{"id":199284,"date":"2025-03-11T03:48:39","date_gmt":"2025-03-11T03:48:39","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=199284"},"modified":"2025-03-11T03:48:41","modified_gmt":"2025-03-11T03:48:41","slug":"a-runner-accelerates-at-a-constant-0-20-m-s2-to-the-north","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/03\/11\/a-runner-accelerates-at-a-constant-0-20-m-s2-to-the-north\/","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 utilize the kinematic equations:<\/p>\n\n\n\n

    \n
  1. Final velocity (( v ))<\/strong>: [ v = v_0 + at ]<\/li>\n\n\n\n
  2. Displacement (( s ))<\/strong>: [ s = v_0 t + \\frac{1}{2} a t^2 ]<\/li>\n\n\n\n
  3. Velocity-displacement relation<\/strong>: [ v^2 = v_0^2 + 2a s ]<\/li>\n<\/ol>\n\n\n\n

    where:<\/p>\n\n\n\n

      \n
    • ( v_0 ) is the initial velocity<\/li>\n\n\n\n
    • ( a ) is the constant acceleration<\/li>\n\n\n\n
    • ( t ) is the time elapsed<\/li>\n\n\n\n
    • ( s ) is the displacement<\/li>\n<\/ul>\n\n\n\n

      1. Reaching 7.0 m\/s from rest with ( a = 0.20 \\, \\text{m\/s}^2 ):<\/strong><\/p>\n\n\n\n

        \n
      • Time to reach 7.0 m\/s:<\/strong> Using ( v = v_0 + at ):\ue206 [ t = \\frac{v – v_0}{a} = \\frac{7.0 \\, \\text{m\/s} – 0}{0.20 \\, \\text{m\/s}^2} = 35 \\, \\text{s} ]<\/li>\n\n\n\n
      • Distance covered in this time:<\/strong> Using ( s = v_0 t + \\frac{1}{2} a t^2 ): [ s = 0 \\times 35 \\, \\text{s} + \\frac{1}{2} \\times 0.20 \\, \\text{m\/s}^2 \\times (35 \\, \\text{s})^2 = 428.75 \\, \\text{m} ]<\/li>\n<\/ul>\n\n\n\n

        2. Covering 50 meters starting at 3.0 m\/s:<\/strong><\/p>\n\n\n\n

          \n
        • Time to cover 50 meters:<\/strong> Using ( s = v_0 t + \\frac{1}{2} a t^2 ):\ue206 [ 50 \\, \\text{m} = 3.0 \\, \\text{m\/s} \\times t + \\frac{1}{2} \\times 0.20 \\, \\text{m\/s}^2 \\times t^2 ]\ue206 This simplifies to:\ue206 [ 0.1 t^2 + 3.0 t – 50 = 0 ]\ue206 Solving this quadratic equation:\ue206 [ t = \\frac{-3.0 \\pm \\sqrt{(3.0)^2 – 4 \\times 0.1 \\times (-50)}}{2 \\times 0.1} ]\ue206 [ t = \\frac{-3.0 \\pm \\sqrt{9 + 20}}{0.2} = \\frac{-3.0 \\pm \\sqrt{29}}{0.2} ] Taking the positive root:\ue206 [ t \\approx \\frac{-3.0 + 5.385}{0.2} = \\frac{2.385}{0.2} \\approx 11.925 \\, \\text{s} ]\ue206<\/li>\n\n\n\n
        • Speed at that time:<\/strong> Using ( v = v_0 + at ):\ue206 [ v = 3.0 \\, \\text{m\/s} + 0.20 \\, \\text{m\/s}^2 \\times 11.925 \\, \\text{s} \\approx 5.385 \\, \\text{m\/s} ]<\/li>\n<\/ul>\n\n\n\n

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

          Using ( v^2 = v_0^2 + 2a s ):<\/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. Returning to the starting position after moving south at 4.0 m\/s:<\/strong><\/p>\n\n\n\n

          Assuming southward direction as negative:<\/p>\n\n\n\n

          Initial velocity, ( v_0 = -4.0 \\, \\text{m\/s} )<\/p>\n\n\n\n

          The runner decelerates until velocity becomes zero, then accelerates northward back to the starting point.<\/p>\n\n\n\n

            \n
          • Time to stop:<\/strong> Using ( v = v_0 + at ):[ 0 = -4.0 \\, \\text{m\/s} + 0.20 \\, \\text{m\/s}^2 \\times t ] [ t = \\frac{4.0 \\, \\text{m\/s}}{0.20 \\, \\text{m\/s}^2} = 20 \\, \\text{s} ]<\/li>\n\n\n\n
          • Distance covered while stopping:<\/strong> Using ( s = v_0 t + \\frac{1}{2} a t^2 ): [ s = -4.0 \\, \\text{m\/s} \\times 20 \\, \\text{s} + \\frac{1}{2} \\times 0.20 \\, \\text{m\/s}^2 \\times (20 \\, \\text{s})^2 ] [ s = -80 \\, \\text{m} + 40 \\, \\text{m} = -40 \\, \\text{m} ]<\/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-199284","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/199284","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=199284"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/199284\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=199284"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=199284"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=199284"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}