{"id":89060,"date":"2025-06-01T07:28:37","date_gmt":"2025-06-01T07:28:37","guid":{"rendered":"https:\/\/exam.pscnotes.com\/mcq\/?p=89060"},"modified":"2025-06-01T07:28:37","modified_gmt":"2025-06-01T07:28:37","slug":"a-vehicle-starts-moving-along-a-straight-line-path-from-rest-in-first","status":"publish","type":"post","link":"https:\/\/exam.pscnotes.com\/mcq\/a-vehicle-starts-moving-along-a-straight-line-path-from-rest-in-first\/","title":{"rendered":"A vehicle starts moving along a straight line path from rest. In first"},"content":{"rendered":"

A vehicle starts moving along a straight line path from rest. In first `t` seconds it moves with an acceleration of 2 m\/s\u00b2 and then in next 10 seconds it moves with an acceleration of 5 m\/s\u00b2. The total distance travelled by the vehicle is 550 m. The value of time `t` is<\/p>\n

[amp_mcq option1=”10 s” option2=”13 s” option3=”20 s” option4=”25 s” correct=”option1″]<\/p>\n

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This question was previously asked in<\/div>\n
UPSC NDA-2 – 2024<\/div>\n<\/div>\n
Download PDF<\/a>Attempt Online<\/a><\/div>\n<\/div>\n
\nThe motion consists of two phases of constant acceleration. We need to find the time ‘t’ in the first phase such that the total distance covered in both phases is 550 m. By calculating the distance in each phase and setting the total distance equal to 550 m, we arrive at a quadratic equation in ‘t’. Solving this equation gives t = 10 s (the positive solution).
\n<\/section>\n
\nFor the first phase (0 to t seconds): Initial velocity (u\u2081) = 0, acceleration (a\u2081) = 2 m\/s\u00b2. Distance s\u2081 = u\u2081t + (1\/2)a\u2081t\u00b2 = 0*t + (1\/2)*2*t\u00b2 = t\u00b2. Velocity at time t (v\u2081) = u\u2081 + a\u2081t = 0 + 2t = 2t m\/s.
\nFor the second phase (t to t+10 seconds): Initial velocity (u\u2082) = v\u2081 = 2t m\/s, acceleration (a\u2082) = 5 m\/s\u00b2, time (t\u2082) = 10 s. Distance s\u2082 = u\u2082t\u2082 + (1\/2)a\u2082t\u2082\u00b2 = (2t)*10 + (1\/2)*5*(10)\u00b2 = 20t + 250 m.
\nTotal distance = s\u2081 + s\u2082 = t\u00b2 + 20t + 250. Given total distance is 550 m, t\u00b2 + 20t + 250 = 550, which simplifies to t\u00b2 + 20t – 300 = 0.
\nFactoring the quadratic: (t + 30)(t – 10) = 0. Possible solutions are t = -30 or t = 10. Since time cannot be negative, t = 10 s.
\n<\/section>\n
\nWe can verify the answer: If t=10 s, s\u2081 = 10\u00b2 = 100 m. Velocity after 10s is v\u2081 = 2*10 = 20 m\/s. In the next 10s, starting at 20 m\/s with acceleration 5 m\/s\u00b2, distance s\u2082 = 20*10 + (1\/2)*5*10\u00b2 = 200 + 250 = 450 m. Total distance = s\u2081 + s\u2082 = 100 + 450 = 550 m, which matches the given information.
\n<\/section>\n","protected":false},"excerpt":{"rendered":"

A vehicle starts moving along a straight line path from rest. In first `t` seconds it moves with an acceleration of 2 m\/s\u00b2 and then in next 10 seconds it moves with an acceleration of 5 m\/s\u00b2. The total distance travelled by the vehicle is 550 m. The value of time `t` is [amp_mcq option1=”10 … <\/p>\n

Detailed SolutionA vehicle starts moving along a straight line path from rest. In first<\/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":[1103,1129,1128],"class_list":["post-89060","post","type-post","status-publish","format-standard","hentry","category-upsc-nda-2","tag-1103","tag-mechanics","tag-physics","no-featured-image-padding"],"yoast_head":"\nA vehicle starts moving along a straight line path from rest. In first<\/title>\n<meta name=\"description\" content=\"The motion consists of two phases of constant acceleration. We need to find the time 't' in the first phase such that the total distance covered in both phases is 550 m. By calculating the distance in each phase and setting the total distance equal to 550 m, we arrive at a quadratic equation in 't'. Solving this equation gives t = 10 s (the positive solution). For the first phase (0 to t seconds): Initial velocity (u\u2081) = 0, acceleration (a\u2081) = 2 m\/s\u00b2. Distance s\u2081 = u\u2081t + (1\/2)a\u2081t\u00b2 = 0*t + (1\/2)*2*t\u00b2 = t\u00b2. Velocity at time t (v\u2081) = u\u2081 + a\u2081t = 0 + 2t = 2t m\/s. For the second phase (t to t+10 seconds): Initial velocity (u\u2082) = v\u2081 = 2t m\/s, acceleration (a\u2082) = 5 m\/s\u00b2, time (t\u2082) = 10 s. Distance s\u2082 = u\u2082t\u2082 + (1\/2)a\u2082t\u2082\u00b2 = (2t)*10 + (1\/2)*5*(10)\u00b2 = 20t + 250 m. Total distance = s\u2081 + s\u2082 = t\u00b2 + 20t + 250. Given total distance is 550 m, t\u00b2 + 20t + 250 = 550, which simplifies to t\u00b2 + 20t - 300 = 0. Factoring the quadratic: (t + 30)(t - 10) = 0. Possible solutions are t = -30 or t = 10. Since time cannot be negative, t = 10 s.\" \/>\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-vehicle-starts-moving-along-a-straight-line-path-from-rest-in-first\/\" \/>\n<meta property=\"og:locale\" content=\"en_US\" \/>\n<meta property=\"og:type\" content=\"article\" \/>\n<meta property=\"og:title\" content=\"A vehicle starts moving along a straight line path from rest. In first\" \/>\n<meta property=\"og:description\" content=\"The motion consists of two phases of constant acceleration. We need to find the time 't' in the first phase such that the total distance covered in both phases is 550 m. By calculating the distance in each phase and setting the total distance equal to 550 m, we arrive at a quadratic equation in 't'. Solving this equation gives t = 10 s (the positive solution). For the first phase (0 to t seconds): Initial velocity (u\u2081) = 0, acceleration (a\u2081) = 2 m\/s\u00b2. Distance s\u2081 = u\u2081t + (1\/2)a\u2081t\u00b2 = 0*t + (1\/2)*2*t\u00b2 = t\u00b2. Velocity at time t (v\u2081) = u\u2081 + a\u2081t = 0 + 2t = 2t m\/s. For the second phase (t to t+10 seconds): Initial velocity (u\u2082) = v\u2081 = 2t m\/s, acceleration (a\u2082) = 5 m\/s\u00b2, time (t\u2082) = 10 s. Distance s\u2082 = u\u2082t\u2082 + (1\/2)a\u2082t\u2082\u00b2 = (2t)*10 + (1\/2)*5*(10)\u00b2 = 20t + 250 m. Total distance = s\u2081 + s\u2082 = t\u00b2 + 20t + 250. Given total distance is 550 m, t\u00b2 + 20t + 250 = 550, which simplifies to t\u00b2 + 20t - 300 = 0. Factoring the quadratic: (t + 30)(t - 10) = 0. Possible solutions are t = -30 or t = 10. Since time cannot be negative, t = 10 s.\" \/>\n<meta property=\"og:url\" content=\"https:\/\/exam.pscnotes.com\/mcq\/a-vehicle-starts-moving-along-a-straight-line-path-from-rest-in-first\/\" \/>\n<meta property=\"og:site_name\" content=\"MCQ and Quiz for Exams\" \/>\n<meta property=\"article:published_time\" content=\"2025-06-01T07:28:37+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 vehicle starts moving along a straight line path from rest. In first","description":"The motion consists of two phases of constant acceleration. We need to find the time 't' in the first phase such that the total distance covered in both phases is 550 m. By calculating the distance in each phase and setting the total distance equal to 550 m, we arrive at a quadratic equation in 't'. Solving this equation gives t = 10 s (the positive solution). For the first phase (0 to t seconds): Initial velocity (u\u2081) = 0, acceleration (a\u2081) = 2 m\/s\u00b2. Distance s\u2081 = u\u2081t + (1\/2)a\u2081t\u00b2 = 0*t + (1\/2)*2*t\u00b2 = t\u00b2. Velocity at time t (v\u2081) = u\u2081 + a\u2081t = 0 + 2t = 2t m\/s. For the second phase (t to t+10 seconds): Initial velocity (u\u2082) = v\u2081 = 2t m\/s, acceleration (a\u2082) = 5 m\/s\u00b2, time (t\u2082) = 10 s. Distance s\u2082 = u\u2082t\u2082 + (1\/2)a\u2082t\u2082\u00b2 = (2t)*10 + (1\/2)*5*(10)\u00b2 = 20t + 250 m. Total distance = s\u2081 + s\u2082 = t\u00b2 + 20t + 250. Given total distance is 550 m, t\u00b2 + 20t + 250 = 550, which simplifies to t\u00b2 + 20t - 300 = 0. Factoring the quadratic: (t + 30)(t - 10) = 0. Possible solutions are t = -30 or t = 10. Since time cannot be negative, t = 10 s.","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-vehicle-starts-moving-along-a-straight-line-path-from-rest-in-first\/","og_locale":"en_US","og_type":"article","og_title":"A vehicle starts moving along a straight line path from rest. In first","og_description":"The motion consists of two phases of constant acceleration. We need to find the time 't' in the first phase such that the total distance covered in both phases is 550 m. By calculating the distance in each phase and setting the total distance equal to 550 m, we arrive at a quadratic equation in 't'. Solving this equation gives t = 10 s (the positive solution). For the first phase (0 to t seconds): Initial velocity (u\u2081) = 0, acceleration (a\u2081) = 2 m\/s\u00b2. Distance s\u2081 = u\u2081t + (1\/2)a\u2081t\u00b2 = 0*t + (1\/2)*2*t\u00b2 = t\u00b2. Velocity at time t (v\u2081) = u\u2081 + a\u2081t = 0 + 2t = 2t m\/s. For the second phase (t to t+10 seconds): Initial velocity (u\u2082) = v\u2081 = 2t m\/s, acceleration (a\u2082) = 5 m\/s\u00b2, time (t\u2082) = 10 s. Distance s\u2082 = u\u2082t\u2082 + (1\/2)a\u2082t\u2082\u00b2 = (2t)*10 + (1\/2)*5*(10)\u00b2 = 20t + 250 m. Total distance = s\u2081 + s\u2082 = t\u00b2 + 20t + 250. Given total distance is 550 m, t\u00b2 + 20t + 250 = 550, which simplifies to t\u00b2 + 20t - 300 = 0. Factoring the quadratic: (t + 30)(t - 10) = 0. Possible solutions are t = -30 or t = 10. Since time cannot be negative, t = 10 s.","og_url":"https:\/\/exam.pscnotes.com\/mcq\/a-vehicle-starts-moving-along-a-straight-line-path-from-rest-in-first\/","og_site_name":"MCQ and Quiz for Exams","article_published_time":"2025-06-01T07:28:37+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-vehicle-starts-moving-along-a-straight-line-path-from-rest-in-first\/","url":"https:\/\/exam.pscnotes.com\/mcq\/a-vehicle-starts-moving-along-a-straight-line-path-from-rest-in-first\/","name":"A vehicle starts moving along a straight line path from rest. In first","isPartOf":{"@id":"https:\/\/exam.pscnotes.com\/mcq\/#website"},"datePublished":"2025-06-01T07:28:37+00:00","dateModified":"2025-06-01T07:28:37+00:00","author":{"@id":"https:\/\/exam.pscnotes.com\/mcq\/#\/schema\/person\/5807dafeb27d2ec82344d6cbd6c3d209"},"description":"The motion consists of two phases of constant acceleration. We need to find the time 't' in the first phase such that the total distance covered in both phases is 550 m. By calculating the distance in each phase and setting the total distance equal to 550 m, we arrive at a quadratic equation in 't'. Solving this equation gives t = 10 s (the positive solution). For the first phase (0 to t seconds): Initial velocity (u\u2081) = 0, acceleration (a\u2081) = 2 m\/s\u00b2. Distance s\u2081 = u\u2081t + (1\/2)a\u2081t\u00b2 = 0*t + (1\/2)*2*t\u00b2 = t\u00b2. Velocity at time t (v\u2081) = u\u2081 + a\u2081t = 0 + 2t = 2t m\/s. For the second phase (t to t+10 seconds): Initial velocity (u\u2082) = v\u2081 = 2t m\/s, acceleration (a\u2082) = 5 m\/s\u00b2, time (t\u2082) = 10 s. Distance s\u2082 = u\u2082t\u2082 + (1\/2)a\u2082t\u2082\u00b2 = (2t)*10 + (1\/2)*5*(10)\u00b2 = 20t + 250 m. Total distance = s\u2081 + s\u2082 = t\u00b2 + 20t + 250. Given total distance is 550 m, t\u00b2 + 20t + 250 = 550, which simplifies to t\u00b2 + 20t - 300 = 0. Factoring the quadratic: (t + 30)(t - 10) = 0. Possible solutions are t = -30 or t = 10. Since time cannot be negative, t = 10 s.","breadcrumb":{"@id":"https:\/\/exam.pscnotes.com\/mcq\/a-vehicle-starts-moving-along-a-straight-line-path-from-rest-in-first\/#breadcrumb"},"inLanguage":"en-US","potentialAction":[{"@type":"ReadAction","target":["https:\/\/exam.pscnotes.com\/mcq\/a-vehicle-starts-moving-along-a-straight-line-path-from-rest-in-first\/"]}]},{"@type":"BreadcrumbList","@id":"https:\/\/exam.pscnotes.com\/mcq\/a-vehicle-starts-moving-along-a-straight-line-path-from-rest-in-first\/#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":"A vehicle starts moving along a straight line path from rest. In first"}]},{"@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\/89060","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=89060"}],"version-history":[{"count":0,"href":"https:\/\/exam.pscnotes.com\/mcq\/wp-json\/wp\/v2\/posts\/89060\/revisions"}],"wp:attachment":[{"href":"https:\/\/exam.pscnotes.com\/mcq\/wp-json\/wp\/v2\/media?parent=89060"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/exam.pscnotes.com\/mcq\/wp-json\/wp\/v2\/categories?post=89060"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/exam.pscnotes.com\/mcq\/wp-json\/wp\/v2\/tags?post=89060"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}