{"id":199648,"date":"2025-03-12T08:03:33","date_gmt":"2025-03-12T08:03:33","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=199648"},"modified":"2025-03-12T08:03:36","modified_gmt":"2025-03-12T08:03:36","slug":"a-runner-accelerates-at-a-constant-0-20-m-s2-to-the-north-3","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/03\/12\/a-runner-accelerates-at-a-constant-0-20-m-s2-to-the-north-3\/","title":{"rendered":"A runner accelerates at a constant 0.20 m\/s2 to the north"},"content":{"rendered":"\n

A runner accelerates at a constant 0.20 m\/s2 to the north.
If she starts from rest, how long does it take her to reach 7.0 m\/s? How far does she run in this time?
If she was initially running at 3.0 m\/s to the north, how long does it take her to run 50 meters? What is her speed at that time?
If she was initially running at 3.0 m\/s to the north, how fast is she moving after running 30 meters?
If she was initially moving at 4.0 m\/s to the South, how long does it take her to return to her starting position?
If she was initially moving at 4.0 m\/s to the South, how long does it take her to reach a point 60 meters South of her starting position?<\/p>\n\n\n\n

The correct answer and explanation is :<\/strong><\/mark><\/p>\n\n\n\n

To analyze the runner’s motion under constant acceleration, we employ the kinematic equations:<\/p>\n\n\n\n

    \n
  1. ( v = u + at )<\/li>\n\n\n\n
  2. ( s = ut + \\frac{1}{2}at^2 )<\/li>\n\n\n\n
  3. ( v^2 = u^2 + 2as )<\/li>\n<\/ol>\n\n\n\n

    Where:<\/p>\n\n\n\n

      \n
    • ( u ) is the initial velocity<\/li>\n\n\n\n
    • ( v ) is the final velocity<\/li>\n\n\n\n
    • ( a ) is the acceleration<\/li>\n\n\n\n
    • ( t ) is the time\ue206<\/li>\n\n\n\n
    • ( s ) is the displacement<\/li>\n<\/ul>\n\n\n\n

      1. Starting from rest, reaching 7.0 m\/s:<\/strong><\/p>\n\n\n\n

        \n
      • Time to reach 7.0 m\/s:<\/strong> Using ( v = u + at ): ( 7.0\\, \\text{m\/s} = 0 + (0.20\\, \\text{m\/s}^2) \\times t ) Solving for ( t ): ( t = \\frac{7.0\\, \\text{m\/s}}{0.20\\, \\text{m\/s}^2} = 35\\, \\text{seconds} )<\/li>\n\n\n\n
      • Distance covered in this time:<\/strong> Using ( s = ut + \\frac{1}{2}at^2 ):\ue206 ( s = 0 + \\frac{1}{2}(0.20\\, \\text{m\/s}^2)(35\\, \\text{s})^2 ) ( s = 0.10 \\times 1,225 = 122.5\\, \\text{meters} )<\/li>\n<\/ul>\n\n\n\n

        2. Initially running at 3.0 m\/s, covering 50 meters:<\/strong><\/p>\n\n\n\n

          \n
        • Time to cover 50 meters:<\/strong> Using ( s = ut + \\frac{1}{2}at^2 ): ( 50\\, \\text{m} = (3.0\\, \\text{m\/s})t + \\frac{1}{2}(0.20\\, \\text{m\/s}^2)t^2 ) This simplifies to a quadratic equation: ( 0.1t^2 + 3.0t – 50 = 0 ) Solving for ( t ) using the quadratic formula: ( t = \\frac{-3.0 \\pm \\sqrt{(3.0)^2 – 4 \\times 0.1 \\times (-50)}}{2 \\times 0.1} ) ( t = \\frac{-3.0 \\pm \\sqrt{9 + 20}}{0.2} ) ( t = \\frac{-3.0 \\pm \\sqrt{29}}{0.2} )Taking the positive root:( t = \\frac{-3.0 + 5.385}{0.2} \\approx 11.925\\, \\text{seconds} )<\/li>\n\n\n\n
        • Speed at that time:<\/strong> Using ( v = u + at ): ( v = 3.0\\, \\text{m\/s} + (0.20\\, \\text{m\/s}^2) \\times 11.925\\, \\text{s} ) ( v \\approx 3.0 + 2.385 = 5.385\\, \\text{m\/s} )<\/li>\n<\/ul>\n\n\n\n

          3. Initially running at 3.0 m\/s, covering 30 meters:<\/strong><\/p>\n\n\n\n

            \n
          • Final speed after 30 meters:<\/strong> Using ( v^2 = u^2 + 2as ):\ue206 ( v^2 = (3.0\\, \\text{m\/s})^2 + 2(0.20\\, \\text{m\/s}^2)(30\\, \\text{m}) )\ue206 ( v^2 = 9 + 12 )\ue206 ( v = \\sqrt{21} \\approx 4.582\\, \\text{m\/s} )\ue206<\/li>\n<\/ul>\n\n\n\n

            4. Initially moving at 4.0 m\/s to the South, returning to starting position:<\/strong><\/p>\n\n\n\n

            Since the runner is moving southward initially, to return to the starting position, she must decelerate to a stop and then accelerate northward back to the start.\ue206<\/p>\n\n\n\n

              \n
            • Time to stop:<\/strong> Using ( v = u + at ) with ( v = 0 ):\ue206 ( 0 = 4.0\\, \\text{m\/s} – (0.20\\, \\text{m\/s}^2)t )\ue206 ( t = \\frac{4.0\\, \\text{m\/s}}{0.20\\, \\text{m\/s}^2} = 20\\, \\text{seconds} )\ue206<\/li>\n\n\n\n
            • Distance covered while stopping:<\/strong> Using ( s = ut + \\frac{1}{2}at^2 ):\ue206 ( s = (4.0\\, \\text{m\/s})(20\\, \\text{s}) – \\frac{1}{2}(0.20\\, \\text{m\/s}^2)(20\\, \\text{s})^2 )\ue206 ( s = 80 – 40 = 40\\, \\text{meters} )\ue206 The runner has moved 40 meters south while stopping. To return to the starting position, she needs to cover 40 meters north.\ue206<\/li>\n\n\n\n
            • Time to accelerate northward back to the start:<\/strong> Using ( s = ut + \\frac{1}{2}at^2 ) with<\/li>\n<\/ul>\n","protected":false},"excerpt":{"rendered":"

              A runner accelerates at a constant 0.20 m\/s2 to the north.If she starts from rest, how long does it take her to reach 7.0 m\/s? How far does she run in this time?If she was initially running at 3.0 m\/s to the north, how long does it take her to run 50 meters? What is […]<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"closed","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"site-sidebar-layout":"default","site-content-layout":"","ast-site-content-layout":"default","site-content-style":"default","site-sidebar-style":"default","ast-global-header-display":"","ast-banner-title-visibility":"","ast-main-header-display":"","ast-hfb-above-header-display":"","ast-hfb-below-header-display":"","ast-hfb-mobile-header-display":"","site-post-title":"","ast-breadcrumbs-content":"","ast-featured-img":"","footer-sml-layout":"","ast-disable-related-posts":"","theme-transparent-header-meta":"","adv-header-id-meta":"","stick-header-meta":"","header-above-stick-meta":"","header-main-stick-meta":"","header-below-stick-meta":"","astra-migrate-meta-layouts":"default","ast-page-background-enabled":"default","ast-page-background-meta":{"desktop":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"tablet":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"mobile":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""}},"ast-content-background-meta":{"desktop":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"tablet":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"mobile":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""}},"footnotes":""},"categories":[25],"tags":[],"class_list":["post-199648","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/199648","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/comments?post=199648"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/199648\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=199648"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=199648"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=199648"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}