{"id":802,"date":"2025-06-20T17:13:38","date_gmt":"2025-06-20T17:13:38","guid":{"rendered":"https:\/\/content.one.lumenlearning.com\/calculus2\/?post_type=chapter&#038;p=802"},"modified":"2025-09-18T12:12:27","modified_gmt":"2025-09-18T12:12:27","slug":"basics-of-differential-equations-apply-it","status":"publish","type":"chapter","link":"https:\/\/content.one.lumenlearning.com\/calculus2\/chapter\/basics-of-differential-equations-apply-it\/","title":{"raw":"Basics of Differential Equations: Apply It","rendered":"Basics of Differential Equations: Apply It"},"content":{"raw":"<section class=\"textbox learningGoals\" aria-label=\"Learning Goals\">\r\n<ul>\r\n \t<li>Determine the order of a differential equation<\/li>\r\n \t<li>Tell the difference between a general solution and a particular solution<\/li>\r\n \t<li>Identify what makes a problem an initial-value problem<\/li>\r\n \t<li>Check if a function actually solves a given differential equation or initial-value problem<\/li>\r\n<\/ul>\r\n<\/section>\r\n<h2>Applications in Physics: Projectile Motion<\/h2>\r\n<p class=\"whitespace-normal break-words\">In physics and engineering, we analyze forces acting on objects to predict their motion. This approach is fundamental to understanding everything from falling objects to rocket trajectories. When an object moves near Earth's surface, gravity is typically the dominant force we need to consider.<\/p>\r\n<p class=\"whitespace-normal break-words\">The beauty of this analysis lies in how we can use calculus and differential equations to model real-world phenomena. By applying Newton's second law ([latex]F = ma[\/latex], where [latex]F[\/latex] is force, [latex]m[\/latex] is mass, and [latex]a[\/latex] is acceleration), we can derive equations that precisely describe motion under the influence of gravity.<\/p>\r\nConsider a baseball falling through air, with gravity as the only acting force (we'll ignore air resistance for now). At Earth's surface, gravitational acceleration [latex]g[\/latex] is approximately [latex]9.8 \\text{ m\/s}^2[\/latex].\r\n<figure id=\"CNX_Calc_Figure_08_01_003\"><figcaption><\/figcaption>\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"325\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/4175\/2019\/04\/11233917\/CNX_Calc_Figure_08_01_003.jpg\" alt=\"A picture of a baseball with an arrow underneath it pointing down. The arrow is labeled g = -9.8 m\/sec ^ 2.\" width=\"325\" height=\"212\" data-media-type=\"image\/jpeg\" \/> Figure 3. For a baseball falling in air, the only force acting on it is gravity (neglecting air resistance).[\/caption]<\/figure>\r\n<p class=\"whitespace-normal break-words\">We establish a reference frame where Earth's surface is at height 0 meters. Let [latex]v(t)[\/latex] represent the object's velocity in meters per second:<\/p>\r\n\r\n<ul class=\"[&amp;:not(:last-child)_ul]:pb-1 [&amp;:not(:last-child)_ol]:pb-1 list-disc space-y-1.5 pl-7\">\r\n \t<li class=\"whitespace-normal break-words\">If [latex]v(t) &gt; 0[\/latex], the object is rising<\/li>\r\n \t<li class=\"whitespace-normal break-words\">If [latex]v(t) &lt; 0[\/latex], the object is falling<\/li>\r\n<\/ul>\r\n[caption id=\"attachment_2343\" align=\"aligncenter\" width=\"275\"]<img class=\"size-medium wp-image-2343\" src=\"https:\/\/content-cdn.one.lumenlearning.com\/wp-content\/uploads\/sites\/53\/2025\/06\/18121147\/Screenshot-2025-09-18-081131-275x300.png\" alt=\"A picture of a baseball with an arrow above it pointing up and an arrow below it pointing down. The arrow pointing up is labeled v(t) &gt; 0, and the arrow pointing down is labeled v(t) &lt; 0.\" width=\"275\" height=\"300\" \/> Figure 4. Possible velocities for the rising\/falling baseball.[\/caption]\r\n<p class=\"whitespace-normal break-words\">Our goal is to find velocity [latex]v(t)[\/latex] at any time [latex]t[\/latex]. We start with Newton's second law. The force on the baseball equals mass times acceleration: [latex]F = mv'(t)[\/latex] (since acceleration is the derivative of velocity). Gravity exerts a downward force of [latex]F_g = -mg[\/latex] (negative because it acts downward).<\/p>\r\n<p class=\"whitespace-normal break-words\">Setting these forces equal:<\/p>\r\n<p class=\"whitespace-normal break-words\" style=\"text-align: center;\">[latex]mv'(t) = -mg[\/latex]<\/p>\r\n<p class=\"whitespace-normal break-words\">Dividing both sides by [latex]m[\/latex]:<\/p>\r\n<p class=\"whitespace-normal break-words\" style=\"text-align: center;\">[latex]v'(t) = -g[\/latex]<\/p>\r\n\r\n<section class=\"textbox proTip\" aria-label=\"Pro Tip\">Notice that mass cancels out completely. This means all objects fall at the same rate in a vacuum, regardless of their mass.<\/section>We need an initial condition to solve this differential equation. Since we're finding velocity, we specify the initial velocity: [latex]v(0) = v_0[\/latex].\r\n\r\n<section class=\"textbox example\" aria-label=\"Example\">\r\n<div id=\"fs-id1170571263710\" data-type=\"problem\">\r\n<p id=\"fs-id1170571273705\">A baseball is thrown upward from a height of [latex]3[\/latex] meters above Earth\u2019s surface with an initial velocity of [latex]10\\text{m\/s}[\/latex], and the only force acting on it is gravity. The ball has a mass of [latex]0.15\\text{kg}[\/latex] at Earth\u2019s surface.<\/p>\r\n\r\n<ol id=\"fs-id1170571216089\" type=\"a\">\r\n \t<li>Find the velocity [latex]v\\left(t\\right)[\/latex] of the baseball at time [latex]t[\/latex].<\/li>\r\n \t<li>What is its velocity after [latex]2[\/latex] seconds?<\/li>\r\n<\/ol>\r\n<\/div>\r\n[reveal-answer q=\"44558799\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"44558799\"]\r\n<div id=\"fs-id1170571270205\" data-type=\"solution\">\r\n<ol id=\"fs-id1170571270207\" type=\"a\">\r\n \t<li>From the preceding discussion, the differential equation that applies in this situation is<span data-type=\"newline\">\r\n<\/span>\r\n<div id=\"fs-id1170573401342\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]{v}^{\\prime }\\left(t\\right)=\\text{-}g[\/latex],<\/div>\r\n<span data-type=\"newline\">\r\n<\/span>\r\nwhere [latex]g=9.8{\\text{m\/s}}^{2}[\/latex]. The initial condition is [latex]v\\left(0\\right)={v}_{0}[\/latex], where [latex]{v}_{0}=10\\text{m\/s}\\text{.}[\/latex] Therefore the initial-value problem is [latex]{v}^{\\prime }\\left(t\\right)=-9.8{\\text{m\/s}}^{2},v\\left(0\\right)=10\\text{m\/s}\\text{.}[\/latex] <span data-type=\"newline\">\r\n<\/span>\r\nThe first step in solving this initial-value problem is to take the antiderivative of both sides of the differential equation. This gives<span data-type=\"newline\">\r\n<\/span>\r\n<div id=\"fs-id1170573289010\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{ccc}\\hfill {\\displaystyle\\int {v}^{\\prime }\\left(t\\right)dt}&amp; =\\hfill &amp; {\\displaystyle\\int -9.8dt}\\hfill \\\\ \\hfill v\\left(t\\right)&amp; =\\hfill &amp; -9.8t+C.\\hfill \\end{array}[\/latex]<\/div>\r\n<span data-type=\"newline\">\r\n<\/span>\r\nThe next step is to solve for [latex]C[\/latex]. To do this, substitute [latex]t=0[\/latex] and [latex]v\\left(0\\right)=10\\text{:}[\/latex] <span data-type=\"newline\">\r\n<\/span>\r\n<div id=\"fs-id1170573623436\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{ccc}\\hfill v\\left(t\\right)&amp; =\\hfill &amp; -9.8t+C\\hfill \\\\ \\hfill v\\left(0\\right)&amp; =\\hfill &amp; -9.8\\left(0\\right)+C\\hfill \\\\ \\hfill 10&amp; =\\hfill &amp; C.\\hfill \\end{array}[\/latex]<\/div>\r\n<span data-type=\"newline\">\r\n<\/span>\r\nTherefore [latex]C=10[\/latex] and the velocity function is given by [latex]v\\left(t\\right)=-9.8t+10[\/latex].<\/li>\r\n \t<li>To find the velocity after [latex]2[\/latex] seconds, substitute [latex]t=2[\/latex] into [latex]v\\left(t\\right)[\/latex]. <span data-type=\"newline\">\r\n<\/span>\r\n<div id=\"fs-id1170571152980\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{ccc}\\hfill v\\left(t\\right)&amp; =\\hfill &amp; -9.8t+10\\hfill \\\\ \\hfill v\\left(2\\right)&amp; =\\hfill &amp; -9.8\\left(2\\right)+10\\hfill \\\\ \\hfill v\\left(2\\right)&amp; =\\hfill &amp; -9.6.\\hfill \\end{array}[\/latex]<\/div>\r\n<span data-type=\"newline\">\r\n<\/span>\r\nThe units of velocity are meters per second. Since the answer is negative, the object is falling at a speed of [latex]9.6\\text{m\/s}\\text{.}[\/latex]<\/li>\r\n<\/ol>\r\n<\/div>\r\n[\/hidden-answer]\r\n\r\n<\/section><section class=\"textbox example\" aria-label=\"Example\">\r\n<div id=\"fs-id1170573352098\" data-type=\"problem\">\r\n<p id=\"fs-id1170573352100\">Suppose a rock falls from rest from a height of [latex]100[\/latex] meters and the only force acting on it is gravity. Find an equation for the velocity [latex]v\\left(t\\right)[\/latex] as a function of time, measured in meters per second.<\/p>\r\n\r\n<\/div>\r\n[reveal-answer q=\"44558599\"]Hint[\/reveal-answer]\r\n[hidden-answer a=\"44558599\"]\r\n<div id=\"fs-id1170573318642\" data-type=\"commentary\" data-element-type=\"hint\">\r\n\r\nWhat is the initial velocity of the rock? Use this with the differential equation in the example: Velocity of a Moving Baseball to form an initial-value problem, then solve for [latex]v\\left(t\\right)[\/latex].\r\n\r\n<\/div>\r\n[\/hidden-answer]\r\n\r\n[reveal-answer q=\"44558699\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"44558699\"]\r\n<div id=\"fs-id1170573493548\" data-type=\"solution\">\r\n<p id=\"fs-id1170573400886\">[latex]v\\left(t\\right)=-9.8t[\/latex]<\/p>\r\n\r\n<\/div>\r\n[\/hidden-answer]\r\n\r\n<\/section>\r\n<p class=\"whitespace-normal break-words\">Once you know how fast an object is moving, the next logical question is: where will it be at any given time? This is where we connect velocity to position.<\/p>\r\n<p class=\"whitespace-normal break-words\">Let [latex]s(t)[\/latex] represent the height of the object above Earth's surface, measured in meters. Since velocity is the rate of change of position, we have the fundamental relationship:<\/p>\r\n<p class=\"whitespace-normal break-words\" style=\"text-align: center;\">[latex]s'(t) = v(t)[\/latex]<\/p>\r\n\r\n<section class=\"textbox recall\" aria-label=\"Recall\">Remember that velocity is the derivative of position with respect to time. This means position is the antiderivative of velocity.<\/section>\r\n<p class=\"whitespace-normal break-words\">To solve for the position function, we need an initial condition. The most natural choice is the object's starting height: [latex]s(0) = s_0[\/latex]. This gives us the initial-value problem:<\/p>\r\n<p class=\"whitespace-normal break-words\" style=\"text-align: center;\">[latex]s'(t) = v(t), \\quad s(0) = s_0[\/latex]<\/p>\r\n<p class=\"whitespace-normal break-words\">With a known velocity function [latex]v(t)[\/latex], we can find the position function [latex]s(t)[\/latex] by taking the antiderivative and applying the initial condition.<\/p>\r\n\r\n<section class=\"textbox example\" aria-label=\"Example\">\r\n<div id=\"fs-id1170571046598\" data-type=\"problem\">\r\n<p id=\"fs-id1170573207847\">A baseball is thrown upward from a height of [latex]3[\/latex] meters above Earth\u2019s surface with an initial velocity of [latex]10\\text{m\/s}[\/latex], and the only force acting on it is gravity. The ball has a mass of [latex]0.15[\/latex] kilogram at Earth\u2019s surface.<\/p>\r\n\r\n<ol id=\"fs-id1170573719641\" type=\"a\">\r\n \t<li>Find the position [latex]s\\left(t\\right)[\/latex] of the baseball at time [latex]t[\/latex].<\/li>\r\n \t<li>What is its height after [latex]2[\/latex] seconds?<\/li>\r\n<\/ol>\r\n<\/div>\r\n[reveal-answer q=\"44458899\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"44458899\"]\r\n<div id=\"fs-id1170571232747\" data-type=\"solution\">\r\n<ol id=\"fs-id1170571232749\" type=\"a\">\r\n \t<li>We already know the velocity function for this problem is [latex]v\\left(t\\right)=-9.8t+10[\/latex]. The initial height of the baseball is [latex]3[\/latex] meters, so [latex]{s}_{0}=3[\/latex]. Therefore the initial-value problem for this example is<span data-type=\"newline\">\r\n<\/span>\r\nTo solve the initial-value problem, we first find the antiderivatives:<span data-type=\"newline\">\r\n<\/span>\r\n<div id=\"fs-id1170571207187\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{ccc}\\hfill {\\displaystyle\\int {s}^{\\prime }\\left(t\\right)dt}&amp; =\\hfill &amp; {\\displaystyle\\int -9.8t+10dt}\\hfill \\\\ \\hfill s\\left(t\\right)&amp; =\\hfill &amp; -4.9{t}^{2}+10t+C.\\hfill \\end{array}[\/latex]<\/div>\r\n<span data-type=\"newline\">\r\n<\/span>\r\nNext we substitute [latex]t=0[\/latex] and solve for [latex]C\\text{:}[\/latex] <span data-type=\"newline\">\r\n<\/span>\r\n<div id=\"fs-id1170573384332\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{ccc}\\hfill s\\left(t\\right)&amp; =\\hfill &amp; -4.9{t}^{2}+10t+C\\hfill \\\\ \\hfill s\\left(0\\right)&amp; =\\hfill &amp; -4.9{\\left(0\\right)}^{2}+10\\left(0\\right)+C\\hfill \\\\ \\hfill 3&amp; =\\hfill &amp; C.\\hfill \\end{array}[\/latex]<\/div>\r\n<span data-type=\"newline\">\r\n<\/span>\r\nTherefore the position function is [latex]s\\left(t\\right)=-4.9{t}^{2}+10t+3[\/latex].<\/li>\r\n \t<li>The height of the baseball after [latex]2\\text{s}[\/latex] is given by [latex]s\\left(2\\right)\\text{:}[\/latex] <span data-type=\"newline\">\r\n<\/span>\r\n<div id=\"fs-id1170571064335\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{cc}s\\left(2\\right)\\hfill &amp; =-4.9{\\left(2\\right)}^{2}+10\\left(2\\right)+3\\hfill \\\\ &amp; =-4.9\\left(4\\right)+23\\hfill \\\\ &amp; =3.4.\\hfill \\end{array}[\/latex]<\/div>\r\n<span data-type=\"newline\">\r\n<\/span>\r\nTherefore the baseball is [latex]3.4[\/latex] meters above Earth\u2019s surface after [latex]2[\/latex] seconds. It is worth noting that the mass of the ball cancelled out completely in the process of solving the problem.<\/li>\r\n<\/ol>\r\n<\/div>\r\n[\/hidden-answer]\r\n\r\n<\/section>","rendered":"<section class=\"textbox learningGoals\" aria-label=\"Learning Goals\">\n<ul>\n<li>Determine the order of a differential equation<\/li>\n<li>Tell the difference between a general solution and a particular solution<\/li>\n<li>Identify what makes a problem an initial-value problem<\/li>\n<li>Check if a function actually solves a given differential equation or initial-value problem<\/li>\n<\/ul>\n<\/section>\n<h2>Applications in Physics: Projectile Motion<\/h2>\n<p class=\"whitespace-normal break-words\">In physics and engineering, we analyze forces acting on objects to predict their motion. This approach is fundamental to understanding everything from falling objects to rocket trajectories. When an object moves near Earth&#8217;s surface, gravity is typically the dominant force we need to consider.<\/p>\n<p class=\"whitespace-normal break-words\">The beauty of this analysis lies in how we can use calculus and differential equations to model real-world phenomena. By applying Newton&#8217;s second law ([latex]F = ma[\/latex], where [latex]F[\/latex] is force, [latex]m[\/latex] is mass, and [latex]a[\/latex] is acceleration), we can derive equations that precisely describe motion under the influence of gravity.<\/p>\n<p>Consider a baseball falling through air, with gravity as the only acting force (we&#8217;ll ignore air resistance for now). At Earth&#8217;s surface, gravitational acceleration [latex]g[\/latex] is approximately [latex]9.8 \\text{ m\/s}^2[\/latex].<\/p>\n<figure id=\"CNX_Calc_Figure_08_01_003\"><figcaption><\/figcaption><figure style=\"width: 325px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/4175\/2019\/04\/11233917\/CNX_Calc_Figure_08_01_003.jpg\" alt=\"A picture of a baseball with an arrow underneath it pointing down. The arrow is labeled g = -9.8 m\/sec ^ 2.\" width=\"325\" height=\"212\" data-media-type=\"image\/jpeg\" \/><figcaption class=\"wp-caption-text\">Figure 3. For a baseball falling in air, the only force acting on it is gravity (neglecting air resistance).<\/figcaption><\/figure>\n<\/figure>\n<p class=\"whitespace-normal break-words\">We establish a reference frame where Earth&#8217;s surface is at height 0 meters. Let [latex]v(t)[\/latex] represent the object&#8217;s velocity in meters per second:<\/p>\n<ul class=\"[&amp;:not(:last-child)_ul]:pb-1 [&amp;:not(:last-child)_ol]:pb-1 list-disc space-y-1.5 pl-7\">\n<li class=\"whitespace-normal break-words\">If [latex]v(t) > 0[\/latex], the object is rising<\/li>\n<li class=\"whitespace-normal break-words\">If [latex]v(t) < 0[\/latex], the object is falling<\/li>\n<\/ul>\n<figure id=\"attachment_2343\" aria-describedby=\"caption-attachment-2343\" style=\"width: 275px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" class=\"size-medium wp-image-2343\" src=\"https:\/\/content-cdn.one.lumenlearning.com\/wp-content\/uploads\/sites\/53\/2025\/06\/18121147\/Screenshot-2025-09-18-081131-275x300.png\" alt=\"A picture of a baseball with an arrow above it pointing up and an arrow below it pointing down. The arrow pointing up is labeled v(t) &gt; 0, and the arrow pointing down is labeled v(t) &lt; 0.\" width=\"275\" height=\"300\" srcset=\"https:\/\/content-cdn.one.lumenlearning.com\/wp-content\/uploads\/sites\/53\/2025\/06\/18121147\/Screenshot-2025-09-18-081131-275x300.png 275w, https:\/\/content-cdn.one.lumenlearning.com\/wp-content\/uploads\/sites\/53\/2025\/06\/18121147\/Screenshot-2025-09-18-081131-65x71.png 65w, https:\/\/content-cdn.one.lumenlearning.com\/wp-content\/uploads\/sites\/53\/2025\/06\/18121147\/Screenshot-2025-09-18-081131-225x246.png 225w, https:\/\/content-cdn.one.lumenlearning.com\/wp-content\/uploads\/sites\/53\/2025\/06\/18121147\/Screenshot-2025-09-18-081131.png 318w\" sizes=\"(max-width: 275px) 100vw, 275px\" \/><figcaption id=\"caption-attachment-2343\" class=\"wp-caption-text\">Figure 4. Possible velocities for the rising\/falling baseball.<\/figcaption><\/figure>\n<p class=\"whitespace-normal break-words\">Our goal is to find velocity [latex]v(t)[\/latex] at any time [latex]t[\/latex]. We start with Newton&#8217;s second law. The force on the baseball equals mass times acceleration: [latex]F = mv'(t)[\/latex] (since acceleration is the derivative of velocity). Gravity exerts a downward force of [latex]F_g = -mg[\/latex] (negative because it acts downward).<\/p>\n<p class=\"whitespace-normal break-words\">Setting these forces equal:<\/p>\n<p class=\"whitespace-normal break-words\" style=\"text-align: center;\">[latex]mv'(t) = -mg[\/latex]<\/p>\n<p class=\"whitespace-normal break-words\">Dividing both sides by [latex]m[\/latex]:<\/p>\n<p class=\"whitespace-normal break-words\" style=\"text-align: center;\">[latex]v'(t) = -g[\/latex]<\/p>\n<section class=\"textbox proTip\" aria-label=\"Pro Tip\">Notice that mass cancels out completely. This means all objects fall at the same rate in a vacuum, regardless of their mass.<\/section>\n<p>We need an initial condition to solve this differential equation. Since we&#8217;re finding velocity, we specify the initial velocity: [latex]v(0) = v_0[\/latex].<\/p>\n<section class=\"textbox example\" aria-label=\"Example\">\n<div id=\"fs-id1170571263710\" data-type=\"problem\">\n<p id=\"fs-id1170571273705\">A baseball is thrown upward from a height of [latex]3[\/latex] meters above Earth\u2019s surface with an initial velocity of [latex]10\\text{m\/s}[\/latex], and the only force acting on it is gravity. The ball has a mass of [latex]0.15\\text{kg}[\/latex] at Earth\u2019s surface.<\/p>\n<ol id=\"fs-id1170571216089\" type=\"a\">\n<li>Find the velocity [latex]v\\left(t\\right)[\/latex] of the baseball at time [latex]t[\/latex].<\/li>\n<li>What is its velocity after [latex]2[\/latex] seconds?<\/li>\n<\/ol>\n<\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q44558799\">Show Solution<\/button><\/p>\n<div id=\"q44558799\" class=\"hidden-answer\" style=\"display: none\">\n<div id=\"fs-id1170571270205\" data-type=\"solution\">\n<ol id=\"fs-id1170571270207\" type=\"a\">\n<li>From the preceding discussion, the differential equation that applies in this situation is<span data-type=\"newline\"><br \/>\n<\/span><\/p>\n<div id=\"fs-id1170573401342\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]{v}^{\\prime }\\left(t\\right)=\\text{-}g[\/latex],<\/div>\n<p><span data-type=\"newline\"><br \/>\n<\/span><br \/>\nwhere [latex]g=9.8{\\text{m\/s}}^{2}[\/latex]. The initial condition is [latex]v\\left(0\\right)={v}_{0}[\/latex], where [latex]{v}_{0}=10\\text{m\/s}\\text{.}[\/latex] Therefore the initial-value problem is [latex]{v}^{\\prime }\\left(t\\right)=-9.8{\\text{m\/s}}^{2},v\\left(0\\right)=10\\text{m\/s}\\text{.}[\/latex] <span data-type=\"newline\"><br \/>\n<\/span><br \/>\nThe first step in solving this initial-value problem is to take the antiderivative of both sides of the differential equation. This gives<span data-type=\"newline\"><br \/>\n<\/span><\/p>\n<div id=\"fs-id1170573289010\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{ccc}\\hfill {\\displaystyle\\int {v}^{\\prime }\\left(t\\right)dt}& =\\hfill & {\\displaystyle\\int -9.8dt}\\hfill \\\\ \\hfill v\\left(t\\right)& =\\hfill & -9.8t+C.\\hfill \\end{array}[\/latex]<\/div>\n<p><span data-type=\"newline\"><br \/>\n<\/span><br \/>\nThe next step is to solve for [latex]C[\/latex]. To do this, substitute [latex]t=0[\/latex] and [latex]v\\left(0\\right)=10\\text{:}[\/latex] <span data-type=\"newline\"><br \/>\n<\/span><\/p>\n<div id=\"fs-id1170573623436\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{ccc}\\hfill v\\left(t\\right)& =\\hfill & -9.8t+C\\hfill \\\\ \\hfill v\\left(0\\right)& =\\hfill & -9.8\\left(0\\right)+C\\hfill \\\\ \\hfill 10& =\\hfill & C.\\hfill \\end{array}[\/latex]<\/div>\n<p><span data-type=\"newline\"><br \/>\n<\/span><br \/>\nTherefore [latex]C=10[\/latex] and the velocity function is given by [latex]v\\left(t\\right)=-9.8t+10[\/latex].<\/li>\n<li>To find the velocity after [latex]2[\/latex] seconds, substitute [latex]t=2[\/latex] into [latex]v\\left(t\\right)[\/latex]. <span data-type=\"newline\"><br \/>\n<\/span><\/p>\n<div id=\"fs-id1170571152980\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{ccc}\\hfill v\\left(t\\right)& =\\hfill & -9.8t+10\\hfill \\\\ \\hfill v\\left(2\\right)& =\\hfill & -9.8\\left(2\\right)+10\\hfill \\\\ \\hfill v\\left(2\\right)& =\\hfill & -9.6.\\hfill \\end{array}[\/latex]<\/div>\n<p><span data-type=\"newline\"><br \/>\n<\/span><br \/>\nThe units of velocity are meters per second. Since the answer is negative, the object is falling at a speed of [latex]9.6\\text{m\/s}\\text{.}[\/latex]<\/li>\n<\/ol>\n<\/div>\n<\/div>\n<\/div>\n<\/section>\n<section class=\"textbox example\" aria-label=\"Example\">\n<div id=\"fs-id1170573352098\" data-type=\"problem\">\n<p id=\"fs-id1170573352100\">Suppose a rock falls from rest from a height of [latex]100[\/latex] meters and the only force acting on it is gravity. Find an equation for the velocity [latex]v\\left(t\\right)[\/latex] as a function of time, measured in meters per second.<\/p>\n<\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q44558599\">Hint<\/button><\/p>\n<div id=\"q44558599\" class=\"hidden-answer\" style=\"display: none\">\n<div id=\"fs-id1170573318642\" data-type=\"commentary\" data-element-type=\"hint\">\n<p>What is the initial velocity of the rock? Use this with the differential equation in the example: Velocity of a Moving Baseball to form an initial-value problem, then solve for [latex]v\\left(t\\right)[\/latex].<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q44558699\">Show Solution<\/button><\/p>\n<div id=\"q44558699\" class=\"hidden-answer\" style=\"display: none\">\n<div id=\"fs-id1170573493548\" data-type=\"solution\">\n<p id=\"fs-id1170573400886\">[latex]v\\left(t\\right)=-9.8t[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/section>\n<p class=\"whitespace-normal break-words\">Once you know how fast an object is moving, the next logical question is: where will it be at any given time? This is where we connect velocity to position.<\/p>\n<p class=\"whitespace-normal break-words\">Let [latex]s(t)[\/latex] represent the height of the object above Earth&#8217;s surface, measured in meters. Since velocity is the rate of change of position, we have the fundamental relationship:<\/p>\n<p class=\"whitespace-normal break-words\" style=\"text-align: center;\">[latex]s'(t) = v(t)[\/latex]<\/p>\n<section class=\"textbox recall\" aria-label=\"Recall\">Remember that velocity is the derivative of position with respect to time. This means position is the antiderivative of velocity.<\/section>\n<p class=\"whitespace-normal break-words\">To solve for the position function, we need an initial condition. The most natural choice is the object&#8217;s starting height: [latex]s(0) = s_0[\/latex]. This gives us the initial-value problem:<\/p>\n<p class=\"whitespace-normal break-words\" style=\"text-align: center;\">[latex]s'(t) = v(t), \\quad s(0) = s_0[\/latex]<\/p>\n<p class=\"whitespace-normal break-words\">With a known velocity function [latex]v(t)[\/latex], we can find the position function [latex]s(t)[\/latex] by taking the antiderivative and applying the initial condition.<\/p>\n<section class=\"textbox example\" aria-label=\"Example\">\n<div id=\"fs-id1170571046598\" data-type=\"problem\">\n<p id=\"fs-id1170573207847\">A baseball is thrown upward from a height of [latex]3[\/latex] meters above Earth\u2019s surface with an initial velocity of [latex]10\\text{m\/s}[\/latex], and the only force acting on it is gravity. The ball has a mass of [latex]0.15[\/latex] kilogram at Earth\u2019s surface.<\/p>\n<ol id=\"fs-id1170573719641\" type=\"a\">\n<li>Find the position [latex]s\\left(t\\right)[\/latex] of the baseball at time [latex]t[\/latex].<\/li>\n<li>What is its height after [latex]2[\/latex] seconds?<\/li>\n<\/ol>\n<\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q44458899\">Show Solution<\/button><\/p>\n<div id=\"q44458899\" class=\"hidden-answer\" style=\"display: none\">\n<div id=\"fs-id1170571232747\" data-type=\"solution\">\n<ol id=\"fs-id1170571232749\" type=\"a\">\n<li>We already know the velocity function for this problem is [latex]v\\left(t\\right)=-9.8t+10[\/latex]. The initial height of the baseball is [latex]3[\/latex] meters, so [latex]{s}_{0}=3[\/latex]. Therefore the initial-value problem for this example is<span data-type=\"newline\"><br \/>\n<\/span><br \/>\nTo solve the initial-value problem, we first find the antiderivatives:<span data-type=\"newline\"><br \/>\n<\/span><\/p>\n<div id=\"fs-id1170571207187\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{ccc}\\hfill {\\displaystyle\\int {s}^{\\prime }\\left(t\\right)dt}& =\\hfill & {\\displaystyle\\int -9.8t+10dt}\\hfill \\\\ \\hfill s\\left(t\\right)& =\\hfill & -4.9{t}^{2}+10t+C.\\hfill \\end{array}[\/latex]<\/div>\n<p><span data-type=\"newline\"><br \/>\n<\/span><br \/>\nNext we substitute [latex]t=0[\/latex] and solve for [latex]C\\text{:}[\/latex] <span data-type=\"newline\"><br \/>\n<\/span><\/p>\n<div id=\"fs-id1170573384332\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{ccc}\\hfill s\\left(t\\right)& =\\hfill & -4.9{t}^{2}+10t+C\\hfill \\\\ \\hfill s\\left(0\\right)& =\\hfill & -4.9{\\left(0\\right)}^{2}+10\\left(0\\right)+C\\hfill \\\\ \\hfill 3& =\\hfill & C.\\hfill \\end{array}[\/latex]<\/div>\n<p><span data-type=\"newline\"><br \/>\n<\/span><br \/>\nTherefore the position function is [latex]s\\left(t\\right)=-4.9{t}^{2}+10t+3[\/latex].<\/li>\n<li>The height of the baseball after [latex]2\\text{s}[\/latex] is given by [latex]s\\left(2\\right)\\text{:}[\/latex] <span data-type=\"newline\"><br \/>\n<\/span><\/p>\n<div id=\"fs-id1170571064335\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{cc}s\\left(2\\right)\\hfill & =-4.9{\\left(2\\right)}^{2}+10\\left(2\\right)+3\\hfill \\\\ & =-4.9\\left(4\\right)+23\\hfill \\\\ & =3.4.\\hfill \\end{array}[\/latex]<\/div>\n<p><span data-type=\"newline\"><br \/>\n<\/span><br \/>\nTherefore the baseball is [latex]3.4[\/latex] meters above Earth\u2019s surface after [latex]2[\/latex] seconds. It is worth noting that the mass of the ball cancelled out completely in the process of solving the problem.<\/li>\n<\/ol>\n<\/div>\n<\/div>\n<\/div>\n<\/section>\n","protected":false},"author":15,"menu_order":10,"template":"","meta":{"_candela_citation":"[]","pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"part":669,"module-header":"- Select Header -","content_attributions":[],"internal_book_links":[],"video_content":null,"cc_video_embed_content":{"cc_scripts":"","media_targets":[]},"try_it_collection":null,"_links":{"self":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapters\/802"}],"collection":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/wp\/v2\/users\/15"}],"version-history":[{"count":8,"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapters\/802\/revisions"}],"predecessor-version":[{"id":2346,"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapters\/802\/revisions\/2346"}],"part":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/parts\/669"}],"metadata":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapters\/802\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/wp\/v2\/media?parent=802"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapter-type?post=802"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/wp\/v2\/contributor?post=802"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/wp\/v2\/license?post=802"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}