{"id":85805,"date":"2025-06-01T03:28:30","date_gmt":"2025-06-01T03:28:30","guid":{"rendered":"https:\/\/exam.pscnotes.com\/mcq\/?p=85805"},"modified":"2025-06-01T03:28:30","modified_gmt":"2025-06-01T03:28:30","slug":"a-particle-moves-with-uniform-acceleration-along-a-straight-line-from","status":"publish","type":"post","link":"https:\/\/exam.pscnotes.com\/mcq\/a-particle-moves-with-uniform-acceleration-along-a-straight-line-from\/","title":{"rendered":"A particle moves with uniform acceleration along a straight line from"},"content":{"rendered":"

A particle moves with uniform acceleration along a straight line from rest. The percentage increase in displacement during sixth second compared to that in fifth second is about<\/p>\n

[amp_mcq option1=”11%” option2=”22%” option3=”33%” option4=”44%” correct=”option2″]<\/p>\n

\n
\n
This question was previously asked in<\/div>\n
UPSC CDS-2 – 2018<\/div>\n<\/div>\n
Download PDF<\/a>Attempt Online<\/a><\/div>\n<\/div>\n
\nFor a particle starting from rest (initial velocity u = 0) and moving with uniform acceleration ‘a’, the displacement in the n-th second (S_n) is given by the formula S_n = u + a(n – 1\/2). Since u = 0, S_n = a(n – 1\/2).
\nDisplacement in the fifth second (n=5): S\u2085 = a(5 – 1\/2) = a(4.5).
\nDisplacement in the sixth second (n=6): S\u2086 = a(6 – 1\/2) = a(5.5).
\nThe percentage increase in displacement during the sixth second compared to the fifth second is calculated as [(S\u2086 – S\u2085) \/ S\u2085] * 100%.
\nPercentage Increase = [(a(5.5) – a(4.5)) \/ a(4.5)] * 100% = [a(5.5 – 4.5) \/ a(4.5)] * 100% = [a(1) \/ a(4.5)] * 100% = (1 \/ 4.5) * 100% = (10 \/ 45) * 100% = (2 \/ 9) * 100% \u2248 0.2222 * 100% \u2248 22.22%.
\nThe closest option is 22%.
\n<\/section>\n
\n– Displacement in the n-th second formula: S_n = u + a(n – 1\/2).
\n– Uniform acceleration from rest means u=0.
\n– Percentage increase calculation involves finding the difference, dividing by the original value, and multiplying by 100.
\n<\/section>\n
\nThe total displacement after ‘n’ seconds is given by S = ut + (1\/2)at\u00b2. For a body starting from rest (u=0), S = (1\/2)at\u00b2. The displacement in the n-th second is the difference between the total displacement after n seconds and the total displacement after (n-1) seconds: S_n = S(n) – S(n-1) = (1\/2)an\u00b2 – (1\/2)a(n-1)\u00b2 = (1\/2)a [n\u00b2 – (n\u00b2 – 2n + 1)] = (1\/2)a [n\u00b2 – n\u00b2 + 2n – 1] = (1\/2)a (2n – 1) = a(n – 1\/2). This confirms the formula used.
\n<\/section>\n","protected":false},"excerpt":{"rendered":"

A particle moves with uniform acceleration along a straight line from rest. The percentage increase in displacement during sixth second compared to that in fifth second is about [amp_mcq option1=”11%” option2=”22%” option3=”33%” option4=”44%” correct=”option2″] This question was previously asked in UPSC CDS-2 – 2018 Download PDFAttempt Online For a particle starting from rest (initial velocity … <\/p>\n

Detailed SolutionA particle moves with uniform acceleration along a straight line from<\/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":[1088],"tags":[1114,1129,1128],"class_list":["post-85805","post","type-post","status-publish","format-standard","hentry","category-upsc-cds-2","tag-1114","tag-mechanics","tag-physics","no-featured-image-padding"],"yoast_head":"\nA particle moves with uniform acceleration along a straight line from<\/title>\n<meta name=\"description\" content=\"For a particle starting from rest (initial velocity u = 0) and moving with uniform acceleration 'a', the displacement in the n-th second (S_n) is given by the formula S_n = u + a(n - 1\/2). Since u = 0, S_n = a(n - 1\/2). Displacement in the fifth second (n=5): S\u2085 = a(5 - 1\/2) = a(4.5). Displacement in the sixth second (n=6): S\u2086 = a(6 - 1\/2) = a(5.5). The percentage increase in displacement during the sixth second compared to the fifth second is calculated as [(S\u2086 - S\u2085) \/ S\u2085] * 100%. Percentage Increase = [(a(5.5) - a(4.5)) \/ a(4.5)] * 100% = [a(5.5 - 4.5) \/ a(4.5)] * 100% = [a(1) \/ a(4.5)] * 100% = (1 \/ 4.5) * 100% = (10 \/ 45) * 100% = (2 \/ 9) * 100% \u2248 0.2222 * 100% \u2248 22.22%. The closest option is 22%. - Displacement in the n-th second formula: S_n = u + a(n - 1\/2). - Uniform acceleration from rest means u=0. - Percentage increase calculation involves finding the difference, dividing by the original value, and multiplying by 100.\" \/>\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\/a-particle-moves-with-uniform-acceleration-along-a-straight-line-from\/\" \/>\n<meta property=\"og:locale\" content=\"en_US\" \/>\n<meta property=\"og:type\" content=\"article\" \/>\n<meta property=\"og:title\" content=\"A particle moves with uniform acceleration along a straight line from\" \/>\n<meta property=\"og:description\" content=\"For a particle starting from rest (initial velocity u = 0) and moving with uniform acceleration 'a', the displacement in the n-th second (S_n) is given by the formula S_n = u + a(n - 1\/2). Since u = 0, S_n = a(n - 1\/2). Displacement in the fifth second (n=5): S\u2085 = a(5 - 1\/2) = a(4.5). Displacement in the sixth second (n=6): S\u2086 = a(6 - 1\/2) = a(5.5). The percentage increase in displacement during the sixth second compared to the fifth second is calculated as [(S\u2086 - S\u2085) \/ S\u2085] * 100%. Percentage Increase = [(a(5.5) - a(4.5)) \/ a(4.5)] * 100% = [a(5.5 - 4.5) \/ a(4.5)] * 100% = [a(1) \/ a(4.5)] * 100% = (1 \/ 4.5) * 100% = (10 \/ 45) * 100% = (2 \/ 9) * 100% \u2248 0.2222 * 100% \u2248 22.22%. The closest option is 22%. - Displacement in the n-th second formula: S_n = u + a(n - 1\/2). - Uniform acceleration from rest means u=0. - Percentage increase calculation involves finding the difference, dividing by the original value, and multiplying by 100.\" \/>\n<meta property=\"og:url\" content=\"https:\/\/exam.pscnotes.com\/mcq\/a-particle-moves-with-uniform-acceleration-along-a-straight-line-from\/\" \/>\n<meta property=\"og:site_name\" content=\"MCQ and Quiz for Exams\" \/>\n<meta property=\"article:published_time\" content=\"2025-06-01T03:28:30+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=\"1 minute\" \/>\n<!-- \/ Yoast SEO Premium plugin. -->","yoast_head_json":{"title":"A particle moves with uniform acceleration along a straight line from","description":"For a particle starting from rest (initial velocity u = 0) and moving with uniform acceleration 'a', the displacement in the n-th second (S_n) is given by the formula S_n = u + a(n - 1\/2). Since u = 0, S_n = a(n - 1\/2). Displacement in the fifth second (n=5): S\u2085 = a(5 - 1\/2) = a(4.5). Displacement in the sixth second (n=6): S\u2086 = a(6 - 1\/2) = a(5.5). The percentage increase in displacement during the sixth second compared to the fifth second is calculated as [(S\u2086 - S\u2085) \/ S\u2085] * 100%. Percentage Increase = [(a(5.5) - a(4.5)) \/ a(4.5)] * 100% = [a(5.5 - 4.5) \/ a(4.5)] * 100% = [a(1) \/ a(4.5)] * 100% = (1 \/ 4.5) * 100% = (10 \/ 45) * 100% = (2 \/ 9) * 100% \u2248 0.2222 * 100% \u2248 22.22%. The closest option is 22%. - Displacement in the n-th second formula: S_n = u + a(n - 1\/2). - Uniform acceleration from rest means u=0. - Percentage increase calculation involves finding the difference, dividing by the original value, and multiplying by 100.","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\/a-particle-moves-with-uniform-acceleration-along-a-straight-line-from\/","og_locale":"en_US","og_type":"article","og_title":"A particle moves with uniform acceleration along a straight line from","og_description":"For a particle starting from rest (initial velocity u = 0) and moving with uniform acceleration 'a', the displacement in the n-th second (S_n) is given by the formula S_n = u + a(n - 1\/2). Since u = 0, S_n = a(n - 1\/2). Displacement in the fifth second (n=5): S\u2085 = a(5 - 1\/2) = a(4.5). Displacement in the sixth second (n=6): S\u2086 = a(6 - 1\/2) = a(5.5). The percentage increase in displacement during the sixth second compared to the fifth second is calculated as [(S\u2086 - S\u2085) \/ S\u2085] * 100%. Percentage Increase = [(a(5.5) - a(4.5)) \/ a(4.5)] * 100% = [a(5.5 - 4.5) \/ a(4.5)] * 100% = [a(1) \/ a(4.5)] * 100% = (1 \/ 4.5) * 100% = (10 \/ 45) * 100% = (2 \/ 9) * 100% \u2248 0.2222 * 100% \u2248 22.22%. The closest option is 22%. - Displacement in the n-th second formula: S_n = u + a(n - 1\/2). - Uniform acceleration from rest means u=0. - Percentage increase calculation involves finding the difference, dividing by the original value, and multiplying by 100.","og_url":"https:\/\/exam.pscnotes.com\/mcq\/a-particle-moves-with-uniform-acceleration-along-a-straight-line-from\/","og_site_name":"MCQ and Quiz for Exams","article_published_time":"2025-06-01T03:28:30+00:00","author":"rawan239","twitter_card":"summary_large_image","twitter_misc":{"Written by":"rawan239","Est. reading time":"1 minute"},"schema":{"@context":"https:\/\/schema.org","@graph":[{"@type":"WebPage","@id":"https:\/\/exam.pscnotes.com\/mcq\/a-particle-moves-with-uniform-acceleration-along-a-straight-line-from\/","url":"https:\/\/exam.pscnotes.com\/mcq\/a-particle-moves-with-uniform-acceleration-along-a-straight-line-from\/","name":"A particle moves with uniform acceleration along a straight line from","isPartOf":{"@id":"https:\/\/exam.pscnotes.com\/mcq\/#website"},"datePublished":"2025-06-01T03:28:30+00:00","dateModified":"2025-06-01T03:28:30+00:00","author":{"@id":"https:\/\/exam.pscnotes.com\/mcq\/#\/schema\/person\/5807dafeb27d2ec82344d6cbd6c3d209"},"description":"For a particle starting from rest (initial velocity u = 0) and moving with uniform acceleration 'a', the displacement in the n-th second (S_n) is given by the formula S_n = u + a(n - 1\/2). Since u = 0, S_n = a(n - 1\/2). Displacement in the fifth second (n=5): S\u2085 = a(5 - 1\/2) = a(4.5). Displacement in the sixth second (n=6): S\u2086 = a(6 - 1\/2) = a(5.5). The percentage increase in displacement during the sixth second compared to the fifth second is calculated as [(S\u2086 - S\u2085) \/ S\u2085] * 100%. Percentage Increase = [(a(5.5) - a(4.5)) \/ a(4.5)] * 100% = [a(5.5 - 4.5) \/ a(4.5)] * 100% = [a(1) \/ a(4.5)] * 100% = (1 \/ 4.5) * 100% = (10 \/ 45) * 100% = (2 \/ 9) * 100% \u2248 0.2222 * 100% \u2248 22.22%. The closest option is 22%. - Displacement in the n-th second formula: S_n = u + a(n - 1\/2). - Uniform acceleration from rest means u=0. - Percentage increase calculation involves finding the difference, dividing by the original value, and multiplying by 100.","breadcrumb":{"@id":"https:\/\/exam.pscnotes.com\/mcq\/a-particle-moves-with-uniform-acceleration-along-a-straight-line-from\/#breadcrumb"},"inLanguage":"en-US","potentialAction":[{"@type":"ReadAction","target":["https:\/\/exam.pscnotes.com\/mcq\/a-particle-moves-with-uniform-acceleration-along-a-straight-line-from\/"]}]},{"@type":"BreadcrumbList","@id":"https:\/\/exam.pscnotes.com\/mcq\/a-particle-moves-with-uniform-acceleration-along-a-straight-line-from\/#breadcrumb","itemListElement":[{"@type":"ListItem","position":1,"name":"Home","item":"https:\/\/exam.pscnotes.com\/mcq\/"},{"@type":"ListItem","position":2,"name":"UPSC CDS-2","item":"https:\/\/exam.pscnotes.com\/mcq\/category\/upsc-cds-2\/"},{"@type":"ListItem","position":3,"name":"A particle moves with uniform acceleration along a straight line from"}]},{"@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\/85805","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=85805"}],"version-history":[{"count":0,"href":"https:\/\/exam.pscnotes.com\/mcq\/wp-json\/wp\/v2\/posts\/85805\/revisions"}],"wp:attachment":[{"href":"https:\/\/exam.pscnotes.com\/mcq\/wp-json\/wp\/v2\/media?parent=85805"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/exam.pscnotes.com\/mcq\/wp-json\/wp\/v2\/categories?post=85805"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/exam.pscnotes.com\/mcq\/wp-json\/wp\/v2\/tags?post=85805"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}