><strong>s = vt \u2212 \u00bdat\u00b2<\/strong><\/td><td style=\"padding:10px;border:1px solid #D9CFB8;\">s, v, a, t<\/td><td style=\"padding:10px;border:1px solid #D9CFB8;\">u<\/td><td style=\"padding:10px;border:1px solid #D9CFB8;\">initial velocity<\/td><\/tr>\n<\/tbody>\n<\/table>\n<\/div>\n<p>One row earns its keep more than the rest. When a problem gives you no time and asks for none \u2014 &#8220;how far to stop?&#8221;, &#8220;how fast at the bottom?&#8221; \u2014 reach for <strong>v\u00b2 = u\u00b2 + 2as<\/strong>.<\/p>\n<h2>How the SUVAT Equations Work<\/h2>\n<p>Where do five equations come from? Not from on high. From a single straight line on a velocity\u2013time graph.<\/p>\n<p>When acceleration is constant, a graph of velocity against time is a straight line. It begins at the initial velocity u and climbs (or falls) steadily to the final velocity v. Two features of that line give us everything.<\/p>\n<svg viewBox=\"0 0 640 420\" role=\"img\" aria-label=\"Velocity\u2013time graph for constant acceleration: a straight line rising from initial velocity u to final velocity v. The gradient of the line equals the acceleration a, and the shaded trapezium area beneath it equals the displacement s.\" xmlns=\"http:\/\/www.w3.org\/2000\/svg\" style=\"width:100%;height:auto;max-width:640px;display:block;margin:24px auto;background:#F5F2EA;border:1px solid #D9CFB8;border-radius:6px;\">\n<polygon points=\"80,360 80,300 520,120 520,360\" fill=\"#C8932A\" fill-opacity=\"0.18\"><\/polygon>\n<line x1=\"80\" y1=\"40\" x2=\"80\" y2=\"360\" stroke=\"#0A1628\" stroke-width=\"2\"><\/line>\n<line x1=\"80\" y1=\"360\" x2=\"610\" y2=\"360\" stroke=\"#0A1628\" stroke-width=\"2\"><\/line>\n<polygon points=\"80,34 75,46 85,46\" fill=\"#0A1628\"><\/polygon>\n<polygon points=\"618,360 606,355 606,365\" fill=\"#0A1628\"><\/polygon>\n<line x1=\"80\" y1=\"120\" x2=\"520\" y2=\"120\" stroke=\"#0A1628\" stroke-width=\"1\" stroke-dasharray=\"4 4\" opacity=\"0.5\"><\/line>\n<line x1=\"520\" y1=\"120\" x2=\"520\" y2=\"360\" stroke=\"#0A1628\" stroke-width=\"1\" stroke-dasharray=\"4 4\" opacity=\"0.5\"><\/line>\n<line x1=\"80\" y1=\"300\" x2=\"520\" y2=\"120\" stroke=\"#7A1F2B\" stroke-width=\"3\"><\/line>\n<circle cx=\"80\" cy=\"300\" r=\"4.5\" fill=\"#7A1F2B\"><\/circle>\n<circle cx=\"520\" cy=\"120\" r=\"4.5\" fill=\"#7A1F2B\"><\/circle>\n<text x=\"66\" y=\"306\" font-family=\"Georgia, serif\" font-size=\"20\" font-style=\"italic\" fill=\"#0A1628\" text-anchor=\"end\">u<\/text>\n<text x=\"66\" y=\"126\" font-family=\"Georgia, serif\" font-size=\"20\" font-style=\"italic\" fill=\"#0A1628\" text-anchor=\"end\">v<\/text>\n<text x=\"520\" y=\"382\" font-family=\"Georgia, serif\" font-size=\"20\" font-style=\"italic\" fill=\"#0A1628\" text-anchor=\"middle\">t<\/text>\n<text x=\"72\" y=\"378\" font-family=\"Manrope, Arial, sans-serif\" font-size=\"14\" fill=\"#0A1628\" text-anchor=\"end\">0<\/text>\n<text x=\"300\" y=\"314\" font-family=\"Manrope, Arial, sans-serif\" font-size=\"15\" fill=\"#7A1F2B\" text-anchor=\"middle\" font-weight=\"700\">area = displacement s<\/text>\n<text x=\"318\" y=\"178\" font-family=\"Manrope, Arial, sans-serif\" font-size=\"14\" fill=\"#0A1628\" text-anchor=\"middle\">gradient = acceleration a<\/text>\n<text x=\"32\" y=\"200\" font-family=\"Manrope, Arial, sans-serif\" font-size=\"14\" fill=\"#0A1628\" text-anchor=\"middle\" transform=\"rotate(-90 32 200)\">velocity (m\/s)<\/text>\n<text x=\"345\" y=\"408\" font-family=\"Manrope, Arial, sans-serif\" font-size=\"14\" fill=\"#0A1628\" text-anchor=\"middle\">time (s)<\/text>\n<\/svg>\n<p style=\"text-align:center;font-size:13px;color:#1F2E47;font-style:italic;margin-top:-8px;\">For constant acceleration the line is straight. Its gradient is the acceleration a; the shaded area beneath it is the displacement s \u2014 a trapezium of area \u00bd(u + v) \u00d7 t, which is exactly s = \u00bd(u + v)t.<\/p>\n<p>First, the <strong>gradient<\/strong>. The steepness of the line is how fast velocity changes per second \u2014 that is the acceleration. Rearrange &#8220;gradient = (v \u2212 u) \/ t&#8221; and the first SUVAT equation, v = u + at, falls straight out.<\/p>\n<p>Second, the <strong>area<\/strong> under the line. On any velocity\u2013time graph, the area beneath the curve equals the displacement. Here that area is a trapezium with parallel sides u and v and width t.<\/p>\n<p>A trapezium&#8217;s area is its average height times its width \u2014 \u00bd(u + v) \u00d7 t. That is the equation s = \u00bd(u + v)t, read directly off the geometry.<\/p>\n<p>The remaining three equations are algebra, not new physics. Substitute v = u + at into s = \u00bd(u + v)t and you get s = ut + \u00bdat\u00b2. Eliminate the time between those two, and the time-free equation v\u00b2 = u\u00b2 + 2as appears. The full step-by-step working is laid out clearly in <a href=\"https:\/\/physics.info\/motion-equations\/\" target=\"_blank\" rel=\"noopener\">The Physics Hypertextbook<\/a>.<\/p>\n<p>One quiet assumption does all the work: that the average velocity sits exactly halfway between u and v. That is true <em>only<\/em> when acceleration is constant \u2014 a neat result first written down at Merton College, Oxford, back in 1335.<\/p>\n<p>Change the starting velocity, the acceleration and the clock below, and watch all five quantities update together.<\/p>\n<div class=\"pf-sim-slot\"><div class=\"pf-sim-slot-header\"><span class=\"icon-dot\"><\/span><span class=\"label\">SUVAT Equations Lab<\/span><\/div><div class=\"pf-sim-slot-body\"><style>.pf-sim-frame{width:100%;border:none;height:600px}@media(max-width:760px){.pf-sim-frame{height:1000px}}<\/style><iframe src=\"\/labs\/suvat-equations.html\" class=\"pf-sim-frame\" loading=\"lazy\"><\/iframe><\/div><\/div>\n<h2>When Can You Use the SUVAT Equations?<\/h2>\n<p>The SUVAT equations are powerful, but they come with strict conditions. Break one and the answers quietly turn wrong.<\/p>\n<p><strong>Acceleration must be constant.<\/strong> This is the headline rule. If the acceleration changes during the motion, no single SUVAT equation describes the whole journey. The fix is simple: split the motion into stages where the acceleration <em>is<\/em> constant, and apply SUVAT to each stage in turn.<\/p>\n<p><strong>Motion must be in a straight line.<\/strong> SUVAT is one-dimensional. For two-dimensional motion such as <a href=\"https:\/\/physicsfundamentalsinfo.com\/blog\/mechanics\/projectile-motion-guide\/\">projectile motion<\/a>, you split the motion into horizontal and vertical directions and apply SUVAT to each axis separately.<\/p>\n<p><strong>Keep your units consistent.<\/strong> Use metres, seconds, metres per second and metres per second squared throughout. A speed given in km\/h or a distance in centimetres must be converted first.<\/p>\n<p><strong>Mind the signs.<\/strong> Velocity and acceleration are vectors, so direction matters \u2014 which is part of <a href=\"https:\/\/physicsfundamentalsinfo.com\/blog\/kinematics\/velocity-vs-speed\/\">the difference between velocity and speed<\/a>. Choose one direction as positive, then anything pointing the other way (a deceleration, or gravity on a rising ball) takes a negative sign.<\/p>\n<p>These conditions, with many fully worked cases, are covered in <a href=\"https:\/\/phys.libretexts.org\/Bookshelves\/College_Physics\/College_Physics_1e_(OpenStax)\/02:_Kinematics\/2.05:_Motion_Equations_for_Constant_Acceleration_in_One_Dimension\" target=\"_blank\" rel=\"noopener\">OpenStax College Physics<\/a>. As a sanity check: a family car pulling away manages roughly 2\u20134 m\/s\u00b2; if your answer says 50 m\/s\u00b2, something is off.<\/p>\n<h2>Real-World Examples of the SUVAT Equations<\/h2>\n<p>The equations aren&#8217;t just exam fodder. They quietly run through dozens of everyday situations.<\/p>\n<p><strong>A dropped object.<\/strong> Let go of your phone and it falls under gravity alone, so a = g \u2248 9.81 m\/s\u00b2 downward and the initial velocity u is zero. With those two values, s = ut + \u00bdat\u00b2 tells you how far it has fallen after any time t.<\/p>\n<p><strong>Braking a car.<\/strong> Here v = 0 when the car stops. Because v\u00b2 = u\u00b2 + 2as, the stopping distance grows with the <em>square<\/em> of the speed \u2014 double your speed and you need roughly four times the distance to stop. That single equation is the physics behind every speed-limit sign near a school.<\/p>\n<p><strong>An aircraft taking off.<\/strong> A jet must reach a minimum lift-off speed before the runway runs out. Knowing its acceleration and that target velocity, v\u00b2 = u\u00b2 + 2as gives the runway length required \u2014 which is why heavier loads need longer runways.<\/p>\n<figure style=\"margin:32px auto;max-width:640px;text-align:center;\">\n  <img decoding=\"async\" src=\"https:\/\/physicsfundamentalsinfo.com\/blog\/wp-content\/uploads\/2026\/06\/airplane_landing.webp\"\n       alt=\"Aircraft accelerating along a runway, a real-world example of the SUVAT equations\"\n       loading=\"lazy\"\n       style=\"width:100%;height:auto;border-radius:4px;\" \/>\n  <figcaption style=\"font-size:13px;color:#1F2E47;font-style:italic;margin-top:8px;\">A jet uses v\u00b2 = u\u00b2 + 2as in reverse: a fixed lift-off speed and acceleration set the runway length it needs.<\/figcaption>\n<\/figure>\n<p><strong>A ball thrown upward.<\/strong> At the very top of its flight the ball is momentarily still, so v = 0 \u2014 the trick that lets you find the maximum height. On the way down the same equations run in reverse, with gravity now speeding it up.<\/p>\n<h2>Common Misconceptions About the SUVAT Equations<\/h2>\n<p>A handful of misunderstandings trip up almost every learner. Clear these and SUVAT becomes far more reliable.<\/p>\n<p><strong>&#8220;SUVAT works for any motion.&#8221;<\/strong> It doesn&#8217;t. The equations assume constant acceleration. If the acceleration itself changes with time, you need calculus, not SUVAT \u2014 or you break the motion into constant-acceleration stages.<\/p>\n<p><strong>&#8220;Deceleration needs a different formula.&#8221;<\/strong> No. Slowing down is simply a negative acceleration in the very same five equations. There is no separate &#8220;deceleration equation&#8221; \u2014 just substitute a negative value of a once you&#8217;ve fixed a positive direction.<\/p>\n<p><strong>&#8220;g is always \u22129.81.&#8221;<\/strong> The sign of gravity depends on your choice of positive direction. If you call upward positive, then g = \u22129.81 m\/s\u00b2. If you call downward positive, g = +9.81 m\/s\u00b2. Both are correct; what matters is being consistent within the problem.<\/p>\n<p><strong>&#8220;Average velocity is always halfway between the two.&#8221;<\/strong> Only when acceleration is constant. The formula \u00bd(u + v) for average velocity is a special case, not a universal truth \u2014 it&#8217;s exactly the assumption that makes SUVAT work.<\/p>\n<h2>How SUVAT Connects to Newton&#8217;s Laws, Energy and Projectiles<\/h2>\n<p>SUVAT doesn&#8217;t sit in isolation. It&#8217;s one node in a web of mechanics, and seeing the links makes each part stick.<\/p>\n<p><strong>It pairs with force.<\/strong> SUVAT tells you <em>how<\/em> a constantly accelerating object moves, but not <em>why<\/em> it accelerates. That answer comes from <a href=\"https:\/\/physicsfundamentalsinfo.com\/blog\/mechanics\/newtons-second-law\/\">Newton&#8217;s second law<\/a>, F = ma. A common workflow is to find the acceleration from the forces, then feed that a straight into a SUVAT equation. The broader picture of forces and motion lives in <a href=\"https:\/\/physicsfundamentalsinfo.com\/blog\/mechanics\/newtons-laws-of-motion\/\">Newton&#8217;s laws of motion<\/a>.<\/p>\n<p><strong>It echoes energy.<\/strong> Multiply v\u00b2 = u\u00b2 + 2as by \u00bdm and rearrange, and you get \u00bdmv\u00b2 \u2212 \u00bdmu\u00b2 = (ma)s \u2014 the change in <a href=\"https:\/\/physicsfundamentalsinfo.com\/blog\/mechanics\/kinetic-energy-formula\/\">kinetic energy<\/a> equals force times distance. The kinematic equation and the work\u2013energy theorem are two views of the same motion.<\/p>\n<p><strong>It extends to two dimensions.<\/strong> Throw a ball at an angle and SUVAT still works \u2014 you simply apply it twice, once to the horizontal motion and once to the vertical. That is the whole basis of projectile analysis.<\/p>\n<p>And when acceleration genuinely varies? Calculus takes over, with integration generalising the same ideas. SUVAT is the constant-acceleration corner of a much larger toolkit.<\/p>\n<h2>Worked Problems<\/h2>\n<div class=\"pf-problem\"><div class=\"pf-problem-num\">Problem 1<\/div><div class=\"pf-problem-question\">A car starts from rest and accelerates uniformly at 3.0 m\/s\u00b2 for 5.0 s. What is its final velocity?<\/div><details><summary>Show Solution<\/summary><div class=\"pf-problem-solution\">\n\n<strong>Solution:<\/strong>\n\nStep 1: Known: u = 0 m\/s, a = 3.0 m\/s\u00b2, t = 5.0 s. Unknown: v. No displacement is involved, so use v = u + at.\n\nStep 2: Substitute with units: v = 0 + (3.0 m\/s\u00b2)(5.0 s).\n\nStep 3: v = 15 m\/s.\n\n<strong>Answer: 15 m\/s<\/strong>\n\n<\/div><\/details><\/div>\n<div class=\"pf-problem\"><div class=\"pf-problem-num\">Problem 2<\/div><div class=\"pf-problem-question\">For the same car (from rest, a = 3.0 m\/s\u00b2, t = 5.0 s), how far does it travel in those 5.0 s?<\/div><details><summary>Show Solution<\/summary><div class=\"pf-problem-solution\">\n\n<strong>Solution:<\/strong>\n\nStep 1: Known: u = 0 m\/s, a = 3.0 m\/s\u00b2, t = 5.0 s. Unknown: s. Use s = ut + \u00bdat\u00b2.\n\nStep 2: Substitute: s = (0)(5.0) + \u00bd(3.0 m\/s\u00b2)(5.0 s)\u00b2.\n\nStep 3: s = \u00bd \u00d7 3.0 \u00d7 25 = 37.5 m.\n\n<strong>Answer: 37.5 m (\u2248 38 m)<\/strong>\n\n<\/div><\/details><\/div>\n<div class=\"pf-problem\"><div class=\"pf-problem-num\">Problem 3<\/div><div class=\"pf-problem-question\">A ball is thrown straight up at 20 m\/s. Taking g = 9.81 m\/s\u00b2, how high does it rise before stopping momentarily?<\/div><details><summary>Show Solution<\/summary><div class=\"pf-problem-solution\">\n\n<strong>Solution:<\/strong>\n\nStep 1: Known: u = 20 m\/s, v = 0 m\/s (at the highest point), a = \u22129.81 m\/s\u00b2 (gravity opposes the upward motion). No time is given, so use v\u00b2 = u\u00b2 + 2as.\n\nStep 2: Substitute: 0\u00b2 = (20)\u00b2 + 2(\u22129.81)s \u2192 0 = 400 \u2212 19.62 s.\n\nStep 3: s = 400 \/ 19.62 = 20.4 m.\n\n<strong>Answer: 20.4 m (3 s.f.)<\/strong>\n\n<\/div><\/details><\/div>\n<div class=\"pf-problem\"><div class=\"pf-problem-num\">Problem 4<\/div><div class=\"pf-problem-question\">A cyclist slows from 12 m\/s to 4.0 m\/s over a distance of 40 m. What is the constant acceleration?<\/div><details><summary>Show Solution<\/summary><div class=\"pf-problem-solution\">\n\n<strong>Solution:<\/strong>\n\nStep 1: Known: u = 12 m\/s, v = 4.0 m\/s, s = 40 m. Unknown: a. No time is given, so use v\u00b2 = u\u00b2 + 2as.\n\nStep 2: Substitute: (4.0)\u00b2 = (12)\u00b2 + 2a(40) \u2192 16 = 144 + 80a.\n\nStep 3: 80a = 16 \u2212 144 = \u2212128 \u2192 a = \u22121.6 m\/s\u00b2.\n\n<strong>Answer: \u22121.6 m\/s\u00b2 (a deceleration of 1.6 m\/s\u00b2)<\/strong>\n\n<\/div><\/details><\/div>\n<div class=\"pf-problem\"><div class=\"pf-problem-num\">Problem 5<\/div><div class=\"pf-problem-question\">A stone is released from rest at the top of a 45 m cliff. Taking g = 9.81 m\/s\u00b2, how long does it take to reach the ground?<\/div><details><summary>Show Solution<\/summary><div class=\"pf-problem-solution\">\n\n<strong>Solution:<\/strong>\n\nStep 1: Known: u = 0 m\/s, a = 9.81 m\/s\u00b2 (taking downward as positive), s = 45 m. Unknown: t. Use s = ut + \u00bdat\u00b2.\n\nStep 2: Substitute: 45 = (0)t + \u00bd(9.81)t\u00b2 \u2192 45 = 4.905 t\u00b2.\n\nStep 3: t\u00b2 = 45 \/ 4.905 = 9.174 \u2192 t = 3.03 s.\n\n<strong>Answer: 3.03 s (3 s.f.)<\/strong>\n\n<\/div><\/details><\/div>\n<div class=\"pf-problem\"><div class=\"pf-problem-num\">Problem 6<\/div><div class=\"pf-problem-question\">A train decelerates uniformly from 30 m\/s to rest in 20 s. Find its acceleration and the distance it travels while stopping.<\/div><details><summary>Show Solution<\/summary><div class=\"pf-problem-solution\">\n\n<strong>Solution:<\/strong>\n\nStep 1: Known: u = 30 m\/s, v = 0 m\/s, t = 20 s. For the acceleration, use v = u + at.\n\nStep 2: 0 = 30 + a(20) \u2192 a = \u221230 \/ 20 = \u22121.5 m\/s\u00b2.\n\nStep 3: For the distance, use s = \u00bd(u + v)t = \u00bd(30 + 0)(20) = 300 m.\n\n<strong>Answer: a = \u22121.5 m\/s\u00b2; s = 300 m<\/strong>\n\n<\/div><\/details><\/div>\n<div class=\"pf-problem\"><div class=\"pf-problem-num\">Problem 7<\/div><div class=\"pf-problem-question\">A car travelling at 8.0 m\/s accelerates uniformly and covers 100 m in 8.0 s. Find its acceleration and its final velocity.<\/div><details><summary>Show Solution<\/summary><div class=\"pf-problem-solution\">\n\n<strong>Solution:<\/strong>\n\nStep 1: Known: u = 8.0 m\/s, s = 100 m, t = 8.0 s. For the acceleration, use s = ut + \u00bdat\u00b2.\n\nStep 2: 100 = (8.0)(8.0) + \u00bda(8.0)\u00b2 \u2192 100 = 64 + 32a \u2192 a = 36 \/ 32 = 1.125 m\/s\u00b2.\n\nStep 3: For the final velocity, use v = u + at = 8.0 + (1.125)(8.0) = 17 m\/s.\n\n<strong>Answer: a = 1.125 m\/s\u00b2 (\u2248 1.13 m\/s\u00b2); v = 17 m\/s<\/strong>\n\n<\/div><\/details><\/div>\n<div class=\"pf-problem\"><div class=\"pf-problem-num\">Problem 8<\/div><div class=\"pf-problem-question\">A rocket sled starts from rest and accelerates at 5.0 m\/s\u00b2 for 8.0 s, then decelerates at 4.0 m\/s\u00b2 until it stops. What total distance does it cover?<\/div><details><summary>Show Solution<\/summary><div class=\"pf-problem-solution\">\n\n<strong>Solution:<\/strong>\n\nStep 1: Treat each phase separately, because the acceleration differs \u2014 SUVAT needs a constant acceleration within each stage.\n\nStep 2 (Phase 1, speeding up): u = 0, a = 5.0 m\/s\u00b2, t = 8.0 s. Final speed v = u + at = 0 + (5.0)(8.0) = 40 m\/s. Distance s\u2081 = \u00bd(u + v)t = \u00bd(0 + 40)(8.0) = 160 m.\n\nStep 3 (Phase 2, stopping): u = 40 m\/s, v = 0, a = \u22124.0 m\/s\u00b2. Using v\u00b2 = u\u00b2 + 2as \u2192 0 = 40\u00b2 + 2(\u22124.0)s\u2082 \u2192 s\u2082 = 1600 \/ 8.0 = 200 m.\n\n<strong>Answer: total distance = 160 m + 200 m = 360 m<\/strong>\n\n<\/div><\/details><\/div>\n<h2>Frequently Asked Questions<\/h2>\n<details class=\"pf-faq-item\"><summary>What does SUVAT stand for?<\/summary><div class=\"pf-faq-item-answer\">\n\nSUVAT stands for the five symbols used in the constant-acceleration equations of motion: s (displacement), u (initial velocity), v (final velocity), a (acceleration) and t (time). It is a memory aid used mainly in UK physics courses; elsewhere the same equations are called the kinematic equations or the equations of motion.\n\n<\/div><\/details>\n<details class=\"pf-faq-item\"><summary>What are the five SUVAT equations?<\/summary><div class=\"pf-faq-item-answer\">\n\nThe five SUVAT equations are: v = u + at; s = ut + \u00bdat\u00b2; s = \u00bd(u + v)t; v\u00b2 = u\u00b2 + 2as; and s = vt \u2212 \u00bdat\u00b2. Each links four of the five quantities and leaves one out, so you choose the equation that contains your known values and the unknown you want to find.\n\n<\/div><\/details>\n<details class=\"pf-faq-item\"><summary>When can you use the SUVAT equations?<\/summary><div class=\"pf-faq-item-answer\">\n\nYou can use the SUVAT equations only when the acceleration is constant and the motion is along a straight line. If the acceleration changes, the equations no longer apply to the motion as a whole \u2014 though you can usually split the journey into separate stages, each with its own constant acceleration, and apply SUVAT to each in turn.\n\n<\/div><\/details>\n<details class=\"pf-faq-item\"><summary>Can SUVAT be used for vertical motion and falling objects?<\/summary><div class=\"pf-faq-item-answer\">\n\nYes. For an object in free fall the acceleration is gravity, a = g \u2248 9.81 m\/s\u00b2 directed downward, so all five SUVAT equations apply. Choose one direction as positive and keep your signs consistent: if upward is positive, then g is \u22129.81 m\/s\u00b2. An object simply dropped from rest has u = 0.\n\n<\/div><\/details>\n<details class=\"pf-faq-item\"><summary>Why are there five SUVAT equations?<\/summary><div class=\"pf-faq-item-answer\">\n\nThere are five SUVAT equations because there are five quantities (s, u, v, a, t) and each equation deliberately leaves one of them out. Whatever quantity you don&#8217;t know and don&#8217;t need, there is an equation that avoids it \u2014 so you can always solve a problem using only the three values you are given.\n\n<\/div><\/details>\n<details class=\"pf-faq-item\"><summary>Is acceleration negative when something slows down in SUVAT?<\/summary><div class=\"pf-faq-item-answer\">\n\nWhen an object slows down, its acceleration points opposite to its motion, so it takes a negative value in the SUVAT equations \u2014 there is no separate &#8220;deceleration formula&#8221;. You use the same five equations and simply substitute a negative a, provided you have fixed a positive direction first.\n\n<\/div><\/details>\n<details class=\"pf-faq-item\"><summary>What is the difference between SUVAT and Newton&#039;s second law?<\/summary><div class=\"pf-faq-item-answer\">\n\nSUVAT describes how a constantly accelerating object moves \u2014 its displacement, velocity and time \u2014 while Newton&#8217;s second law, F = ma, explains why it accelerates by linking acceleration to the net force and mass. In practice you often use F = ma to find the acceleration, then feed that value into the SUVAT equations.\n\n<\/div><\/details>\n","protected":false},"excerpt":{"rendered":"<p>A clear guide to the SUVAT equations \u2014 the five equations of motion for constant acceleration. Includes the velocity\u2013time graph derivation, worked examples, common mistakes and an interactive simulation.<\/p>\n","protected":false},"author":1,"featured_media":224,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[7],"tags":[91,90,93,43,92],"class_list":["post-222","post","type-post","status-publish","format-standard","has-post-thumbnail","hentry","category-kinematics","tag-constant-acceleration","tag-equations-of-motion","tag-kinematic-equations","tag-kinematics","tag-suvat"],"_links":{"self":[{"href":"https:\/\/physicsfundamentalsinfo.com\/blog\/wp-json\/wp\/v2\/posts\/222","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/physicsfundamentalsinfo.com\/blog\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/physicsfundamentalsinfo.com\/blog\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/physicsfundamentalsinfo.com\/blog\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/physicsfundamentalsinfo.com\/blog\/wp-json\/wp\/v2\/comments?post=222"}],"version-history":[{"count":1,"href":"https:\/\/physicsfundamentalsinfo.com\/blog\/wp-json\/wp\/v2\/posts\/222\/revisions"}],"predecessor-version":[{"id":225,"href":"https:\/\/physicsfundamentalsinfo.com\/blog\/wp-json\/wp\/v2\/posts\/222\/revisions\/225"}],"wp:featuredmedia":[{"embeddable":true,"href":"https:\/\/physicsfundamentalsinfo.com\/blog\/wp-json\/wp\/v2\/media\/224"}],"wp:attachment":[{"href":"https:\/\/physicsfundamentalsinfo.com\/blog\/wp-json\/wp\/v2\/media?parent=222"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/physicsfundamentalsinfo.com\/blog\/wp-json\/wp\/v2\/categories?post=222"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/physicsfundamentalsinfo.com\/blog\/wp-json\/wp\/v2\/tags?post=222"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}