{"id":188155,"date":"2025-02-06T12:24:30","date_gmt":"2025-02-06T12:24:30","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=188155"},"modified":"2025-02-06T12:24:32","modified_gmt":"2025-02-06T12:24:32","slug":"rectilinear-kinematics-constant-acceleration","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/02\/06\/rectilinear-kinematics-constant-acceleration\/","title":{"rendered":"Rectilinear Kinematics (Constant Acceleration)"},"content":{"rendered":"\n

Rectilinear Kinematics (Constant Acceleration)
Home Work 2-2
Q1) Initially, the car travels along a straight road with a speed of 35 m\/s. If the brakes are applied and the speed of the car is reduced to 10 m\/s in 15 s, determine the constant deceleration of the car.
Q2) A car starts from rest and with constant acceleration achieves a velocity of 15 m\/s when it travels a distance of 200 m. Determine the acceleration of the car and the time required.
Q3) A train starts from rest at a station and travels with a constant acceleration of 1 m\/s\u00b2. Determine the velocity of the train when t = 3s and the distance traveled during this time.
Q4) car is traveling at 15 m\/s, when the traffic light 50 m ahead turns yellow. Determine the required constant deceleration of the car and the time needed to stop the car at the light.
Q5) Car A starts from rest at t = 0 and travels along a straight road with a constant acceleration of 6 ft\/s\u00b2 until it reaches a speed of 80 ft\/s. Afterwards it maintains this speed. Also, when t=0, car B located 6000 ft down the road is traveling towards A at a constant speed of 60 ft\/s. Determine the distance traveled by car A when they pass each other.<\/p>\n\n\n\n

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

The Correct Answer and Explanation is :<\/strong><\/mark><\/p>\n\n\n\n

Let’s go through each question step-by-step:<\/p>\n\n\n\n

Q1) Determine the constant deceleration of the car.<\/h3>\n\n\n\n
    \n
  • Initial speed, vi=35\u2009m\/sv_i = 35 \\, \\text{m\/s}<\/li>\n\n\n\n
  • Final speed, vf=10\u2009m\/sv_f = 10 \\, \\text{m\/s}<\/li>\n\n\n\n
  • Time, t=15\u2009st = 15 \\, \\text{s}<\/li>\n\n\n\n
  • We can use the equation of motion: vf=vi+atv_f = v_i + a t Rearranging for acceleration (or deceleration): a=vf\u2212vita = \\frac{v_f – v_i}{t} Substituting values: a=10\u22123515=\u22122515=\u22121.67\u2009m\/s2a = \\frac{10 – 35}{15} = \\frac{-25}{15} = -1.67 \\, \\text{m\/s}^2 Therefore, the constant deceleration is \u22121.67\u2009m\/s2-1.67 \\, \\text{m\/s}^2.<\/li>\n<\/ul>\n\n\n\n

    Q2) Determine the acceleration and time required for the car to achieve 15 m\/s after traveling 200 m.<\/h3>\n\n\n\n
      \n
    • Initial speed, vi=0\u2009m\/sv_i = 0 \\, \\text{m\/s} (starts from rest)<\/li>\n\n\n\n
    • Final speed, vf=15\u2009m\/sv_f = 15 \\, \\text{m\/s}<\/li>\n\n\n\n
    • Distance, d=200\u2009md = 200 \\, \\text{m}<\/li>\n\n\n\n
    • We can use the kinematic equation: vf2=vi2+2adv_f^2 = v_i^2 + 2 a d Substituting the values: 152=0+2a\u00d720015^2 = 0 + 2 a \\times 200 225=400a225 = 400a a=225400=0.5625\u2009m\/s2a = \\frac{225}{400} = 0.5625 \\, \\text{m\/s}^2 Now, using the equation vf=vi+atv_f = v_i + a t to find time tt: 15=0+0.5625t15 = 0 + 0.5625 t t=150.5625=26.67\u2009st = \\frac{15}{0.5625} = 26.67 \\, \\text{s} Therefore, the acceleration is 0.5625\u2009m\/s20.5625 \\, \\text{m\/s}^2 and the time required is 26.67\u2009s26.67 \\, \\text{s}.<\/li>\n<\/ul>\n\n\n\n

      Q3) Determine the velocity and distance traveled by the train after 3 seconds.<\/h3>\n\n\n\n
        \n
      • Initial velocity, vi=0\u2009m\/sv_i = 0 \\, \\text{m\/s} (starts from rest)<\/li>\n\n\n\n
      • Acceleration, a=1\u2009m\/s2a = 1 \\, \\text{m\/s}^2<\/li>\n\n\n\n
      • Time, t=3\u2009st = 3 \\, \\text{s}<\/li>\n\n\n\n
      • We can calculate the velocity using: v=vi+atv = v_i + a t v=0+1\u00d73=3\u2009m\/sv = 0 + 1 \\times 3 = 3 \\, \\text{m\/s} For the distance, use: d=vit+12at2d = v_i t + \\frac{1}{2} a t^2 d=0+12\u00d71\u00d732=4.5\u2009md = 0 + \\frac{1}{2} \\times 1 \\times 3^2 = 4.5 \\, \\text{m} Therefore, the velocity is 3\u2009m\/s3 \\, \\text{m\/s} and the distance traveled is 4.5\u2009m4.5 \\, \\text{m}.<\/li>\n<\/ul>\n\n\n\n

        Q4) Determine the deceleration and time needed for the car to stop at the light.<\/h3>\n\n\n\n
          \n
        • Initial speed, vi=15\u2009m\/sv_i = 15 \\, \\text{m\/s}<\/li>\n\n\n\n
        • Final speed, vf=0\u2009m\/sv_f = 0 \\, \\text{m\/s}<\/li>\n\n\n\n
        • Distance to the light, d=50\u2009md = 50 \\, \\text{m}<\/li>\n\n\n\n
        • We can use the equation: vf2=vi2+2adv_f^2 = v_i^2 + 2 a d Substituting values: 0=152+2a\u00d7500 = 15^2 + 2a \\times 50 \u2212225=100a-225 = 100a a=\u2212225100=\u22122.25\u2009m\/s2a = \\frac{-225}{100} = -2.25 \\, \\text{m\/s}^2 Now, to find the time, use vf=vi+atv_f = v_i + a t: 0=15+(\u22122.25)t0 = 15 + (-2.25) t t=152.25=6.67\u2009st = \\frac{15}{2.25} = 6.67 \\, \\text{s} Therefore, the required deceleration is \u22122.25\u2009m\/s2-2.25 \\, \\text{m\/s}^2 and the time is 6.67\u2009s6.67 \\, \\text{s}.<\/li>\n<\/ul>\n\n\n\n

          Q5) Determine the distance traveled by car A when they pass each other.<\/h3>\n\n\n\n
            \n
          • Car A starts from rest with acceleration a=6\u2009ft\/s2a = 6 \\, \\text{ft\/s}^2 until it reaches 80\u2009ft\/s80 \\, \\text{ft\/s}.<\/li>\n\n\n\n
          • Car B starts 6000 feet ahead of A and travels at 60\u2009ft\/s60 \\, \\text{ft\/s}.<\/li>\n\n\n\n
          • Car A’s speed at time tt is given by vA=6tv_A = 6t, and the distance it travels by time tt is dA=3t2d_A = 3t^2.<\/li>\n\n\n\n
          • Car B’s position at time tt is dB=6000+60td_B = 6000 + 60t.<\/li>\n<\/ul>\n\n\n\n

            We need to solve for tt when dA=dBd_A = d_B: 3t2=6000+60t3t^2 = 6000 + 60t<\/p>\n\n\n\n

            Rearranging and solving: 3t2\u221260t\u22126000=03t^2 – 60t – 6000 = 0<\/p>\n\n\n\n

            Dividing by 3: t2\u221220t\u22122000=0t^2 – 20t – 2000 = 0<\/p>\n\n\n\n

            Solving using the quadratic formula: t=\u2212(\u221220)\u00b1(\u221220)2\u22124(1)(\u22122000)2(1)=20\u00b1400+80002=20\u00b184002t = \\frac{-(-20) \\pm \\sqrt{(-20)^2 – 4(1)(-2000)}}{2(1)} = \\frac{20 \\pm \\sqrt{400 + 8000}}{2} = \\frac{20 \\pm \\sqrt{8400}}{2} t=20\u00b191.652t = \\frac{20 \\pm 91.65}{2}<\/p>\n\n\n\n

            Taking the positive root: t=111.652=55.83\u2009secondst = \\frac{111.65}{2} = 55.83 \\, \\text{seconds}<\/p>\n\n\n\n

            Finally, the distance traveled by car A is: dA=3(55.83)2=9333.3\u2009ftd_A = 3(55.83)^2 = 9333.3 \\, \\text{ft}<\/p>\n\n\n\n

            So, car A travels 9333.3 feet before they pass each other.<\/p>\n\n\n\n

            This covers all the required calculations and answers to the questions!<\/p>\n","protected":false},"excerpt":{"rendered":"

            Rectilinear Kinematics (Constant Acceleration)Home Work 2-2Q1) Initially, the car travels along a straight road with a speed of 35 m\/s. If the brakes are applied and the speed of the car is reduced to 10 m\/s in 15 s, determine the constant deceleration of the car.Q2) A car starts from rest and with constant acceleration […]<\/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-188155","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/188155","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=188155"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/188155\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=188155"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=188155"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=188155"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}