{"id":2413,"date":"2026-05-27T22:43:38","date_gmt":"2026-05-27T17:13:38","guid":{"rendered":"https:\/\/mymockmate.com\/notes\/?p=2413"},"modified":"2026-05-27T22:43:40","modified_gmt":"2026-05-27T17:13:40","slug":"ncert-class-9-science-motion-chapter-complete-solutions-and-mcqs","status":"publish","type":"post","link":"https:\/\/mymockmate.com\/notes\/ncert-class-9-science-motion-chapter-complete-solutions-and-mcqs\/","title":{"rendered":"NCERT Class 9 Science Motion Chapter Complete Solutions and MCQs"},"content":{"rendered":"
\"image_pdf\"Save as PDF<\/span><\/a>\"image_print\"Print<\/span><\/a> <\/span><\/div>\n

<\/p>\n\n\n\n

<\/p>\n\n\n\n

Short Intro<\/h1>\n\n\n\n

\u201cDescribing Motion Around Us\u201d chapter explains the concepts of motion, distance, displacement, speed, velocity, acceleration, graphs of motion, kinematic equations, and circular motion. This chapter is very important for exams because it contains conceptual questions, graph-based numericals, and formula applications. Below are complete Unicode solutions prepared for uploading on www.mymockmate.com<\/a>.<\/p>\n\n\n\n

Quick Information Box<\/h1>\n\n\n\n
Topic<\/th>Details<\/th><\/tr><\/thead>
Chapter Name<\/td>Describing Motion Around Us<\/td><\/tr>
Subject<\/td>Science<\/td><\/tr>
Grade<\/td>Class 9<\/td><\/tr>
Main Topics<\/td>Motion, Velocity, Acceleration<\/td><\/tr>
Important Concepts<\/td>Graphs & Kinematic Equations<\/td><\/tr>
Exam Importance<\/td>Very High<\/td><\/tr>
Numerical Problems<\/td>Included<\/td><\/tr><\/tbody><\/table><\/figure>\n\n\n\n

Concepts Used (Topics Covered)<\/h1>\n\n\n\n
    \n
  • Distance and displacement<\/li>\n\n\n\n
  • Speed and velocity<\/li>\n\n\n\n
  • Average velocity<\/li>\n\n\n\n
  • Average acceleration<\/li>\n\n\n\n
  • Position-time graph<\/li>\n\n\n\n
  • Velocity-time graph<\/li>\n\n\n\n
  • Uniform motion<\/li>\n\n\n\n
  • Non-uniform motion<\/li>\n\n\n\n
  • Kinematic equations<\/li>\n\n\n\n
  • Uniform circular motion<\/li>\n\n\n\n
  • Motion in a plane<\/li>\n<\/ul>\n\n\n\n

    Important Formulas<\/h1>\n\n\n\n

    Average Speed Formula<\/h2>\n\n\n\n

    Average Speed<\/mtext>=<\/mo>Total Distance<\/mtext>Total Time<\/mtext><\/mfrac><\/mrow>\\text{Average Speed} = \\frac{\\text{Total Distance}}{\\text{Total Time}}<\/annotation><\/semantics><\/math><\/p>\n\n\n\n

    Average Velocity Formula<\/h2>\n\n\n\n

    v<\/mi>a<\/mi>v<\/mi><\/mrow><\/msub>=<\/mo>s<\/mi>t<\/mi><\/mfrac><\/mrow>v_{av}=\\frac{s}{t}<\/annotation><\/semantics><\/math><\/p>\n\n\n\n

    Acceleration Formula<\/h2>\n\n\n\n

    a<\/mi>=<\/mo>v<\/mi>\u2212<\/mo>u<\/mi><\/mrow>t<\/mi><\/mfrac><\/mrow>a=\\frac{v-u}{t}<\/annotation><\/semantics><\/math><\/p>\n\n\n\n

    First Equation of Motion<\/h2>\n\n\n\n

    v<\/mi>=<\/mo>u<\/mi>+<\/mo>a<\/mi>t<\/mi><\/mrow>v=u+at<\/annotation><\/semantics><\/math><\/p>\n\n\n\n

    Second Equation of Motion<\/h2>\n\n\n\n

    s<\/mi>=<\/mo>u<\/mi>t<\/mi>+<\/mo>1<\/mn>2<\/mn><\/mfrac>a<\/mi>t<\/mi>2<\/mn><\/msup><\/mrow>s=ut+\\frac{1}{2}at^2<\/annotation><\/semantics><\/math><\/p>\n\n\n\n

    Third Equation of Motion<\/h2>\n\n\n\n

    v<\/mi>2<\/mn><\/msup>=<\/mo>u<\/mi>2<\/mn><\/msup>+<\/mo>2<\/mn>a<\/mi>s<\/mi><\/mrow>v^2=u^2+2as<\/annotation><\/semantics><\/math><\/p>\n\n\n\n

    Questions & Step-by-step Solutions with Explanation<\/h1>\n\n\n\n

    Question 1<\/h1>\n\n\n\n

    My father went to a shop from home which is located at a distance
    of 250 m on a straight road. On reaching there, he discovered that
    he forgot to carry a cloth bag. He came home to take it, went to the
    shop again, bought provisions and came back home. How much was
    the total distance travelled by him? What was his displacement from
    home?<\/p>\n\n\n\n

    Total distance and displacement of father<\/h3>\n\n\n\n

    Given<\/h3>\n\n\n\n

    Distance between home and shop = 250 m<\/p>\n\n\n\n

    Solution<\/h3>\n\n\n\n

    Father travelled:<\/p>\n\n\n\n

      \n
    • Home \u2192 Shop = 250 m<\/li>\n\n\n\n
    • Shop \u2192 Home = 250 m<\/li>\n\n\n\n
    • Home \u2192 Shop = 250 m<\/li>\n\n\n\n
    • Shop \u2192 Home = 250 m<\/li>\n<\/ul>\n\n\n\n

      Total Distance<\/h3>\n\n\n\n

      = 250 + 250 + 250 + 250
      = 1000 m<\/p>\n\n\n\n

      Displacement<\/h3>\n\n\n\n

      Initial and final positions are same.<\/p>\n\n\n\n

      Answer<\/h3>\n\n\n\n
        \n
      • Total distance travelled = 1000 m<\/li>\n\n\n\n
      • Displacement = 0 m<\/li>\n<\/ul>\n\n\n\n
        \n\n\n\n

        Question 2<\/h1>\n\n\n\n

        A student runs from the ground floor to the fourth floor of a school
        building to collect a book and then comes down to their classroom on
        the second floor. If the height of each floor is 3 m, find:
        (i) the total vertical distance travelled, and
        (ii) their displacement from the starting point<\/p>\n\n\n\n

        Vertical distance and displacement<\/h3>\n\n\n\n

        Given<\/h3>\n\n\n\n

        Height of each floor = 3 m<\/p>\n\n\n\n

        Solution<\/h3>\n\n\n\n

        Ground floor to fourth floor:
        = 4 \u00d7 3 = 12 m<\/p>\n\n\n\n

        Fourth floor to second floor:
        = 2 \u00d7 3 = 6 m<\/p>\n\n\n\n

        Total Distance<\/h3>\n\n\n\n

        = 12 + 6
        = 18 m<\/p>\n\n\n\n

        Displacement<\/h3>\n\n\n\n

        From ground floor to second floor:
        = 2 \u00d7 3
        = 6 m upward<\/p>\n\n\n\n

        Answer<\/h3>\n\n\n\n
          \n
        • Total vertical distance = 18 m<\/li>\n\n\n\n
        • Displacement = 6 m upward<\/li>\n<\/ul>\n\n\n\n
          \n\n\n\n

          Question 3<\/h1>\n\n\n\n

          A girl is riding her scooter and finds that its speedometer reading is
          constant. Is it possible for her scooter to be accelerating and if so, how?<\/p>\n\n\n\n

          Can scooter accelerate with constant speedometer reading?<\/h3>\n\n\n\n

          Answer<\/h3>\n\n\n\n

          Yes.<\/p>\n\n\n\n

          Explanation<\/h3>\n\n\n\n

          If the scooter changes direction while moving at constant speed, velocity changes and acceleration occurs.<\/p>\n\n\n\n

          Example:<\/p>\n\n\n\n

            \n
          • Moving on a circular road<\/li>\n\n\n\n
          • Taking a turn<\/li>\n<\/ul>\n\n\n\n
            \n\n\n\n

            Question 4<\/h1>\n\n\n\n

            A car starts from rest and its velocity reaches 24 m s\u20131 in 6 s. Find the
            average acceleration and the distance travelled in these 6 s.<\/p>\n\n\n\n

            Find acceleration and distance travelled<\/h3>\n\n\n\n

            Given<\/h3>\n\n\n\n

            Initial velocity:
            u<\/mi>=<\/mo>0<\/mn><\/mrow>u=0<\/annotation><\/semantics><\/math><\/p>\n\n\n\n

            Final velocity:
            v<\/mi>=<\/mo>24<\/mn>\u2009<\/mtext>m<\/mi>\/<\/mi>s<\/mi><\/mrow>v=24\\,m\/s<\/annotation><\/semantics><\/math><\/p>\n\n\n\n

            Time:
            t<\/mi>=<\/mo>6<\/mn>\u2009<\/mtext>s<\/mi><\/mrow>t=6\\,s<\/annotation><\/semantics><\/math><\/p>\n\n\n\n

            Step 1: Acceleration<\/h2>\n\n\n\n

            Using formula:
            a<\/mi>=<\/mo>v<\/mi>\u2212<\/mo>u<\/mi><\/mrow>t<\/mi><\/mfrac><\/mrow>a=\\frac{v-u}{t}<\/annotation><\/semantics><\/math><\/p>\n\n\n\n

            Substitute values:
            a<\/mi>=<\/mo>24<\/mn>\u2212<\/mo>0<\/mn><\/mrow>6<\/mn><\/mfrac>=<\/mo>4<\/mn>\u2009<\/mtext>m<\/mi>\/<\/mi>s<\/mi>2<\/mn><\/msup><\/mrow>a=\\frac{24-0}{6}=4\\,m\/s^2<\/annotation><\/semantics><\/math><\/p>\n\n\n\n

            Step 2: Distance<\/h2>\n\n\n\n

            Using:
            s<\/mi>=<\/mo>u<\/mi>t<\/mi>+<\/mo>1<\/mn>2<\/mn><\/mfrac>a<\/mi>t<\/mi>2<\/mn><\/msup><\/mrow>s=ut+\\frac{1}{2}at^2<\/annotation><\/semantics><\/math><\/p>\n\n\n\n

            Substitute values:
            s<\/mi>=<\/mo>0<\/mn>+<\/mo>1<\/mn>2<\/mn><\/mfrac>(<\/mo>4<\/mn>)<\/mo>(<\/mo>6<\/mn>2<\/mn><\/msup>)<\/mo>=<\/mo>72<\/mn>\u2009<\/mtext>m<\/mi><\/mrow>s=0+\\frac{1}{2}(4)(6^2)=72\\,m<\/annotation><\/semantics><\/math><\/p>\n\n\n\n

            Answer<\/h3>\n\n\n\n
              \n
            • Acceleration = 4 m\/s\u00b2<\/li>\n\n\n\n
            • Distance travelled = 72 m<\/li>\n<\/ul>\n\n\n\n
              \n\n\n\n

              Question 5<\/h1>\n\n\n\n

              A motorbike moving with initial velocity 28 m s\u20131<\/sup> and constant
              acceleration stops after travelling 98 m. Find the acceleration of the
              motorbike and the time taken to come to a stop.<\/p>\n\n\n\n

              Motorbike stopping problem<\/h3>\n\n\n\n

              Given<\/h3>\n\n\n\n

              Initial velocity:
              u<\/mi>=<\/mo>28<\/mn>\u2009<\/mtext>m<\/mi>\/<\/mi>s<\/mi><\/mrow>u=28\\,m\/s<\/annotation><\/semantics><\/math><\/p>\n\n\n\n

              Final velocity:
              v<\/mi>=<\/mo>0<\/mn><\/mrow>v=0<\/annotation><\/semantics><\/math><\/p>\n\n\n\n

              Distance:
              s<\/mi>=<\/mo>98<\/mn>\u2009<\/mtext>m<\/mi><\/mrow>s=98\\,m<\/annotation><\/semantics><\/math><\/p>\n\n\n\n

              Step 1: Find acceleration<\/h2>\n\n\n\n

              Using:
              v<\/mi>2<\/mn><\/msup>=<\/mo>u<\/mi>2<\/mn><\/msup>+<\/mo>2<\/mn>a<\/mi>s<\/mi><\/mrow>v^2=u^2+2as<\/annotation><\/semantics><\/math><\/p>\n\n\n\n

              Substitute values:
              0<\/mn>=<\/mo>(<\/mo>28<\/mn>)<\/mo>2<\/mn><\/msup>+<\/mo>2<\/mn>(<\/mo>a<\/mi>)<\/mo>(<\/mo>98<\/mn>)<\/mo><\/mrow>0=(28)^2+2(a)(98)<\/annotation><\/semantics><\/math><\/p>\n\n\n\n

              a<\/mi>=<\/mo>\u2212<\/mo>4<\/mn>\u2009<\/mtext>m<\/mi>\/<\/mi>s<\/mi>2<\/mn><\/msup><\/mrow>a=-4\\,m\/s^2<\/annotation><\/semantics><\/math><\/p>\n\n\n\n

              Step 2: Find time<\/h2>\n\n\n\n

              Using:
              v<\/mi>=<\/mo>u<\/mi>+<\/mo>a<\/mi>t<\/mi><\/mrow>v=u+at<\/annotation><\/semantics><\/math><\/p>\n\n\n\n

              Substitute:
              0<\/mn>=<\/mo>28<\/mn>\u2212<\/mo>4<\/mn>t<\/mi><\/mrow>0=28-4t<\/annotation><\/semantics><\/math><\/p>\n\n\n\n

              t<\/mi>=<\/mo>7<\/mn>\u2009<\/mtext>s<\/mi><\/mrow>t=7\\,s<\/annotation><\/semantics><\/math><\/p>\n\n\n\n

              Answer<\/h3>\n\n\n\n
                \n
              • Acceleration = \u20134 m\/s\u00b2<\/li>\n\n\n\n
              • Time taken = 7 s<\/li>\n<\/ul>\n\n\n\n
                \n\n\n\n

                Question 6<\/h1>\n\n\n\n

                Following Figure shows a position-time graph of two objects A and B that are
                moving along the parallel tracks in the same direction. Do objects
                A and B ever have equal velocity? Justify your answer<\/p>\n\n\n\n

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

                Equal velocity of objects A and B<\/h3>\n\n\n\n

                Answer<\/h3>\n\n\n\n

                No.<\/p>\n\n\n\n

                Explanation<\/h3>\n\n\n\n

                The slopes of both position-time graphs are different. Different slopes indicate different velocities.<\/p>\n\n\n\n

                \n\n\n\n

                Question 7<\/h1>\n\n\n\n

                A graph in Fig. 4.28 shows the change in position with time for two
                objects A and B moving in a straight line from 0 to 10 seconds. Choose
                the correct option(s).
                (i) The average velocity of both over the 10 s time interval is equal
                since they have the same initial and final positions.
                (ii) The average speeds of both over the 10 s time interval are equal
                since both cover equal distance in equal time.
                (iii) The average speed of A over the 10 s time interval is lower than
                that of B since it covers a shorter distance than B in 10 seconds.
                (iv) The average speed of A over the 10 s time interval is greater than
                that of B since B\u2019s speed is lower than A\u2019s in some segments.<\/p>\n\n\n\n

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

                Correct options from graph<\/h3>\n\n\n\n

                Answer<\/h3>\n\n\n\n

                Correct options:<\/p>\n\n\n\n

                  \n
                • (i)<\/li>\n\n\n\n
                • (iii)<\/li>\n<\/ul>\n\n\n\n

                  Explanation<\/h3>\n\n\n\n

                  Average velocity depends on displacement, while average speed depends on total distance travelled.<\/p>\n\n\n\n

                  \n\n\n\n

                  Question 8<\/h1>\n\n\n\n

                  A truck driver driving at the speed of 54 km h\u20131<\/sup> notices a road sign
                  with a speed limit of 40 km h\u20131<\/sup> (Fig. 4.29) for trucks. He slows down
                  to 36 km h\u20131<\/sup> in 36 s. What was the distance travelled by him during this
                  time? Assume the acceleration to be constant while slowing down.<\/p>\n\n\n\n

                  Truck slowing down<\/h3>\n\n\n\n

                  Given<\/h3>\n\n\n\n

                  Initial velocity:
                  54 km\/h = 15 m\/s<\/p>\n\n\n\n

                  Final velocity:
                  36 km\/h = 10 m\/s<\/p>\n\n\n\n

                  Time:
                  36 s<\/p>\n\n\n\n

                  Distance<\/h2>\n\n\n\n

                  Using:
                  s<\/mi>=<\/mo>(<\/mo>u<\/mi>+<\/mo>v<\/mi>)<\/mo><\/mrow>2<\/mn><\/mfrac>t<\/mi><\/mrow>s=\\frac{(u+v)}{2}t<\/annotation><\/semantics><\/math><\/p>\n\n\n\n

                  Substitute:
                  s<\/mi>=<\/mo>(<\/mo>15<\/mn>+<\/mo>10<\/mn>)<\/mo><\/mrow>2<\/mn><\/mfrac>\u00d7<\/mo>36<\/mn>=<\/mo>450<\/mn>\u2009<\/mtext>m<\/mi><\/mrow>s=\\frac{(15+10)}{2}\\times36=450\\,m<\/annotation><\/semantics><\/math><\/p>\n\n\n\n

                  Answer<\/h3>\n\n\n\n

                  Distance travelled = 450 m<\/p>\n\n\n\n

                  \n\n\n\n

                  Question 9<\/h1>\n\n\n\n

                  A car starts from rest and accelerates uniformly to 20 m s\u20131 in
                  5 seconds. It then travels at 20 m s\u20131<\/sup> for 10 seconds and finally
                  applies the brake (with uniform acceleration) to stop in 6 seconds.
                  Find the total distance travelled.<\/p>\n\n\n\n

                  Total distance travelled by car<\/h3>\n\n\n\n

                  Part 1: Acceleration phase<\/h2>\n\n\n\n

                  u<\/mi>=<\/mo>0<\/mn>,<\/mo> <\/mtext>v<\/mi>=<\/mo>20<\/mn>\u2009<\/mtext>m<\/mi>\/<\/mi>s<\/mi>,<\/mo> <\/mtext>t<\/mi>=<\/mo>5<\/mn>\u2009<\/mtext>s<\/mi><\/mrow>u=0,\\ v=20\\,m\/s,\\ t=5\\,s<\/annotation><\/semantics><\/math><\/p>\n\n\n\n

                  Distance:
                  s<\/mi>1<\/mn><\/msub>=<\/mo>(<\/mo>0<\/mn>+<\/mo>20<\/mn>)<\/mo><\/mrow>2<\/mn><\/mfrac>\u00d7<\/mo>5<\/mn>=<\/mo>50<\/mn>\u2009<\/mtext>m<\/mi><\/mrow>s_1=\\frac{(0+20)}{2}\\times5=50\\,m<\/annotation><\/semantics><\/math><\/p>\n\n\n\n

                  Part 2: Constant velocity<\/h2>\n\n\n\n

                  s<\/mi>2<\/mn><\/msub>=<\/mo>20<\/mn>\u00d7<\/mo>10<\/mn>=<\/mo>200<\/mn>\u2009<\/mtext>m<\/mi><\/mrow>s_2=20\\times10=200\\,m<\/annotation><\/semantics><\/math><\/p>\n\n\n\n

                  Part 3: Braking<\/h2>\n\n\n\n

                  s<\/mi>3<\/mn><\/msub>=<\/mo>(<\/mo>20<\/mn>+<\/mo>0<\/mn>)<\/mo><\/mrow>2<\/mn><\/mfrac>\u00d7<\/mo>6<\/mn>=<\/mo>60<\/mn>\u2009<\/mtext>m<\/mi><\/mrow>s_3=\\frac{(20+0)}{2}\\times6=60\\,m<\/annotation><\/semantics><\/math><\/p>\n\n\n\n

                  Total Distance<\/h2>\n\n\n\n

                  s<\/mi>=<\/mo>50<\/mn>+<\/mo>200<\/mn>+<\/mo>60<\/mn>=<\/mo>310<\/mn>\u2009<\/mtext>m<\/mi><\/mrow>s=50+200+60=310\\,m<\/annotation><\/semantics><\/math><\/p>\n\n\n\n

                  Answer<\/h3>\n\n\n\n

                  Total distance travelled = 310 m<\/p>\n\n\n\n

                  \n\n\n\n

                  Question 10<\/h1>\n\n\n\n

                  A bus is travelling at 36 km h\u20131<\/sup> when the driver sees an obstacle
                  30 m ahead. The driver takes 0.5 seconds to react before pressing the
                  brake. Once the brake is applied, the velocity of the bus reduces with
                  constant acceleration of 2.5 m s\u20132<\/sup>. Will the bus be able to stop before
                  reaching the obstacle?<\/p>\n\n\n\n

                  Will bus stop before obstacle?<\/h3>\n\n\n\n

                  Given<\/h3>\n\n\n\n

                  Initial velocity:
                  36 km\/h = 10 m\/s<\/p>\n\n\n\n

                  Reaction time:
                  0.5 s<\/p>\n\n\n\n

                  Distance during reaction time<\/h2>\n\n\n\n

                  s<\/mi>=<\/mo>v<\/mi>t<\/mi>=<\/mo>10<\/mn>\u00d7<\/mo>0.5<\/mn>=<\/mo>5<\/mn>\u2009<\/mtext>m<\/mi><\/mrow>s=vt=10\\times0.5=5\\,m<\/annotation><\/semantics><\/math><\/p>\n\n\n\n

                  Braking distance<\/h2>\n\n\n\n

                  Using:
                  v<\/mi>2<\/mn><\/msup>=<\/mo>u<\/mi>2<\/mn><\/msup>+<\/mo>2<\/mn>a<\/mi>s<\/mi><\/mrow>v^2=u^2+2as<\/annotation><\/semantics><\/math><\/p>\n\n\n\n

                  Substitute:
                  0<\/mn>=<\/mo>(<\/mo>10<\/mn>)<\/mo>2<\/mn><\/msup>+<\/mo>2<\/mn>(<\/mo>\u2212<\/mo>2.5<\/mn>)<\/mo>s<\/mi><\/mrow>0=(10)^2+2(-2.5)s<\/annotation><\/semantics><\/math><\/p>\n\n\n\n

                  s<\/mi>=<\/mo>20<\/mn>\u2009<\/mtext>m<\/mi><\/mrow>s=20\\,m<\/annotation><\/semantics><\/math><\/p>\n\n\n\n

                  Total stopping distance<\/h2>\n\n\n\n

                  5<\/mn>+<\/mo>20<\/mn>=<\/mo>25<\/mn>\u2009<\/mtext>m<\/mi><\/mrow>5+20=25\\,m<\/annotation><\/semantics><\/math><\/p>\n\n\n\n

                  Answer<\/h3>\n\n\n\n

                  Yes, the bus stops before obstacle because stopping distance is 25 m which is less than 30 m.<\/p>\n\n\n\n

                  Question 11 <\/h2>\n\n\n\n

                  A student said, \u201cThe Earth moves around the Sun\u201d. In this context,
                  discuss whether an object kept on the Earth can be considered to be
                  at rest.<\/p>\n\n\n\n

                  Solution<\/p>\n\n\n\n

                  Yes, an object kept on Earth can be considered at rest with respect to Earth<\/strong>, because it does not change its position relative to Earth.<\/p>\n\n\n\n

                  But with respect to the Sun, the object is in motion<\/strong>, because Earth itself moves around the Sun.<\/p>\n\n\n\n

                  Conclusion:<\/strong> Rest and motion are relative terms. An object may be at rest for one observer and in motion for another.<\/p>\n\n\n\n

                  \n\n\n\n

                  Question 12 <\/h2>\n\n\n\n

                  The velocity-time graph from 0 s to 120 s for a cyclist is shown in Following Figure
                  Shade the areas (in different colours) representing the displacement of
                  the cyclist
                  (i) while cyclist is moving with constant velocity.
                  (ii) when the velocity of cyclist is decreasing.
                  Also, calculate the displacement and average acceleration in the 120 s
                  time interval<\/p>\n\n\n\n

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

                  Solution<\/p>\n\n\n\n

                  (i) Constant velocity<\/h3>\n\n\n\n

                  From 20 s to 100 s<\/strong>, velocity is constant.<\/p>\n\n\n\n

                  (ii) Decreasing velocity<\/h3>\n\n\n\n

                  From 100 s to 120 s<\/strong>, velocity is decreasing.<\/p>\n\n\n\n

                  Displacement<\/h3>\n\n\n\n

                  Area under velocity-time graph:<\/p>\n\n\n\n

                  From 0 to 20 s:
                  = 1\/2 \u00d7 20 \u00d7 3 = 30 m<\/strong><\/p>\n\n\n\n

                  From 20 to 100 s:
                  = 80 \u00d7 3 = 240 m<\/strong><\/p>\n\n\n\n

                  From 100 to 120 s:
                  = 1\/2 \u00d7 (3 + 2) \u00d7 20 = 50 m<\/strong><\/p>\n\n\n\n

                  Total displacement:
                  = 30 + 240 + 50 = 320 m<\/strong><\/p>\n\n\n\n

                  Average acceleration<\/h3>\n\n\n\n

                  Initial velocity = 0 m\/s
                  Final velocity = 2 m\/s
                  Time = 120 s<\/p>\n\n\n\n

                  Average acceleration = (2 – 0) \/ 120
                  = 1\/60 m\/s\u00b2<\/strong>
                  = 0.0167 m\/s\u00b2<\/strong><\/p>\n\n\n\n

                  \n\n\n\n

                  Question 13 <\/h2>\n\n\n\n

                  A girl is preparing for her first marathon by running on a straight
                  road. She uses a smartwatch to calculate her running speed at
                  different intervals. The graph (Shown in Following Figure) depicts her velocity versus
                  time. Estimate the distance she ran based on the graph.<\/p>\n\n\n\n

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

                  Solution<\/p>\n\n\n\n

                  Distance = area under velocity-time graph.<\/p>\n\n\n\n

                  Approximate area from graph:<\/p>\n\n\n\n

                  0 to 2 h:
                  = 1\/2 \u00d7 (7 + 7.5) \u00d7 2 = 14.5 km<\/strong><\/p>\n\n\n\n

                  2 to 4 h:
                  = 1\/2 \u00d7 (7.5 + 7) \u00d7 2 = 14.5 km<\/strong><\/p>\n\n\n\n

                  4 to 6 h:
                  = 1\/2 \u00d7 (7 + 5.5) \u00d7 2 = 12.5 km<\/strong><\/p>\n\n\n\n

                  6 to 7 h:
                  = 1\/2 \u00d7 (5.5 + 4.5) \u00d7 1 = 5 km<\/strong><\/p>\n\n\n\n

                  Total distance \u2248 46.5 km<\/strong><\/p>\n\n\n\n

                  So, the girl runs approximately 46 km to 47 km<\/strong>.<\/p>\n\n\n\n

                  \n\n\n\n

                  Question 14 <\/h2>\n\n\n\n

                  On entering a state highway, a car continues to move with a constant
                  velocity of 6 m s\u20131<\/sup> for 2 minutes and then accelerates with a constant
                  acceleration 1 m s\u20132<\/sup> for 6 seconds. Find the displacement of the car
                  on the state highway in the 2 min 6 s time interval by drawing a
                  velocity-time graph for its motion.<\/p>\n\n\n\n

                  Solution<\/p>\n\n\n\n

                  Given:
                  Initial velocity = 6 m\/s
                  Time at constant velocity = 2 min = 120 s<\/p>\n\n\n\n

                  Distance in first 120 s:
                  = 6 \u00d7 120 = 720 m<\/strong><\/p>\n\n\n\n

                  Now acceleration = 1 m\/s\u00b2
                  Time = 6 s<\/p>\n\n\n\n

                  Distance in next 6 s:
                  s = ut + 1\/2 at\u00b2
                  = 6 \u00d7 6 + 1\/2 \u00d7 1 \u00d7 6\u00b2
                  = 36 + 18
                  = 54 m<\/strong><\/p>\n\n\n\n

                  Total displacement:
                  = 720 + 54
                  = 774 m<\/strong><\/p>\n\n\n\n

                  Velocity-time graph<\/h3>\n\n\n\n