{"id":86180,"date":"2025-06-01T03:39:39","date_gmt":"2025-06-01T03:39:39","guid":{"rendered":"https:\/\/exam.pscnotes.com\/mcq\/?p=86180"},"modified":"2025-06-01T03:39:39","modified_gmt":"2025-06-01T03:39:39","slug":"a-bus-starting-from-a-bus-stand-and-moving-with-uniform-acceleration-a","status":"publish","type":"post","link":"https:\/\/exam.pscnotes.com\/mcq\/a-bus-starting-from-a-bus-stand-and-moving-with-uniform-acceleration-a\/","title":{"rendered":"A bus starting from a bus-stand and moving with uniform acceleration a"},"content":{"rendered":"

A bus starting from a bus-stand and moving with uniform acceleration attains a speed of 20 km\/h in 10 minutes. What is its acceleration?<\/p>\n

[amp_mcq option1=”200 km\/h2<\/sup>” option2=”120 km\/h2<\/sup>” option3=”100 km\/h2<\/sup>” option4=”240 km\/h2<\/sup>” correct=”option2″]<\/p>\n

\n
\n
This question was previously asked in<\/div>\n
UPSC CDS-2 – 2021<\/div>\n<\/div>\n
Download PDF<\/a>Attempt Online<\/a><\/div>\n<\/div>\n
\nThe bus starts from rest, so its initial velocity (u) is 0 km\/h. Its final velocity (v) is 20 km\/h, attained in time (t) = 10 minutes. To find acceleration (a) in km\/h\u00b2, we need to convert time to hours: 10 minutes = 10\/60 hours = 1\/6 hours. Using the equation of motion v = u + at, we get 20 km\/h = 0 km\/h + a \u00d7 (1\/6) hours. Solving for a: a = 20 \u00d7 6 = 120 km\/h\u00b2.
\n<\/section>\n
\nAcceleration is the rate of change of velocity (a = (v-u)\/t). Ensure that all units are consistent before performing calculations. If velocity is in km\/h and time in hours, acceleration will be in km\/h\u00b2.
\n<\/section>\n
\nThe problem specifies uniform acceleration, allowing the use of standard kinematic equations. Other equations of motion like s = ut + \u00bdat\u00b2 and v\u00b2 = u\u00b2 + 2as can also be used depending on the given and required variables.
\n<\/section>\n","protected":false},"excerpt":{"rendered":"

A bus starting from a bus-stand and moving with uniform acceleration attains a speed of 20 km\/h in 10 minutes. What is its acceleration? [amp_mcq option1=”200 km\/h2” option2=”120 km\/h2” option3=”100 km\/h2” option4=”240 km\/h2” correct=”option2″] This question was previously asked in UPSC CDS-2 – 2021 Download PDFAttempt Online The bus starts from rest, so its initial … <\/p>\n

Detailed SolutionA bus starting from a bus-stand and moving with uniform acceleration a<\/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":[1110,1129,1128],"class_list":["post-86180","post","type-post","status-publish","format-standard","hentry","category-upsc-cds-2","tag-1110","tag-mechanics","tag-physics","no-featured-image-padding"],"yoast_head":"\nA bus starting from a bus-stand and moving with uniform acceleration a<\/title>\n<meta name=\"description\" content=\"The bus starts from rest, so its initial velocity (u) is 0 km\/h. Its final velocity (v) is 20 km\/h, attained in time (t) = 10 minutes. To find acceleration (a) in km\/h\u00b2, we need to convert time to hours: 10 minutes = 10\/60 hours = 1\/6 hours. Using the equation of motion v = u + at, we get 20 km\/h = 0 km\/h + a \u00d7 (1\/6) hours. Solving for a: a = 20 \u00d7 6 = 120 km\/h\u00b2. Acceleration is the rate of change of velocity (a = (v-u)\/t). Ensure that all units are consistent before performing calculations. If velocity is in km\/h and time in hours, acceleration will be in km\/h\u00b2.\" \/>\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-bus-starting-from-a-bus-stand-and-moving-with-uniform-acceleration-a\/\" \/>\n<meta property=\"og:locale\" content=\"en_US\" \/>\n<meta property=\"og:type\" content=\"article\" \/>\n<meta property=\"og:title\" content=\"A bus starting from a bus-stand and moving with uniform acceleration a\" \/>\n<meta property=\"og:description\" content=\"The bus starts from rest, so its initial velocity (u) is 0 km\/h. Its final velocity (v) is 20 km\/h, attained in time (t) = 10 minutes. To find acceleration (a) in km\/h\u00b2, we need to convert time to hours: 10 minutes = 10\/60 hours = 1\/6 hours. Using the equation of motion v = u + at, we get 20 km\/h = 0 km\/h + a \u00d7 (1\/6) hours. Solving for a: a = 20 \u00d7 6 = 120 km\/h\u00b2. Acceleration is the rate of change of velocity (a = (v-u)\/t). Ensure that all units are consistent before performing calculations. If velocity is in km\/h and time in hours, acceleration will be in km\/h\u00b2.\" \/>\n<meta property=\"og:url\" content=\"https:\/\/exam.pscnotes.com\/mcq\/a-bus-starting-from-a-bus-stand-and-moving-with-uniform-acceleration-a\/\" \/>\n<meta property=\"og:site_name\" content=\"MCQ and Quiz for Exams\" \/>\n<meta property=\"article:published_time\" content=\"2025-06-01T03:39:39+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 bus starting from a bus-stand and moving with uniform acceleration a","description":"The bus starts from rest, so its initial velocity (u) is 0 km\/h. Its final velocity (v) is 20 km\/h, attained in time (t) = 10 minutes. To find acceleration (a) in km\/h\u00b2, we need to convert time to hours: 10 minutes = 10\/60 hours = 1\/6 hours. Using the equation of motion v = u + at, we get 20 km\/h = 0 km\/h + a \u00d7 (1\/6) hours. Solving for a: a = 20 \u00d7 6 = 120 km\/h\u00b2. Acceleration is the rate of change of velocity (a = (v-u)\/t). Ensure that all units are consistent before performing calculations. If velocity is in km\/h and time in hours, acceleration will be in km\/h\u00b2.","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-bus-starting-from-a-bus-stand-and-moving-with-uniform-acceleration-a\/","og_locale":"en_US","og_type":"article","og_title":"A bus starting from a bus-stand and moving with uniform acceleration a","og_description":"The bus starts from rest, so its initial velocity (u) is 0 km\/h. Its final velocity (v) is 20 km\/h, attained in time (t) = 10 minutes. To find acceleration (a) in km\/h\u00b2, we need to convert time to hours: 10 minutes = 10\/60 hours = 1\/6 hours. Using the equation of motion v = u + at, we get 20 km\/h = 0 km\/h + a \u00d7 (1\/6) hours. Solving for a: a = 20 \u00d7 6 = 120 km\/h\u00b2. Acceleration is the rate of change of velocity (a = (v-u)\/t). Ensure that all units are consistent before performing calculations. If velocity is in km\/h and time in hours, acceleration will be in km\/h\u00b2.","og_url":"https:\/\/exam.pscnotes.com\/mcq\/a-bus-starting-from-a-bus-stand-and-moving-with-uniform-acceleration-a\/","og_site_name":"MCQ and Quiz for Exams","article_published_time":"2025-06-01T03:39:39+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-bus-starting-from-a-bus-stand-and-moving-with-uniform-acceleration-a\/","url":"https:\/\/exam.pscnotes.com\/mcq\/a-bus-starting-from-a-bus-stand-and-moving-with-uniform-acceleration-a\/","name":"A bus starting from a bus-stand and moving with uniform acceleration a","isPartOf":{"@id":"https:\/\/exam.pscnotes.com\/mcq\/#website"},"datePublished":"2025-06-01T03:39:39+00:00","dateModified":"2025-06-01T03:39:39+00:00","author":{"@id":"https:\/\/exam.pscnotes.com\/mcq\/#\/schema\/person\/5807dafeb27d2ec82344d6cbd6c3d209"},"description":"The bus starts from rest, so its initial velocity (u) is 0 km\/h. 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If velocity is in km\/h and time in hours, acceleration will be in km\/h\u00b2.","breadcrumb":{"@id":"https:\/\/exam.pscnotes.com\/mcq\/a-bus-starting-from-a-bus-stand-and-moving-with-uniform-acceleration-a\/#breadcrumb"},"inLanguage":"en-US","potentialAction":[{"@type":"ReadAction","target":["https:\/\/exam.pscnotes.com\/mcq\/a-bus-starting-from-a-bus-stand-and-moving-with-uniform-acceleration-a\/"]}]},{"@type":"BreadcrumbList","@id":"https:\/\/exam.pscnotes.com\/mcq\/a-bus-starting-from-a-bus-stand-and-moving-with-uniform-acceleration-a\/#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 bus starting from a bus-stand and moving with uniform acceleration a"}]},{"@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\/86180","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=86180"}],"version-history":[{"count":0,"href":"https:\/\/exam.pscnotes.com\/mcq\/wp-json\/wp\/v2\/posts\/86180\/revisions"}],"wp:attachment":[{"href":"https:\/\/exam.pscnotes.com\/mcq\/wp-json\/wp\/v2\/media?parent=86180"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/exam.pscnotes.com\/mcq\/wp-json\/wp\/v2\/categories?post=86180"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/exam.pscnotes.com\/mcq\/wp-json\/wp\/v2\/tags?post=86180"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}