{"id":89000,"date":"2025-06-01T07:26:43","date_gmt":"2025-06-01T07:26:43","guid":{"rendered":"https:\/\/exam.pscnotes.com\/mcq\/?p=89000"},"modified":"2025-06-01T07:26:43","modified_gmt":"2025-06-01T07:26:43","slug":"which-one-of-the-following-graphs-represents-the-equation-of-motion-v","status":"publish","type":"post","link":"https:\/\/exam.pscnotes.com\/mcq\/which-one-of-the-following-graphs-represents-the-equation-of-motion-v\/","title":{"rendered":"Which one of the following graphs represents the equation of motion v"},"content":{"rendered":"

Which one of the following graphs represents the equation of motion v = u + at; where all quantities are non-zero and symbols carry their usual meanings ?<\/p>\n

[amp_mcq option1=”(a)” option2=”(b)” option3=”(c)” option4=”(d)” correct=”option3″]<\/p>\n

\n
\n
This question was previously asked in<\/div>\n
UPSC NDA-2 – 2023<\/div>\n<\/div>\n
Download PDF<\/a>Attempt Online<\/a><\/div>\n<\/div>\n
\nThe equation of motion is given by v = u + at, where:
\n– v is the final velocity
\n– u is the initial velocity
\n– a is the acceleration (constant)
\n– t is the time
\nThis equation is in the form of a linear equation y = mx + c, where v is the dependent variable (y), t is the independent variable (x), ‘a’ is the slope (m), and ‘u’ is the y-intercept (c). The problem states that all quantities (u, a, v, t) are non-zero, except t=0 at the start.
\nA graph of v versus t for this equation should be a straight line.
\nSince ‘u’ is non-zero, the line will intersect the v-axis (at t=0) at a point other than the origin.
\nSince ‘a’ is non-zero, the slope of the line will not be zero (the line is not horizontal).
\nConsidering these points, graph (c) represents a straight line with a positive y-intercept (u > 0) and a positive slope (a > 0). While ‘a’ could be negative (negative slope) or ‘u’ could be negative (negative intercept), option (c) is the only graph that shows a straight line with a non-zero intercept and a non-zero slope, fitting the general form v = u + at with non-zero u and a. Graphs (a) and (d) are not straight lines, and graph (b) is a straight line passing through the origin, which would represent v = at (i.e., u=0), contrary to the condition that u is non-zero.
\n<\/section>\n
\n– The equation v = u + at represents linear motion with constant acceleration.
\n– A graph of velocity (v) versus time (t) for this equation is a straight line.
\n– The y-intercept of the line is the initial velocity (u).
\n– The slope of the line is the acceleration (a).
\n<\/section>\n
\nIf ‘a’ were zero, the graph would be a horizontal line at v = u. If ‘u’ were zero, the line would pass through the origin (v = at). The shape of the line (upward or downward slope) depends on the sign of acceleration ‘a’. Option (c) shows a positive slope (acceleration).
\n<\/section>\n","protected":false},"excerpt":{"rendered":"

Which one of the following graphs represents the equation of motion v = u + at; where all quantities are non-zero and symbols carry their usual meanings ? [amp_mcq option1=”(a)” option2=”(b)” option3=”(c)” option4=”(d)” correct=”option3″] This question was previously asked in UPSC NDA-2 – 2023 Download PDFAttempt Online The equation of motion is given by v … <\/p>\n

Detailed SolutionWhich one of the following graphs represents the equation of motion v<\/span><\/a><\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"closed","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[1094],"tags":[1105,1129,1128],"class_list":["post-89000","post","type-post","status-publish","format-standard","hentry","category-upsc-nda-2","tag-1105","tag-mechanics","tag-physics","no-featured-image-padding"],"yoast_head":"\nWhich one of the following graphs represents the equation of motion v<\/title>\n<meta name=\"description\" content=\"The equation of motion is given by v = u + at, where: - v is the final velocity - u is the initial velocity - a is the acceleration (constant) - t is the time This equation is in the form of a linear equation y = mx + c, where v is the dependent variable (y), t is the independent variable (x), 'a' is the slope (m), and 'u' is the y-intercept (c). The problem states that all quantities (u, a, v, t) are non-zero, except t=0 at the start. A graph of v versus t for this equation should be a straight line. Since 'u' is non-zero, the line will intersect the v-axis (at t=0) at a point other than the origin. Since 'a' is non-zero, the slope of the line will not be zero (the line is not horizontal). Considering these points, graph (c) represents a straight line with a positive y-intercept (u > 0) and a positive slope (a > 0). While 'a' could be negative (negative slope) or 'u' could be negative (negative intercept), option (c) is the only graph that shows a straight line with a non-zero intercept and a non-zero slope, fitting the general form v = u + at with non-zero u and a. Graphs (a) and (d) are not straight lines, and graph (b) is a straight line passing through the origin, which would represent v = at (i.e., u=0), contrary to the condition that u is non-zero. - The equation v = u + at represents linear motion with constant acceleration. - A graph of velocity (v) versus time (t) for this equation is a straight line. - The y-intercept of the line is the initial velocity (u). - The slope of the line is the acceleration (a).\" \/>\n<meta name=\"robots\" content=\"index, follow, max-snippet:-1, max-image-preview:large, max-video-preview:-1\" \/>\n<link rel=\"canonical\" href=\"https:\/\/exam.pscnotes.com\/mcq\/which-one-of-the-following-graphs-represents-the-equation-of-motion-v\/\" \/>\n<meta property=\"og:locale\" content=\"en_US\" \/>\n<meta property=\"og:type\" content=\"article\" \/>\n<meta property=\"og:title\" content=\"Which one of the following graphs represents the equation of motion v\" \/>\n<meta property=\"og:description\" content=\"The equation of motion is given by v = u + at, where: - v is the final velocity - u is the initial velocity - a is the acceleration (constant) - t is the time This equation is in the form of a linear equation y = mx + c, where v is the dependent variable (y), t is the independent variable (x), 'a' is the slope (m), and 'u' is the y-intercept (c). The problem states that all quantities (u, a, v, t) are non-zero, except t=0 at the start. A graph of v versus t for this equation should be a straight line. Since 'u' is non-zero, the line will intersect the v-axis (at t=0) at a point other than the origin. Since 'a' is non-zero, the slope of the line will not be zero (the line is not horizontal). Considering these points, graph (c) represents a straight line with a positive y-intercept (u > 0) and a positive slope (a > 0). While 'a' could be negative (negative slope) or 'u' could be negative (negative intercept), option (c) is the only graph that shows a straight line with a non-zero intercept and a non-zero slope, fitting the general form v = u + at with non-zero u and a. Graphs (a) and (d) are not straight lines, and graph (b) is a straight line passing through the origin, which would represent v = at (i.e., u=0), contrary to the condition that u is non-zero. - The equation v = u + at represents linear motion with constant acceleration. - A graph of velocity (v) versus time (t) for this equation is a straight line. - The y-intercept of the line is the initial velocity (u). - The slope of the line is the acceleration (a).\" \/>\n<meta property=\"og:url\" content=\"https:\/\/exam.pscnotes.com\/mcq\/which-one-of-the-following-graphs-represents-the-equation-of-motion-v\/\" \/>\n<meta property=\"og:site_name\" content=\"MCQ and Quiz for Exams\" \/>\n<meta property=\"article:published_time\" content=\"2025-06-01T07:26:43+00:00\" \/>\n<meta name=\"author\" content=\"rawan239\" \/>\n<meta name=\"twitter:card\" content=\"summary_large_image\" \/>\n<meta name=\"twitter:label1\" content=\"Written by\" \/>\n\t<meta name=\"twitter:data1\" content=\"rawan239\" \/>\n\t<meta name=\"twitter:label2\" content=\"Est. reading time\" \/>\n\t<meta name=\"twitter:data2\" content=\"2 minutes\" \/>\n<!-- \/ Yoast SEO Premium plugin. -->","yoast_head_json":{"title":"Which one of the following graphs represents the equation of motion v","description":"The equation of motion is given by v = u + at, where: - v is the final velocity - u is the initial velocity - a is the acceleration (constant) - t is the time This equation is in the form of a linear equation y = mx + c, where v is the dependent variable (y), t is the independent variable (x), 'a' is the slope (m), and 'u' is the y-intercept (c). The problem states that all quantities (u, a, v, t) are non-zero, except t=0 at the start. A graph of v versus t for this equation should be a straight line. Since 'u' is non-zero, the line will intersect the v-axis (at t=0) at a point other than the origin. Since 'a' is non-zero, the slope of the line will not be zero (the line is not horizontal). Considering these points, graph (c) represents a straight line with a positive y-intercept (u > 0) and a positive slope (a > 0). While 'a' could be negative (negative slope) or 'u' could be negative (negative intercept), option (c) is the only graph that shows a straight line with a non-zero intercept and a non-zero slope, fitting the general form v = u + at with non-zero u and a. Graphs (a) and (d) are not straight lines, and graph (b) is a straight line passing through the origin, which would represent v = at (i.e., u=0), contrary to the condition that u is non-zero. - The equation v = u + at represents linear motion with constant acceleration. - A graph of velocity (v) versus time (t) for this equation is a straight line. - The y-intercept of the line is the initial velocity (u). - The slope of the line is the acceleration (a).","robots":{"index":"index","follow":"follow","max-snippet":"max-snippet:-1","max-image-preview":"max-image-preview:large","max-video-preview":"max-video-preview:-1"},"canonical":"https:\/\/exam.pscnotes.com\/mcq\/which-one-of-the-following-graphs-represents-the-equation-of-motion-v\/","og_locale":"en_US","og_type":"article","og_title":"Which one of the following graphs represents the equation of motion v","og_description":"The equation of motion is given by v = u + at, where: - v is the final velocity - u is the initial velocity - a is the acceleration (constant) - t is the time This equation is in the form of a linear equation y = mx + c, where v is the dependent variable (y), t is the independent variable (x), 'a' is the slope (m), and 'u' is the y-intercept (c). The problem states that all quantities (u, a, v, t) are non-zero, except t=0 at the start. A graph of v versus t for this equation should be a straight line. Since 'u' is non-zero, the line will intersect the v-axis (at t=0) at a point other than the origin. Since 'a' is non-zero, the slope of the line will not be zero (the line is not horizontal). Considering these points, graph (c) represents a straight line with a positive y-intercept (u > 0) and a positive slope (a > 0). While 'a' could be negative (negative slope) or 'u' could be negative (negative intercept), option (c) is the only graph that shows a straight line with a non-zero intercept and a non-zero slope, fitting the general form v = u + at with non-zero u and a. Graphs (a) and (d) are not straight lines, and graph (b) is a straight line passing through the origin, which would represent v = at (i.e., u=0), contrary to the condition that u is non-zero. - The equation v = u + at represents linear motion with constant acceleration. - A graph of velocity (v) versus time (t) for this equation is a straight line. - The y-intercept of the line is the initial velocity (u). - The slope of the line is the acceleration (a).","og_url":"https:\/\/exam.pscnotes.com\/mcq\/which-one-of-the-following-graphs-represents-the-equation-of-motion-v\/","og_site_name":"MCQ and Quiz for Exams","article_published_time":"2025-06-01T07:26:43+00:00","author":"rawan239","twitter_card":"summary_large_image","twitter_misc":{"Written by":"rawan239","Est. reading time":"2 minutes"},"schema":{"@context":"https:\/\/schema.org","@graph":[{"@type":"WebPage","@id":"https:\/\/exam.pscnotes.com\/mcq\/which-one-of-the-following-graphs-represents-the-equation-of-motion-v\/","url":"https:\/\/exam.pscnotes.com\/mcq\/which-one-of-the-following-graphs-represents-the-equation-of-motion-v\/","name":"Which one of the following graphs represents the equation of motion v","isPartOf":{"@id":"https:\/\/exam.pscnotes.com\/mcq\/#website"},"datePublished":"2025-06-01T07:26:43+00:00","dateModified":"2025-06-01T07:26:43+00:00","author":{"@id":"https:\/\/exam.pscnotes.com\/mcq\/#\/schema\/person\/5807dafeb27d2ec82344d6cbd6c3d209"},"description":"The equation of motion is given by v = u + at, where: - v is the final velocity - u is the initial velocity - a is the acceleration (constant) - t is the time This equation is in the form of a linear equation y = mx + c, where v is the dependent variable (y), t is the independent variable (x), 'a' is the slope (m), and 'u' is the y-intercept (c). The problem states that all quantities (u, a, v, t) are non-zero, except t=0 at the start. A graph of v versus t for this equation should be a straight line. Since 'u' is non-zero, the line will intersect the v-axis (at t=0) at a point other than the origin. Since 'a' is non-zero, the slope of the line will not be zero (the line is not horizontal). Considering these points, graph (c) represents a straight line with a positive y-intercept (u > 0) and a positive slope (a > 0). While 'a' could be negative (negative slope) or 'u' could be negative (negative intercept), option (c) is the only graph that shows a straight line with a non-zero intercept and a non-zero slope, fitting the general form v = u + at with non-zero u and a. Graphs (a) and (d) are not straight lines, and graph (b) is a straight line passing through the origin, which would represent v = at (i.e., u=0), contrary to the condition that u is non-zero. - The equation v = u + at represents linear motion with constant acceleration. - A graph of velocity (v) versus time (t) for this equation is a straight line. - The y-intercept of the line is the initial velocity (u). - The slope of the line is the acceleration (a).","breadcrumb":{"@id":"https:\/\/exam.pscnotes.com\/mcq\/which-one-of-the-following-graphs-represents-the-equation-of-motion-v\/#breadcrumb"},"inLanguage":"en-US","potentialAction":[{"@type":"ReadAction","target":["https:\/\/exam.pscnotes.com\/mcq\/which-one-of-the-following-graphs-represents-the-equation-of-motion-v\/"]}]},{"@type":"BreadcrumbList","@id":"https:\/\/exam.pscnotes.com\/mcq\/which-one-of-the-following-graphs-represents-the-equation-of-motion-v\/#breadcrumb","itemListElement":[{"@type":"ListItem","position":1,"name":"Home","item":"https:\/\/exam.pscnotes.com\/mcq\/"},{"@type":"ListItem","position":2,"name":"UPSC NDA-2","item":"https:\/\/exam.pscnotes.com\/mcq\/category\/upsc-nda-2\/"},{"@type":"ListItem","position":3,"name":"Which one of the following graphs represents the equation of motion v"}]},{"@type":"WebSite","@id":"https:\/\/exam.pscnotes.com\/mcq\/#website","url":"https:\/\/exam.pscnotes.com\/mcq\/","name":"MCQ and Quiz for Exams","description":"","potentialAction":[{"@type":"SearchAction","target":{"@type":"EntryPoint","urlTemplate":"https:\/\/exam.pscnotes.com\/mcq\/?s={search_term_string}"},"query-input":"required name=search_term_string"}],"inLanguage":"en-US"},{"@type":"Person","@id":"https:\/\/exam.pscnotes.com\/mcq\/#\/schema\/person\/5807dafeb27d2ec82344d6cbd6c3d209","name":"rawan239","image":{"@type":"ImageObject","inLanguage":"en-US","@id":"https:\/\/exam.pscnotes.com\/mcq\/#\/schema\/person\/image\/","url":"https:\/\/secure.gravatar.com\/avatar\/761a7274f9cce048fa5b921221e7934820d74514df93ef195a9d22af0c1c9001?s=96&d=mm&r=g","contentUrl":"https:\/\/secure.gravatar.com\/avatar\/761a7274f9cce048fa5b921221e7934820d74514df93ef195a9d22af0c1c9001?s=96&d=mm&r=g","caption":"rawan239"},"sameAs":["https:\/\/exam.pscnotes.com"],"url":"https:\/\/exam.pscnotes.com\/mcq\/author\/rawan239\/"}]}},"amp_enabled":true,"_links":{"self":[{"href":"https:\/\/exam.pscnotes.com\/mcq\/wp-json\/wp\/v2\/posts\/89000","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/exam.pscnotes.com\/mcq\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/exam.pscnotes.com\/mcq\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/exam.pscnotes.com\/mcq\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/exam.pscnotes.com\/mcq\/wp-json\/wp\/v2\/comments?post=89000"}],"version-history":[{"count":0,"href":"https:\/\/exam.pscnotes.com\/mcq\/wp-json\/wp\/v2\/posts\/89000\/revisions"}],"wp:attachment":[{"href":"https:\/\/exam.pscnotes.com\/mcq\/wp-json\/wp\/v2\/media?parent=89000"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/exam.pscnotes.com\/mcq\/wp-json\/wp\/v2\/categories?post=89000"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/exam.pscnotes.com\/mcq\/wp-json\/wp\/v2\/tags?post=89000"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}