{"id":199574,"date":"2025-03-12T06:37:16","date_gmt":"2025-03-12T06:37:16","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=199574"},"modified":"2025-03-12T06:37:19","modified_gmt":"2025-03-12T06:37:19","slug":"a-runner-accelerates-at-a-constant-0-20-m-s2-to-the-north-2","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-2\/","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

\"\"<\/figure>\n\n\n\n

To analyze the runner’s motion under constant acceleration, we utilize 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\ue206<\/li>\n<\/ul>\n\n\n\n

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

        \n
      • Initial velocity (( u )) = 0 m\/s\ue206<\/li>\n\n\n\n
      • Final velocity (( v )) = 7.0 m\/s\ue206<\/li>\n\n\n\n
      • Acceleration (( a )) = 0.20 m\/s\u00b2<\/li>\n<\/ul>\n\n\n\n

        Time to reach 7.0 m\/s:<\/strong><\/p>\n\n\n\n

        Using ( v = u + at ):<\/p>\n\n\n\n

        ( 7.0\\, \\text{m\/s} = 0 + (0.20\\, \\text{m\/s}^2) \\times t )<\/p>\n\n\n\n

        ( t = \\frac{7.0\\, \\text{m\/s}}{0.20\\, \\text{m\/s}^2} = 35\\, \\text{seconds} )<\/p>\n\n\n\n

        Distance covered in this time:<\/strong><\/p>\n\n\n\n

        Using ( s = ut + \\frac{1}{2}at^2 ):<\/p>\n\n\n\n

        ( s = 0 + \\frac{1}{2} \\times 0.20\\, \\text{m\/s}^2 \\times (35\\, \\text{seconds})^2 )<\/p>\n\n\n\n

        ( s = 0.10 \\times 1225 = 122.5\\, \\text{meters} )<\/p>\n\n\n\n

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

          \n
        • Initial velocity (( u )) = 3.0 m\/s<\/li>\n\n\n\n
        • Displacement (( s )) = 50 meters<\/li>\n\n\n\n
        • Acceleration (( a )) = 0.20 m\/s\u00b2<\/li>\n<\/ul>\n\n\n\n

          Time to cover 50 meters:<\/strong><\/p>\n\n\n\n

          Using ( s = ut + \\frac{1}{2}at^2 ):<\/p>\n\n\n\n

          ( 50\\, \\text{m} = (3.0\\, \\text{m\/s}) \\times t + \\frac{1}{2} \\times 0.20\\, \\text{m\/s}^2 \\times t^2 )<\/p>\n\n\n\n

          ( 50 = 3t + 0.1t^2 )<\/p>\n\n\n\n

          Rearranging:<\/p>\n\n\n\n

          ( 0.1t^2 + 3t – 50 = 0 )<\/p>\n\n\n\n

          Solving this quadratic equation:<\/p>\n\n\n\n

          ( t = \\frac{-3 \\pm \\sqrt{(3)^2 – 4 \\times 0.1 \\times (-50)}}{2 \\times 0.1} )<\/p>\n\n\n\n

          ( t = \\frac{-3 \\pm \\sqrt{9 + 20}}{0.2} = \\frac{-3 \\pm \\sqrt{29}}{0.2} )<\/p>\n\n\n\n

          ( t \\approx \\frac{-3 + 5.385}{0.2} = \\frac{2.385}{0.2} \\approx 11.925\\, \\text{seconds} )<\/p>\n\n\n\n

          Speed at that time:<\/strong><\/p>\n\n\n\n

          Using ( v = u + at ):<\/p>\n\n\n\n

          ( v = 3.0\\, \\text{m\/s} + 0.20\\, \\text{m\/s}^2 \\times 11.925\\, \\text{seconds} )<\/p>\n\n\n\n

          ( v \\approx 3.0 + 2.385 = 5.385\\, \\text{m\/s} )<\/p>\n\n\n\n

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

            \n
          • Initial velocity (( u )) = 3.0 m\/s<\/li>\n\n\n\n
          • Displacement (( s )) = 30 meters<\/li>\n\n\n\n
          • Acceleration (( a )) = 0.20 m\/s\u00b2<\/li>\n<\/ul>\n\n\n\n

            Final speed after 30 meters:<\/strong><\/p>\n\n\n\n

            Using ( v^2 = u^2 + 2as ):<\/p>\n\n\n\n

            ( v^2 = (3.0\\, \\text{m\/s})^2 + 2 \\times 0.20\\, \\text{m\/s}^2 \\times 30\\, \\text{m} )<\/p>\n\n\n\n

            ( v^2 = 9 + 12 = 21 )<\/p>\n\n\n\n

            ( v = \\sqrt{21} \\approx 4.58\\, \\text{m\/s} )<\/p>\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

              \n
            • Initial velocity (( u )) = -4.0 m\/s (negative sign indicates southward direction)<\/li>\n\n\n\n
            • Acceleration (( a )) = 0.20 m\/s\u00b2 (northward)<\/li>\n<\/ul>\n\n\n\n

              Time to return to starting position:<\/strong><\/p>\n\n\n\n

              The runner decelerates to a stop and then accelerates northward back to the starting point.<\/p>\n\n\n\n

              Time to stop:<\/strong><\/p>\n\n\n\n

              Using ( v = u + at ):<\/p>\n\n\n\n

              ( 0 = -4.0\\, \\text{m\/s} + 0.20\\, \\text{m\/s}^2 \\times t )<\/p>\n\n\n\n

              ( t = \\frac{4.0<\/p>\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-199574","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/199574","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=199574"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/199574\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=199574"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=199574"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=199574"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}