{"id":1802,"date":"2022-11-18T09:30:24","date_gmt":"2022-11-18T04:00:24","guid":{"rendered":"https:\/\/www.aplustopper.com\/?p=1802"},"modified":"2022-11-19T09:52:52","modified_gmt":"2022-11-19T04:22:52","slug":"equations-of-motion","status":"publish","type":"post","link":"https:\/\/www.aplustopper.com\/equations-of-motion\/","title":{"rendered":"What Are The Equations Of Motion"},"content":{"rendered":"<h2><strong>Equations Of Motion<\/strong><\/h2>\n<p><strong>Motion under uniform acceleration<\/strong><\/p>\n<p><strong>1<sup>st<\/sup>\u00a0Equation of <a href=\"https:\/\/www.aplustopper.com\/motion-and-types-of-motion\/\">motion<\/a><\/strong><br \/>\nFor an object moving with uniform <a href=\"https:\/\/www.aplustopper.com\/difference-between-speed-and-velocity\/\">velocity<\/a>, v, its <a href=\"https:\/\/www.aplustopper.com\/difference-between-distance-and-displacement\/\">displacement<\/a>, s after time, t is given by:<br \/>\n<strong>s = v \u00d7\u00a0t<\/strong><br \/>\nConsider a body having initial velocity &#8216;u&#8217;. Suppose it is subjected to a uniform <a href=\"https:\/\/www.aplustopper.com\/acceleration-and-types-of-acceleration\/\">acceleration<\/a> &#8216;a&#8217; so that after time &#8216;t&#8217; its final velocity becomes &#8216;v&#8217;. Now we know,<br \/>\n\\( \\text{ }\\!\\!~\\!\\!\\text{ Acceleration}=\\frac{\\text{change}\\,\\,\\text{in}\\,\\text{velocity}}{\\text{Time}} \\)<br \/>\n\\( a=\\frac{v-u}{t} \\)<br \/>\n<strong>or \u00a0 \u00a0 v = u + at \u00a0 \u00a0 \u00a0 \u00a0 \u00a0\u2026..(i)<\/strong><\/p>\n<p><strong>2<sup>nd<\/sup> Equation of motion<\/strong><br \/>\nSuppose a body has an initial velocity &#8216;u&#8217; and uniform acceleration &#8216;a&#8217; for time &#8216;t&#8217; so that its final velocity becomes &#8216;v&#8217;. The <a href=\"https:\/\/www.aplustopper.com\/difference-between-distance-and-displacement\/\">distance<\/a> travelled by moving body in time &#8216;t&#8217; is &#8216;s&#8217; then the average velocity = (v + u)\/2.<br \/>\nDistance travelled = Average velocity \u00d7 time<br \/>\n\\( \\text{S}=\\left( \\frac{u+v}{2} \\right)t\\text{ } \\)<br \/>\n\\( \\Rightarrow \\text{S}=\\left( \\frac{u+u+at}{2} \\right)t\\text{ }\\left( as\\text{ }v=u+at \\right)~ \\)<br \/>\n\\( \\Rightarrow \\text{S}=\\left( \\frac{2u+at}{2} \\right)t \\)<br \/>\n\\( \\Rightarrow \\text{S}=\\frac{2ut+a{{t}^{2}}}{2} \\)<br \/>\n\\( \\text{S}=ut+\\frac{1}{2}a{{t}^{2}}\\text{ }&#8230;&#8230;\\text{ (ii)} \\)<\/p>\n<p><strong>3<sup>rd<\/sup> Equation of motion<\/strong><br \/>\nDistance travelled = Average velocity x time<br \/>\n\\( S=\\left( \\frac{u+v}{2} \\right)t\\text{ }&#8230;&#8230;&#8230;\\text{ (iii)} \\)<br \/>\n\\( \\text{from equation }\\left( \\text{i} \\right)\\text{ }t=\\frac{v-u}{a} \\)<br \/>\nSubstituting the value of t in equation (iii),<br \/>\n\\( \\text{we get }S=\\left( \\frac{v-u}{a} \\right)\\left( \\frac{v+u}{2} \\right) \\)<br \/>\n\\( S=\\left( \\frac{{{v}^{2}}-{{u}^{2}}}{2a} \\right) \\)<br \/>\n\u21d2\u00a02as = v<sup>2<\/sup> \u2013 u<sup>2<\/sup> or<br \/>\n<strong>v<sup>2<\/sup> = u<sup>2<\/sup> + 2as \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0\u2026.(iv)<\/strong><\/p>\n<p>The equations of motion under gravity can be obtained by replacing acceleration by acceleration due to gravity (g) and can be written as follows<\/p>\n<ul>\n<li>When the body is coming towards the centre of earth<br \/>\n(a) v = u + gt \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0(b) h = ut + \\(\\frac { 1 }{ 2 }\\) gt<sup>2<\/sup>\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0(c) v<sup>2<\/sup> = u<sup>2<\/sup> + 2gh<\/li>\n<li>When a body is thrown upwards with some initial velocity, then a retardation produced due to attraction of the earth. In equations of motion, a is replaced by (\u2013g) and thus equations become.<br \/>\n(a) v = u \u2013 gt \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 (b) h = ut \u2013 \\(\\frac { 1 }{ 2 }\\) gt<sup>2<\/sup>\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 (c) v<sup>2<\/sup>\u00a0= u<sup>2<\/sup> \u2013 2gh<\/li>\n<\/ul>\n<h3><strong>Body Falling Freely Under Gravity<\/strong><\/h3>\n<p>Assuming u = 0 for a freely falling body<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"alignnone wp-image-1816 size-full\" src=\"https:\/\/www.aplustopper.com\/wp-content\/uploads\/2016\/09\/Body-Falling-Freely-Under-Gravity-2.png\" alt=\"Body-Falling-Freely-Under-Gravity\" width=\"420\" height=\"290\" srcset=\"https:\/\/www.aplustopper.com\/wp-content\/uploads\/2016\/09\/Body-Falling-Freely-Under-Gravity-2.png 420w, https:\/\/www.aplustopper.com\/wp-content\/uploads\/2016\/09\/Body-Falling-Freely-Under-Gravity-2-300x207.png 300w\" sizes=\"(max-width: 420px) 100vw, 420px\" \/><\/p>\n<p>Body is projected vertically up:<br \/>\nTaking initial position as origin and direction of motion (i.e. vertically up) as positive.<br \/>\n(a) At the highest point v = 0<br \/>\n(b) a = \u2013 g<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"alignnone wp-image-1815 size-full\" src=\"https:\/\/www.aplustopper.com\/wp-content\/uploads\/2016\/09\/Body-Falling-Freely-Under-Gravity-1.png\" alt=\"Body-Falling-Freely-Under-Gravity-1\" width=\"418\" height=\"317\" srcset=\"https:\/\/www.aplustopper.com\/wp-content\/uploads\/2016\/09\/Body-Falling-Freely-Under-Gravity-1.png 418w, https:\/\/www.aplustopper.com\/wp-content\/uploads\/2016\/09\/Body-Falling-Freely-Under-Gravity-1-300x228.png 300w\" sizes=\"(max-width: 418px) 100vw, 418px\" \/><br \/>\nIt is clear that in case of motion under gravity<br \/>\n(a) Time taken to go up is equal to the time taken to fall down through the same distance.<br \/>\n(b) The speed with which a body is projected up is equal to the speed with which it comes back to the point of projection.<br \/>\n(c) The body returns to the starting point with the same speed with which it was thrown.<\/p>\n<h2><strong>Equations Of Motion Example Problems With Solutions<\/strong><\/h2>\n<p><strong>Example 1.<\/strong> A car is travelling with a uniform velocity of 80 km h<sup>-1<\/sup> northward from Johor Bahru.<br \/>\nWhat is its displacement after 15 minutes?<br \/>\n<strong>Solution:<\/strong><br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-91175\" src=\"https:\/\/www.aplustopper.com\/wp-content\/uploads\/2016\/09\/What-Are-The-Equations-Of-Motion-1.jpg\" alt=\"What Are The Equations Of Motion 1\" width=\"393\" height=\"142\" srcset=\"https:\/\/www.aplustopper.com\/wp-content\/uploads\/2016\/09\/What-Are-The-Equations-Of-Motion-1.jpg 393w, https:\/\/www.aplustopper.com\/wp-content\/uploads\/2016\/09\/What-Are-The-Equations-Of-Motion-1-300x108.jpg 300w\" sizes=\"(max-width: 393px) 100vw, 393px\" \/><\/p>\n<p><strong>Example 2.<\/strong> A train travelling in a straight line at 30 ms<sup>-1<\/sup>accelerates uniformly to 54 ms<sup>-1<\/sup> in 3.0 seconds. Calculate the distance travelled by the train during that time.<br \/>\n<strong>Solution:<\/strong><br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-91186\" src=\"https:\/\/www.aplustopper.com\/wp-content\/uploads\/2016\/09\/What-Are-The-Equations-Of-Motion-2.jpg\" alt=\"What Are The Equations Of Motion 2\" width=\"386\" height=\"383\" srcset=\"https:\/\/www.aplustopper.com\/wp-content\/uploads\/2016\/09\/What-Are-The-Equations-Of-Motion-2.jpg 386w, https:\/\/www.aplustopper.com\/wp-content\/uploads\/2016\/09\/What-Are-The-Equations-Of-Motion-2-150x150.jpg 150w, https:\/\/www.aplustopper.com\/wp-content\/uploads\/2016\/09\/What-Are-The-Equations-Of-Motion-2-300x298.jpg 300w, https:\/\/www.aplustopper.com\/wp-content\/uploads\/2016\/09\/What-Are-The-Equations-Of-Motion-2-100x100.jpg 100w\" sizes=\"(max-width: 386px) 100vw, 386px\" \/><\/p>\n<p><strong>Example 3.<\/strong> A school bus accelerates with an acceleration of 4.0 m s<sup>-2<\/sup>\u00a0after picking up some students at a bus stop.<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-91198\" src=\"https:\/\/www.aplustopper.com\/wp-content\/uploads\/2016\/09\/What-Are-The-Equations-Of-Motion-3.jpg\" alt=\"What Are The Equations Of Motion 3\" width=\"248\" height=\"121\" \/><br \/>\nCalculate the<br \/>\n(a) velocity of the bus after 5 s.<br \/>\n(b) distance travelled by the bus after 5 s.<br \/>\n<strong>Solution:<\/strong><br \/>\nInitial velocity, u = 0 ms<sup>-1<\/sup><br \/>\nAcceleration, a = 4.0 ms<sup>-2<\/sup><br \/>\nTime, t = 5 s<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-91202\" src=\"https:\/\/www.aplustopper.com\/wp-content\/uploads\/2016\/09\/What-Are-The-Equations-Of-Motion-4.jpg\" alt=\"What Are The Equations Of Motion 4\" width=\"256\" height=\"213\" \/><\/p>\n<p><strong>Example 4.\u00a0<\/strong>A long jumper was running at a velocity of 5 m s<sup>-1<\/sup>\u00a0towards the long jump pit. He needed to achieve a velocity of 10 m s<sup>-1<\/sup>\u00a0after covering a distance of 4.5 m before lifting himself off the ground from the jumping board.<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-91204\" src=\"https:\/\/www.aplustopper.com\/wp-content\/uploads\/2016\/09\/What-Are-The-Equations-Of-Motion-5.jpg\" alt=\"What Are The Equations Of Motion 5\" width=\"358\" height=\"157\" srcset=\"https:\/\/www.aplustopper.com\/wp-content\/uploads\/2016\/09\/What-Are-The-Equations-Of-Motion-5.jpg 358w, https:\/\/www.aplustopper.com\/wp-content\/uploads\/2016\/09\/What-Are-The-Equations-Of-Motion-5-300x132.jpg 300w\" sizes=\"(max-width: 358px) 100vw, 358px\" \/><br \/>\n(a) Calculate the required acceleration for him to do so.<br \/>\n(b) Calculate the time taken for him to cover the horizontal distance of 4.5 m.<br \/>\n<strong>Solution:<\/strong><br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-91232\" src=\"https:\/\/www.aplustopper.com\/wp-content\/uploads\/2016\/09\/What-Are-The-Equations-Of-Motion-6.jpg\" alt=\"What Are The Equations Of Motion 6\" width=\"267\" height=\"287\" \/><br \/>\n<strong>Example 5.\u00a0<\/strong>10 A body starts moving with an initial velocity 50 m\/s and acceleration 20 m\/s<sup>2<\/sup>. How much distance it will cover in 4s ? Also, calculate its average speed during this time interval.<br \/>\n<strong>Solution:<\/strong> \u00a0 \u00a0Given: u = 50 m\/s, a = 20 m\/s<sup>2<\/sup>, \u00a0t = 4s, s = ?<br \/>\ns = ut + \\(\\frac { 1 }{ 2 }\\) at<sup>2<\/sup> = 50 \u00d7 4 + \\(\\frac { 1 }{ 2 }\\) \u00d7 20 \u00d7 (4)<sup>2<\/sup><br \/>\n= 200 + 160 = 360 m<br \/>\nAverage speed during this interval,<br \/>\n\\(\\overline{V}=\\frac{\\text{distance travelled}}{\\text{time interval}}=\\frac{360}{4}=90\\text{ m\/s}\\)<\/p>\n<p><strong>Example 6.\u00a0<\/strong>A body is moving with a speed of 20 m\/s. When certain force is applied, an acceleration of 4 m\/s<sup>2<\/sup> is produced. After how much time its velocity will be 80 m\/s ?<br \/>\n<strong>Solution:<\/strong> \u00a0 \u00a0Given: u = 20 m\/s, a = 4 m\/s<sup>2<\/sup>,\u00a0v = 80 m\/s, t = ?<br \/>\nUsing equation, v = u + at, we get<br \/>\n80 = 20 + 4 \u00d7 t<br \/>\nor \u00a0 \u00a0 4t = 80 \u2013 20 = 60<br \/>\nor \u00a0 \u00a0 \u00a0t = 15 s<br \/>\nTherefore, after 15 seconds, the velocity of the body will be 80 m\/s.<\/p>\n<p><strong>Example 7.\u00a0<\/strong>A body starts from rest and moves with a constant acceleration. It travels a distance s<sub>1<\/sub>\u00a0in first 10 s, and a distance s<sub>2<\/sub> in next 10 s. Find the relation between s<sub>2\u00a0<\/sub>and s<sub>1<\/sub>.<br \/>\n<strong>Solution:<\/strong> \u00a0 \u00a0Given : u = 0, t<sub>1<\/sub> = 10 s<br \/>\n\u2234 Distance travelled in first 10 seconds, is given by<br \/>\ns<sub>1<\/sub> = ut + \\(\\frac { 1 }{ 2 }\\) at<sup>2<\/sup> = 0 + \\(\\frac { 1 }{ 2 }\\) \u00d7 a \u00d7 (10)<sup>2<\/sup><br \/>\ns<sub>1<\/sub> = 50a \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0&#8230;(1)<br \/>\nTo calculate the distance travelled in next 10s, we first calculate distance travelled in 20 s and then subtract distance travelled in first 10 s.<br \/>\ns = ut + \\(\\frac { 1 }{ 2 }\\) at<sup>2<\/sup> = 0 + \\(\\frac { 1 }{ 2 }\\) \u00d7 a \u00d7 (20)<sup>2<\/sup><br \/>\ns = 200a \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 &#8230;(2)<br \/>\n\u2234 Distance travelled in 10th second interval,<br \/>\ns<sub>2<\/sub> = s \u2013 s<sub>1<\/sub> = 200a \u2013 50a \u00a0 \u00a0 \u00a0 \u00a0 \u00a0&#8230;(3)<br \/>\nor \u00a0 \u00a0s<sub>2<\/sub> = 150a<br \/>\n\\(\\text{Now, \u00a0 \u00a0}\\frac{{{s}_{2}}}{{{s}_{1}}}=\\frac{150a}{50a}=\\frac{3}{1}\\)<br \/>\nor \u00a0 \u00a0 \u00a0s<sub>2<\/sub> = 3s<sub>1<\/sub><\/p>\n<p><strong>Example 8.\u00a0<\/strong>A train is moving with a velocity 400 m\/s. With the application of brakes a retardation of 10 m\/s<sup>2<\/sup> is produced. Calculate the following<br \/>\n(i) After how much time it will stop ?<br \/>\n(ii) How much distance will it travel before it stops ?<br \/>\n<strong>Solution:<\/strong> \u00a0 \u00a0<strong>(i)<\/strong> Given: u = 400 m\/s, a = \u201310 m\/s<sup>2<\/sup>, v = 0, t = ?<br \/>\nUsing equation, v = u + at, we get<br \/>\n0 = 400 + (\u201310) \u00d7 t<br \/>\nor t = 40 s<br \/>\n<strong>(ii)<\/strong> For calculating the distance travelled, we use equation,<br \/>\nv<sup>2<\/sup> = u<sup>2<\/sup> + 2as, we get<br \/>\n(0)<sup>2<\/sup> = (400)<sup>2<\/sup> + 2 \u00d7 (\u201310) \u00d7 s<br \/>\nor \u00a0 \u00a0 \u00a020s = 400 \u00d7 400<br \/>\nor \u00a0 \u00a0 \u00a0 s = 8000 m = 8 km<\/p>\n<p><strong>Example 9.\u00a0<\/strong>A body is thrown vertically upwards with an initial velocity of 19.6 m\/s. If g = \u20139.8 m\/s<sup>2<\/sup>. Calculate the following<br \/>\n(i) The maximum height attained by the body.<br \/>\n(ii) After how much time will it come back to the ground ?<br \/>\n<strong>Solution:<\/strong> \u00a0 \u00a0<strong>(i)<\/strong> Given: u = 19.6 m\/s, g = \u20139.8 m\/s<sup>2<\/sup>, v = 0, h = ?<br \/>\nUsing equation v<sup>2<\/sup> = u<sup>2<\/sup> + 2gh, we get<br \/>\n(0)<sup>2<\/sup> = (19.6)<sup>2<\/sup> + 2(\u20139.8) \u00d7 h<br \/>\n\\(h=\\frac{19.6\\times 19.6}{2\\times 9.8}=19.6\\text{ m}\\)<br \/>\n<strong>(ii)<\/strong> Time taken to reach the maximum height can be calculated by the equation,<br \/>\nv = u + gt<br \/>\n0 = 19.6 + (\u20139.8) \u00d7 t<br \/>\nt = 2s<br \/>\nIn the same time, it will come back to its original position.<br \/>\n\u2234 Total time = 2 \u00d7 2 = 4s<\/p>\n<p><strong>Example 10.\u00a0<\/strong>From the top of a tower of height 490 m, a shell is fired horizontally with a velocity 100 m\/s. At what distance from the bottom of the tower, the shell will hit the ground ?<br \/>\n<strong>Solution:<\/strong> \u00a0 \u00a0We know that the horizontal motion and the vertical motion are independent of each other. Now for vertical motion, we have u = 0,<br \/>\nh = 490 m, g = 9.8 m\/s<sup>2<\/sup>, t = ?<br \/>\nUsing equation, h = ut + \\(\\frac { 1 }{ 2 }\\) gt<sup>2<\/sup>, we get<br \/>\n490 = 0 + \\(\\frac { 1 }{ 2 }\\) \u00d7 9.8 \u00d7 t<sup>2<\/sup><br \/>\nor \u00a0 \u00a0 t<sup>2<\/sup> = \\(\\frac { 490 }{ 4.9 }\\) = 100<br \/>\nor \u00a0 \u00a0 t = 10 s<br \/>\n\u2234 It takes 10 seconds to reach the ground.<br \/>\nNow, horizontal distance\u00a0= horizontal velocity \u00d7 time<br \/>\n= 100 m\/s \u00d7 10 s = 1000 m<br \/>\n\u2234 The shell will strike the ground at a distance of 100 m from the bottom of the tower.<\/p>\n","protected":false},"excerpt":{"rendered":"<p>Equations Of Motion Motion under uniform acceleration 1st\u00a0Equation of motion For an object moving with uniform velocity, v, its displacement, s after time, t is given by: s = v \u00d7\u00a0t Consider a body having initial velocity &#8216;u&#8217;. Suppose it is subjected to a uniform acceleration &#8216;a&#8217; so that after time &#8216;t&#8217; its final velocity [&hellip;]<\/p>\n","protected":false},"author":3,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"_genesis_hide_title":false,"_genesis_hide_breadcrumbs":false,"_genesis_hide_singular_image":false,"_genesis_hide_footer_widgets":false,"_genesis_custom_body_class":"","_genesis_custom_post_class":"","_genesis_layout":"","footnotes":""},"categories":[404],"tags":[648,1011,622],"yoast_head":"<!-- This site is optimized with the Yoast SEO Premium plugin v17.8 (Yoast SEO v17.8) - https:\/\/yoast.com\/wordpress\/plugins\/seo\/ -->\n<title>What Are The Equations Of Motion - A Plus Topper<\/title>\n<meta name=\"robots\" content=\"index, follow, max-snippet:-1, max-image-preview:large, max-video-preview:-1\" \/>\n<link rel=\"canonical\" href=\"https:\/\/www.aplustopper.com\/equations-of-motion\/\" \/>\n<meta property=\"og:locale\" content=\"en_US\" \/>\n<meta property=\"og:type\" content=\"article\" \/>\n<meta property=\"og:title\" content=\"What Are The Equations Of Motion\" \/>\n<meta property=\"og:description\" content=\"Equations Of Motion Motion under uniform acceleration 1st\u00a0Equation of motion For an object moving with uniform velocity, v, its displacement, s after time, t is given by: s = v \u00d7\u00a0t Consider a body having initial velocity &#8216;u&#8217;. 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