{"id":85379,"date":"2025-06-01T03:14:59","date_gmt":"2025-06-01T03:14:59","guid":{"rendered":"https:\/\/exam.pscnotes.com\/mcq\/?p=85379"},"modified":"2025-06-01T03:14:59","modified_gmt":"2025-06-01T03:14:59","slug":"a-car-moving-with-a-speed-of-12-m-s-is-subjected-to-brakes-which-produ","status":"publish","type":"post","link":"https:\/\/exam.pscnotes.com\/mcq\/a-car-moving-with-a-speed-of-12-m-s-is-subjected-to-brakes-which-produ\/","title":{"rendered":"A car moving with a speed of 12 m\/s is subjected to brakes which produ"},"content":{"rendered":"

A car moving with a speed of 12 m\/s is subjected to brakes which produces a deceleration of 6 m\/s2<\/sup>. The car takes 2 s to stop after the application of brakes. What is the distance covered by the car after the application of brakes?<\/p>\n

[amp_mcq option1=”12 m” option2=”24 m” option3=”36 m” option4=”48 m” correct=”option1″]<\/p>\n

\n
\n
This question was previously asked in<\/div>\n
UPSC CDS-1 – 2023<\/div>\n<\/div>\n
Download PDF<\/a>Attempt Online<\/a><\/div>\n<\/div>\n
\nGiven:
\nInitial velocity, $u = 12$ m\/s
\nDeceleration, $a = -6$ m\/s\u00b2 (negative because it’s deceleration)
\nFinal velocity, $v = 0$ m\/s (the car stops)
\nTime, $t = 2$ s
\nWe need to find the distance covered, $s$.
\nWe can use the equation of motion $s = ut + (1\/2)at^2$.
\nSubstituting the given values:
\n$s = (12 \\, \\text{m\/s}) \\times (2 \\, \\text{s}) + (1\/2) \\times (-6 \\, \\text{m\/s}^2) \\times (2 \\, \\text{s})^2$
\n$s = 24 \\, \\text{m} + (1\/2) \\times (-6) \\times 4 \\, \\text{m}$
\n$s = 24 \\, \\text{m} – (3) \\times 4 \\, \\text{m}$
\n$s = 24 \\, \\text{m} – 12 \\, \\text{m}$
\n$s = 12 \\, \\text{m}$
\nAlternatively, using $v^2 = u^2 + 2as$:
\n$0^2 = (12)^2 + 2 \\times (-6) \\times s$
\n$0 = 144 – 12s$
\n$12s = 144$
\n$s = 144 \/ 12 = 12$ m.
\n<\/section>\n
\n– Use the appropriate kinematic equations for uniformly accelerated motion.
\n– Deceleration is negative acceleration.
\n<\/section>\n
\nThe kinematic equations for constant acceleration are:
\n1. $v = u + at$
\n2. $s = ut + (1\/2)at^2$
\n3. $v^2 = u^2 + 2as$
\n4. $s = (u+v)t\/2$
\nThese equations are valid only when acceleration is constant.
\n<\/section>\n","protected":false},"excerpt":{"rendered":"

A car moving with a speed of 12 m\/s is subjected to brakes which produces a deceleration of 6 m\/s2. The car takes 2 s to stop after the application of brakes. What is the distance covered by the car after the application of brakes? [amp_mcq option1=”12 m” option2=”24 m” option3=”36 m” option4=”48 m” correct=”option1″] … <\/p>\n

Detailed SolutionA car moving with a speed of 12 m\/s is subjected to brakes which produ<\/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":[1087],"tags":[1105,1129,1128],"class_list":["post-85379","post","type-post","status-publish","format-standard","hentry","category-upsc-cds-1","tag-1105","tag-mechanics","tag-physics","no-featured-image-padding"],"yoast_head":"\nA car moving with a speed of 12 m\/s is subjected to brakes which produ<\/title>\n<meta name=\"description\" content=\"Given: Initial velocity, $u = 12$ m\/s Deceleration, $a = -6$ m\/s\u00b2 (negative because it's deceleration) Final velocity, $v = 0$ m\/s (the car stops) Time, $t = 2$ s We need to find the distance covered, $s$. We can use the equation of motion $s = ut + (1\/2)at^2$. Substituting the given values: $s = (12 , text{m\/s}) times (2 , text{s}) + (1\/2) times (-6 , text{m\/s}^2) times (2 , text{s})^2$ $s = 24 , text{m} + (1\/2) times (-6) times 4 , text{m}$ $s = 24 , text{m} - (3) times 4 , text{m}$ $s = 24 , text{m} - 12 , text{m}$ $s = 12 , text{m}$ Alternatively, using $v^2 = u^2 + 2as$: $0^2 = (12)^2 + 2 times (-6) times s$ $0 = 144 - 12s$ $12s = 144$ $s = 144 \/ 12 = 12$ m. - Use the appropriate kinematic equations for uniformly accelerated motion. - Deceleration is negative acceleration.\" \/>\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-car-moving-with-a-speed-of-12-m-s-is-subjected-to-brakes-which-produ\/\" \/>\n<meta property=\"og:locale\" content=\"en_US\" \/>\n<meta property=\"og:type\" content=\"article\" \/>\n<meta property=\"og:title\" content=\"A car moving with a speed of 12 m\/s is subjected to brakes which produ\" \/>\n<meta property=\"og:description\" content=\"Given: Initial velocity, $u = 12$ m\/s Deceleration, $a = -6$ m\/s\u00b2 (negative because it's deceleration) Final velocity, $v = 0$ m\/s (the car stops) Time, $t = 2$ s We need to find the distance covered, $s$. We can use the equation of motion $s = ut + (1\/2)at^2$. Substituting the given values: $s = (12 , text{m\/s}) times (2 , text{s}) + (1\/2) times (-6 , text{m\/s}^2) times (2 , text{s})^2$ $s = 24 , text{m} + (1\/2) times (-6) times 4 , text{m}$ $s = 24 , text{m} - (3) times 4 , text{m}$ $s = 24 , text{m} - 12 , text{m}$ $s = 12 , text{m}$ Alternatively, using $v^2 = u^2 + 2as$: $0^2 = (12)^2 + 2 times (-6) times s$ $0 = 144 - 12s$ $12s = 144$ $s = 144 \/ 12 = 12$ m. - Use the appropriate kinematic equations for uniformly accelerated motion. - Deceleration is negative acceleration.\" \/>\n<meta property=\"og:url\" content=\"https:\/\/exam.pscnotes.com\/mcq\/a-car-moving-with-a-speed-of-12-m-s-is-subjected-to-brakes-which-produ\/\" \/>\n<meta property=\"og:site_name\" content=\"MCQ and Quiz for Exams\" \/>\n<meta property=\"article:published_time\" content=\"2025-06-01T03:14:59+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 car moving with a speed of 12 m\/s is subjected to brakes which produ","description":"Given: Initial velocity, $u = 12$ m\/s Deceleration, $a = -6$ m\/s\u00b2 (negative because it's deceleration) Final velocity, $v = 0$ m\/s (the car stops) Time, $t = 2$ s We need to find the distance covered, $s$. We can use the equation of motion $s = ut + (1\/2)at^2$. Substituting the given values: $s = (12 , text{m\/s}) times (2 , text{s}) + (1\/2) times (-6 , text{m\/s}^2) times (2 , text{s})^2$ $s = 24 , text{m} + (1\/2) times (-6) times 4 , text{m}$ $s = 24 , text{m} - (3) times 4 , text{m}$ $s = 24 , text{m} - 12 , text{m}$ $s = 12 , text{m}$ Alternatively, using $v^2 = u^2 + 2as$: $0^2 = (12)^2 + 2 times (-6) times s$ $0 = 144 - 12s$ $12s = 144$ $s = 144 \/ 12 = 12$ m. - Use the appropriate kinematic equations for uniformly accelerated motion. - Deceleration is negative acceleration.","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-car-moving-with-a-speed-of-12-m-s-is-subjected-to-brakes-which-produ\/","og_locale":"en_US","og_type":"article","og_title":"A car moving with a speed of 12 m\/s is subjected to brakes which produ","og_description":"Given: Initial velocity, $u = 12$ m\/s Deceleration, $a = -6$ m\/s\u00b2 (negative because it's deceleration) Final velocity, $v = 0$ m\/s (the car stops) Time, $t = 2$ s We need to find the distance covered, $s$. We can use the equation of motion $s = ut + (1\/2)at^2$. Substituting the given values: $s = (12 , text{m\/s}) times (2 , text{s}) + (1\/2) times (-6 , text{m\/s}^2) times (2 , text{s})^2$ $s = 24 , text{m} + (1\/2) times (-6) times 4 , text{m}$ $s = 24 , text{m} - (3) times 4 , text{m}$ $s = 24 , text{m} - 12 , text{m}$ $s = 12 , text{m}$ Alternatively, using $v^2 = u^2 + 2as$: $0^2 = (12)^2 + 2 times (-6) times s$ $0 = 144 - 12s$ $12s = 144$ $s = 144 \/ 12 = 12$ m. - Use the appropriate kinematic equations for uniformly accelerated motion. - Deceleration is negative acceleration.","og_url":"https:\/\/exam.pscnotes.com\/mcq\/a-car-moving-with-a-speed-of-12-m-s-is-subjected-to-brakes-which-produ\/","og_site_name":"MCQ and Quiz for Exams","article_published_time":"2025-06-01T03:14:59+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-car-moving-with-a-speed-of-12-m-s-is-subjected-to-brakes-which-produ\/","url":"https:\/\/exam.pscnotes.com\/mcq\/a-car-moving-with-a-speed-of-12-m-s-is-subjected-to-brakes-which-produ\/","name":"A car moving with a speed of 12 m\/s is subjected to brakes which produ","isPartOf":{"@id":"https:\/\/exam.pscnotes.com\/mcq\/#website"},"datePublished":"2025-06-01T03:14:59+00:00","dateModified":"2025-06-01T03:14:59+00:00","author":{"@id":"https:\/\/exam.pscnotes.com\/mcq\/#\/schema\/person\/5807dafeb27d2ec82344d6cbd6c3d209"},"description":"Given: Initial velocity, $u = 12$ m\/s Deceleration, $a = -6$ m\/s\u00b2 (negative because it's deceleration) Final velocity, $v = 0$ m\/s (the car stops) Time, $t = 2$ s We need to find the distance covered, $s$. We can use the equation of motion $s = ut + (1\/2)at^2$. Substituting the given values: $s = (12 \\, \\text{m\/s}) \\times (2 \\, \\text{s}) + (1\/2) \\times (-6 \\, \\text{m\/s}^2) \\times (2 \\, \\text{s})^2$ $s = 24 \\, \\text{m} + (1\/2) \\times (-6) \\times 4 \\, \\text{m}$ $s = 24 \\, \\text{m} - (3) \\times 4 \\, \\text{m}$ $s = 24 \\, \\text{m} - 12 \\, \\text{m}$ $s = 12 \\, \\text{m}$ Alternatively, using $v^2 = u^2 + 2as$: $0^2 = (12)^2 + 2 \\times (-6) \\times s$ $0 = 144 - 12s$ $12s = 144$ $s = 144 \/ 12 = 12$ m. - Use the appropriate kinematic equations for uniformly accelerated motion. - Deceleration is negative acceleration.","breadcrumb":{"@id":"https:\/\/exam.pscnotes.com\/mcq\/a-car-moving-with-a-speed-of-12-m-s-is-subjected-to-brakes-which-produ\/#breadcrumb"},"inLanguage":"en-US","potentialAction":[{"@type":"ReadAction","target":["https:\/\/exam.pscnotes.com\/mcq\/a-car-moving-with-a-speed-of-12-m-s-is-subjected-to-brakes-which-produ\/"]}]},{"@type":"BreadcrumbList","@id":"https:\/\/exam.pscnotes.com\/mcq\/a-car-moving-with-a-speed-of-12-m-s-is-subjected-to-brakes-which-produ\/#breadcrumb","itemListElement":[{"@type":"ListItem","position":1,"name":"Home","item":"https:\/\/exam.pscnotes.com\/mcq\/"},{"@type":"ListItem","position":2,"name":"UPSC CDS-1","item":"https:\/\/exam.pscnotes.com\/mcq\/category\/upsc-cds-1\/"},{"@type":"ListItem","position":3,"name":"A car moving with a speed of 12 m\/s is subjected to brakes which produ"}]},{"@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\/85379","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=85379"}],"version-history":[{"count":0,"href":"https:\/\/exam.pscnotes.com\/mcq\/wp-json\/wp\/v2\/posts\/85379\/revisions"}],"wp:attachment":[{"href":"https:\/\/exam.pscnotes.com\/mcq\/wp-json\/wp\/v2\/media?parent=85379"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/exam.pscnotes.com\/mcq\/wp-json\/wp\/v2\/categories?post=85379"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/exam.pscnotes.com\/mcq\/wp-json\/wp\/v2\/tags?post=85379"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}