{"id":4139,"date":"2016-05-06T11:49:00","date_gmt":"2016-05-06T11:49:00","guid":{"rendered":"http:\/\/mycbseguide.com\/blog\/ncert-solutions-for-class-9-science-motion-part-2\/"},"modified":"2018-06-07T14:59:55","modified_gmt":"2018-06-07T09:29:55","slug":"ncert-solutions-for-class-9-science-motion-part-2","status":"publish","type":"post","link":"https:\/\/mycbseguide.com\/blog\/ncert-solutions-for-class-9-science-motion-part-2\/","title":{"rendered":"NCERT Solutions for Class 9 Science Motion part 2"},"content":{"rendered":"<div id=\"ez-toc-container\" class=\"ez-toc-v2_0_76 counter-hierarchy ez-toc-counter ez-toc-grey ez-toc-container-direction\">\n<div class=\"ez-toc-title-container\">\n<p class=\"ez-toc-title\" style=\"cursor:inherit\">Table of Contents<\/p>\n<span class=\"ez-toc-title-toggle\"><a href=\"#\" class=\"ez-toc-pull-right ez-toc-btn ez-toc-btn-xs ez-toc-btn-default ez-toc-toggle\" aria-label=\"Toggle Table of Content\"><span class=\"ez-toc-js-icon-con\"><span class=\"\"><span class=\"eztoc-hide\" style=\"display:none;\">Toggle<\/span><span class=\"ez-toc-icon-toggle-span\"><svg style=\"fill: #999;color:#999\" xmlns=\"http:\/\/www.w3.org\/2000\/svg\" class=\"list-377408\" width=\"20px\" height=\"20px\" viewBox=\"0 0 24 24\" fill=\"none\"><path d=\"M6 6H4v2h2V6zm14 0H8v2h12V6zM4 11h2v2H4v-2zm16 0H8v2h12v-2zM4 16h2v2H4v-2zm16 0H8v2h12v-2z\" fill=\"currentColor\"><\/path><\/svg><svg style=\"fill: #999;color:#999\" class=\"arrow-unsorted-368013\" xmlns=\"http:\/\/www.w3.org\/2000\/svg\" width=\"10px\" height=\"10px\" viewBox=\"0 0 24 24\" version=\"1.2\" baseProfile=\"tiny\"><path d=\"M18.2 9.3l-6.2-6.3-6.2 6.3c-.2.2-.3.4-.3.7s.1.5.3.7c.2.2.4.3.7.3h11c.3 0 .5-.1.7-.3.2-.2.3-.5.3-.7s-.1-.5-.3-.7zM5.8 14.7l6.2 6.3 6.2-6.3c.2-.2.3-.5.3-.7s-.1-.5-.3-.7c-.2-.2-.4-.3-.7-.3h-11c-.3 0-.5.1-.7.3-.2.2-.3.5-.3.7s.1.5.3.7z\"\/><\/svg><\/span><\/span><\/span><\/a><\/span><\/div>\n<nav><ul class='ez-toc-list ez-toc-list-level-1 eztoc-toggle-hide-by-default' ><li class='ez-toc-page-1 ez-toc-heading-level-2'><a class=\"ez-toc-link ez-toc-heading-1\" href=\"https:\/\/mycbseguide.com\/blog\/ncert-solutions-for-class-9-science-motion-part-2\/#NCERT_Class_9_Science_Chapter_wise_Solutions\" >NCERT Class 9 Science Chapter wise Solutions<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-2'><a class=\"ez-toc-link ez-toc-heading-2\" href=\"https:\/\/mycbseguide.com\/blog\/ncert-solutions-for-class-9-science-motion-part-2\/#NCERT_Solutions_for_Class_9_Science_Motion_part_2\" >NCERT Solutions for Class 9 Science Motion part 2<\/a><ul class='ez-toc-list-level-6' ><li class='ez-toc-heading-level-6'><ul class='ez-toc-list-level-6' ><li class='ez-toc-heading-level-6'><ul class='ez-toc-list-level-6' ><li class='ez-toc-heading-level-6'><ul class='ez-toc-list-level-6' ><li class='ez-toc-heading-level-6'><a class=\"ez-toc-link ez-toc-heading-3\" href=\"https:\/\/mycbseguide.com\/blog\/ncert-solutions-for-class-9-science-motion-part-2\/#1_An_athlete_completes_one_round_of_a_circular_track_of_diameter_200_m_in_40_s_What_will_be_the_distance_covered_and_the_displacement_at_the_end_of_2_minutes_20_s\" >1. An athlete completes one round of a circular track of diameter 200 m in 40 s. What will be the distance covered and the displacement at the end of 2 minutes 20 s?<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-6'><a class=\"ez-toc-link ez-toc-heading-4\" href=\"https:\/\/mycbseguide.com\/blog\/ncert-solutions-for-class-9-science-motion-part-2\/#2_Joseph_jogs_from_one_end_A_to_the_other_end_B_of_a_straight_300_m_road_in_2_minutes_50_seconds_and_then_turns_around_and_jogs_100_m_back_to_point_C_in_another_1_minute_What_are_Josephs_average_speeds_and_velocities_in_jogging_a_from_A_to_B_and_b_from_A_to_C\" >2. Joseph jogs from one end A to the other end B of a straight 300 m road in 2 minutes 50 seconds and then turns around and jogs 100 m back to point C in another 1 minute. What are Joseph\u2019s average speeds and velocities in jogging (a) from A to B and (b) from A to C?<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-6'><a class=\"ez-toc-link ez-toc-heading-5\" href=\"https:\/\/mycbseguide.com\/blog\/ncert-solutions-for-class-9-science-motion-part-2\/#3_Abdul_while_driving_to_school_computes_the_average_speed_for_his_trip_to_be_On_his_return_trip_along_the_same_route_there_is_less_traffic_and_the_average_speed_is_What_is_the_average_speed_for_Abduls_trip\" >3. Abdul, while driving to school, computes the average speed for his trip to be. On his return trip along the same route, there is less traffic and the average speed is. What is the average speed for Abdul\u2019s trip?<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-6'><a class=\"ez-toc-link ez-toc-heading-6\" href=\"https:\/\/mycbseguide.com\/blog\/ncert-solutions-for-class-9-science-motion-part-2\/#4_A_motorboat_starting_from_rest_on_a_lake_accelerates_in_a_straight_line_at_a_constant_rate_of_for_80_s_How_far_does_the_boat_travel_during_this_time\" >4. A motorboat starting from rest on a lake accelerates in a straight line at a constant rate of for 8.0 s. How far does the boat travel during this time?<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-6'><a class=\"ez-toc-link ez-toc-heading-7\" href=\"https:\/\/mycbseguide.com\/blog\/ncert-solutions-for-class-9-science-motion-part-2\/#5_A_driver_of_a_car_travelling_at_applies_the_brakes_and_accelerates_uniformly_in_the_opposite_direction_The_car_stops_in_5_s_Another_driver_going_at_in_another_car_applies_his_brakes_slowly_and_stops_in_10_s_On_the_same_graph_paper_plot_the_speed_versus_time_graphs_for_the_two_cars_Which_of_the_two_cars_travelled_farther_after_the_brakes_were_applied\" >5. A driver of a car travelling at  applies the brakes and accelerates uniformly in the opposite direction. The car stops in 5 s. Another driver going at in another car applies his brakes slowly and stops in 10 s. On the same graph paper, plot the speed versus time graphs for the two cars. Which of the two cars travelled farther after the brakes were applied?<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-6'><a class=\"ez-toc-link ez-toc-heading-8\" href=\"https:\/\/mycbseguide.com\/blog\/ncert-solutions-for-class-9-science-motion-part-2\/#6_Fig_811_shows_the_distance-time_graph_of_three_objects_A_B_and_C_Study_the_graph_and_answer_the_following_questions\" >6. Fig 8.11 shows the distance-time graph of three objects A, B and C. Study the graph and answer the following questions:<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-6'><a class=\"ez-toc-link ez-toc-heading-9\" href=\"https:\/\/mycbseguide.com\/blog\/ncert-solutions-for-class-9-science-motion-part-2\/#7_A_ball_is_gently_dropped_from_a_height_of_20_m_If_its_velocity_increases_uniformly_at_the_rate_of_with_what_velocity_will_it_strike_the_ground_After_what_time_will_it_strike_the_ground\" >7. A ball is gently dropped from a height of 20 m. If its velocity increases uniformly at the rate of, with what velocity will it strike the ground? After what time will it strike the ground?<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-6'><a class=\"ez-toc-link ez-toc-heading-10\" href=\"https:\/\/mycbseguide.com\/blog\/ncert-solutions-for-class-9-science-motion-part-2\/#8_The_speed-time_graph_for_a_car_is_shown_is_Fig_812\" >8. The speed-time graph for a car is shown is Fig. 8.12.<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-6'><a class=\"ez-toc-link ez-toc-heading-11\" href=\"https:\/\/mycbseguide.com\/blog\/ncert-solutions-for-class-9-science-motion-part-2\/#9_State_which_of_the_following_situations_are_possible_and_give_an_example_for_each_of_these\" >9. State which of the following situations are possible and give an example for each of these:<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-6'><a class=\"ez-toc-link ez-toc-heading-12\" href=\"https:\/\/mycbseguide.com\/blog\/ncert-solutions-for-class-9-science-motion-part-2\/#10_An_artificial_satellite_is_moving_in_a_circular_orbit_of_radius_42250_km_Calculate_its_speed_if_it_takes_24_hours_to_revolve_around_the_earth\" >10. An artificial satellite is moving in a circular orbit of radius 42250 km. Calculate its speed if it takes 24 hours to revolve around the earth.<\/a><\/li><\/ul><\/li><\/ul><\/li><\/ul><\/li><\/ul><\/li><li class='ez-toc-page-1 ez-toc-heading-level-2'><a class=\"ez-toc-link ez-toc-heading-13\" href=\"https:\/\/mycbseguide.com\/blog\/ncert-solutions-for-class-9-science-motion-part-2\/#NCERT_Solutions_for_Class_9_Science\" >NCERT Solutions for Class 9 Science<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-2'><a class=\"ez-toc-link ez-toc-heading-14\" href=\"https:\/\/mycbseguide.com\/blog\/ncert-solutions-for-class-9-science-motion-part-2\/#CBSE_app_for_Class_9\" >CBSE app for Class 9<\/a><\/li><\/ul><\/nav><\/div>\n<p>NCERT Solutions for Class 9 Science Motion part 2 Class 9 Science book solutions are available in PDF format for free download. These ncert book chapter wise questions and answers are very helpful for CBSE exam. CBSE recommends NCERT books and most of the questions in CBSE exam are asked from NCERT text books. Class 9 Science chapter wise NCERT solution for Science Book for all the chapters can be downloaded from our website and myCBSEguide mobile app for free.<\/p>\n<p style=\"text-align: center;\"><strong>NCERT solutions for Science\u00a0Motion <a class=\"button\" href=\"https:\/\/mycbseguide.com\/downloads\/cbse-class-09-science\/1218\/ncert-solutions\/5\/\">Download as PDF<\/a><\/strong><\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter\" src=\"https:\/\/media-mycbseguide.s3.ap-south-1.amazonaws.com\/images\/blog\/09_Science_Book.jpg\" alt=\"NCERT Solutions for Class 9 Science Motion part 2\" width=\"212\" height=\"285\" \/><\/p>\n<h2><span class=\"ez-toc-section\" id=\"NCERT_Class_9_Science_Chapter_wise_Solutions\"><\/span><span class=\"underline-text\">NCERT Class 9 Science Chapter wise Solutions<\/span><span class=\"ez-toc-section-end\"><\/span><\/h2>\n<ul>\n<li>01 \u2013 \u00a0Matter in Our Surroundings<\/li>\n<li>02 \u2013 Is Mattter Around us Pure<\/li>\n<li>03 \u2013 Atoms and Molecules<\/li>\n<li>04 \u2013 Structure of the Atom<\/li>\n<li>05 \u2013 The Fundamental Unit of Life<\/li>\n<li>06 \u2013 Tissues<\/li>\n<li>07 \u2013 Diversity in Living Organisms<\/li>\n<li>08 \u2013 Motion<\/li>\n<li>09 \u2013 Force and Laws of Motion<\/li>\n<li>10 \u2013 Gravitation<\/li>\n<li>11 \u2013 Word and Energy<\/li>\n<li>12 \u2013 Sound<\/li>\n<li>13 \u2013 Why Do We Fall Ill<\/li>\n<li>14 \u2013 Natural Resources<\/li>\n<li>15 \u2013 Improvement in Food Resources<\/li>\n<\/ul>\n<h2><span class=\"ez-toc-section\" id=\"NCERT_Solutions_for_Class_9_Science_Motion_part_2\"><\/span>NCERT Solutions for Class 9 Science Motion part 2<span class=\"ez-toc-section-end\"><\/span><\/h2>\n<h6 style=\"text-align: justify;\"><span class=\"ez-toc-section\" id=\"1_An_athlete_completes_one_round_of_a_circular_track_of_diameter_200_m_in_40_s_What_will_be_the_distance_covered_and_the_displacement_at_the_end_of_2_minutes_20_s\"><\/span><strong>1. An athlete completes one round of a circular track of diameter 200 m in 40 <em>s<\/em>. What will be the distance covered and the displacement at the end of 2 minutes 20 s?<\/strong><span class=\"ez-toc-section-end\"><\/span><\/h6>\n<p style=\"text-align: justify;\"><strong>Ans. <\/strong>circumference of circular track = 2<img decoding=\"async\" style=\"height: 19px; width: 10px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/science\/ch08\/image058.png\" \/>r<\/p>\n<p style=\"text-align: justify;\"><img decoding=\"async\" style=\"height: 19px; width: 167px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/science\/ch08\/image059.png\" \/><\/p>\n<p style=\"text-align: justify;\"><img decoding=\"async\" style=\"height: 19px; width: 133px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/science\/ch08\/image060.png\" \/>= 4400\/7 m<\/p>\n<p style=\"text-align: justify;\">rounds completed by athlete in 2min20sec = s= 140\/40 = 3.5<\/p>\n<p style=\"text-align: justify;\">therefore, total distance covered <img decoding=\"async\" style=\"height: 19px; width: 105px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/science\/ch08\/image061.png\" \/>= 2200 m<\/p>\n<p style=\"text-align: justify;\"><img decoding=\"async\" id=\"Picture 27\" style=\"height: 200px; width: 200px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/science\/ch08\/image062.png\" alt=\"cbse-science-ix-chapter-8-motion-3.PNG\" \/><\/p>\n<p style=\"text-align: justify;\">Since one complete round of circular track needs 40s so he will complete 3 rounds in 2mins and in next 20s he can complete half round therefore displacement = diameter = 200m.<\/p>\n<hr \/>\n<h6 style=\"text-align: justify;\"><span class=\"ez-toc-section\" id=\"2_Joseph_jogs_from_one_end_A_to_the_other_end_B_of_a_straight_300_m_road_in_2_minutes_50_seconds_and_then_turns_around_and_jogs_100_m_back_to_point_C_in_another_1_minute_What_are_Josephs_average_speeds_and_velocities_in_jogging_a_from_A_to_B_and_b_from_A_to_C\"><\/span><strong>2. Joseph jogs from one end A to the other end B of a straight 300 m road in 2 minutes 50 seconds and then turns around and jogs 100 m back to point C in another 1 minute. What are Joseph\u2019s average speeds and velocities in jogging (a) from A to B and (b) from A to C?<\/strong><span class=\"ez-toc-section-end\"><\/span><\/h6>\n<p style=\"text-align: justify;\"><strong>Ans. <\/strong> <strong>(a)<\/strong> distance = 300m<\/p>\n<p style=\"text-align: justify;\">time = 2min30seconds = 150 seconds<\/p>\n<p style=\"text-align: justify;\">average speed from A to B = average velocity from A to B<\/p>\n<p style=\"text-align: justify;\">= 300m\/150s = 2m\/s<\/p>\n<p style=\"text-align: justify;\"><strong>(b) <\/strong>average speed from A to C = (300+100)m\/(150+60)sec<\/p>\n<p style=\"text-align: justify;\">= 400m\/210s = 1.90m\/s<\/p>\n<p style=\"text-align: justify;\">displacement from A to C = (300 \u2013 100)m =200m<\/p>\n<p style=\"text-align: justify;\">time =2min30sec + 1min = 210s<\/p>\n<p style=\"text-align: justify;\">velocity = displacement\/time = 200m\/210s = 0.95m\/s<\/p>\n<p style=\"text-align: justify;\"><img decoding=\"async\" id=\"Picture 30\" style=\"height: 138px; width: 267px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/science\/ch08\/image063.png\" alt=\"cbse-science-ix-chapter-8-motion-4.PNG\" \/><\/p>\n<hr \/>\n<p style=\"text-align: center;\">NCERT Solutions for Class 9 Science Motion part 2<\/p>\n<h6 style=\"text-align: justify;\"><span class=\"ez-toc-section\" id=\"3_Abdul_while_driving_to_school_computes_the_average_speed_for_his_trip_to_be_On_his_return_trip_along_the_same_route_there_is_less_traffic_and_the_average_speed_is_What_is_the_average_speed_for_Abduls_trip\"><\/span><strong>3. Abdul, while driving to school, computes the average speed for his trip to be<img decoding=\"async\" style=\"height: 22px; width: 72px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/science\/ch08\/image064.png\" \/>. On his return trip along the same route, there is less traffic and the average speed is<img decoding=\"async\" style=\"height: 22px; width: 71px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/science\/ch08\/image065.png\" \/>. What is the average speed for Abdul\u2019s trip?<\/strong><span class=\"ez-toc-section-end\"><\/span><\/h6>\n<p style=\"text-align: justify;\"><strong>Ans. <\/strong>If we suppose that distance from Abdul\u2019s home to school = x kms<\/p>\n<p style=\"text-align: justify;\">while driving to school :-<\/p>\n<p style=\"text-align: justify;\"><img decoding=\"async\" style=\"height: 24px; width: 125px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/science\/ch08\/image066.png\" \/>,<\/p>\n<p style=\"text-align: justify;\">velocity = displacement\/time<\/p>\n<p style=\"text-align: justify;\">20 = x\/t, or, t=x\/20 hr<\/p>\n<p style=\"text-align: justify;\">on his return trip :-<\/p>\n<p style=\"text-align: justify;\">speed = 40 km h<sup>\u20131 <\/sup>,<\/p>\n<p style=\"text-align: justify;\">40= x \/t\u2019<\/p>\n<p style=\"text-align: justify;\">or, t\u2019 =x\/40 hr<\/p>\n<p style=\"text-align: justify;\">total distance travelled = x + x = 2x<\/p>\n<p style=\"text-align: justify;\">total time = t + t\u2019 = x\/20 + x\/40 =(2x + x)\/40 = 3x\/40 hr<\/p>\n<p style=\"text-align: justify;\">average speed for Abdul\u2019s trip = 2x\/(3x\/40) = 80x\/3x = 26.67km\/hr<\/p>\n<hr \/>\n<p style=\"text-align: center;\">NCERT Solutions for Class 9 Science Motion part 2<\/p>\n<h6 style=\"text-align: justify;\"><span class=\"ez-toc-section\" id=\"4_A_motorboat_starting_from_rest_on_a_lake_accelerates_in_a_straight_line_at_a_constant_rate_of_for_80_s_How_far_does_the_boat_travel_during_this_time\"><\/span><strong>4. A motorboat starting from rest on a lake accelerates in a straight line at a constant rate of <img decoding=\"async\" style=\"height: 22px; width: 64px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/science\/ch08\/image067.png\" \/>for 8.0 s. How far does the boat travel during this time?<\/strong><span class=\"ez-toc-section-end\"><\/span><\/h6>\n<p style=\"text-align: justify;\"><strong>Ans. <\/strong>since the motorboat starts from rest so u= 0<\/p>\n<p style=\"text-align: justify;\">time (t) = 8s, <img decoding=\"async\" style=\"height: 22px; width: 82px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/science\/ch08\/image068.png\" \/><\/p>\n<p style=\"text-align: justify;\">distance<img decoding=\"async\" style=\"height: 26px; width: 122px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/science\/ch08\/image069.png\" \/> <img decoding=\"async\" style=\"height: 22px; width: 95px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/science\/ch08\/image070.png\" \/>= 96m<\/p>\n<hr \/>\n<h6 style=\"text-align: justify;\"><span class=\"ez-toc-section\" id=\"5_A_driver_of_a_car_travelling_at_applies_the_brakes_and_accelerates_uniformly_in_the_opposite_direction_The_car_stops_in_5_s_Another_driver_going_at_in_another_car_applies_his_brakes_slowly_and_stops_in_10_s_On_the_same_graph_paper_plot_the_speed_versus_time_graphs_for_the_two_cars_Which_of_the_two_cars_travelled_farther_after_the_brakes_were_applied\"><\/span><strong>5. A driver of a car travelling at <img decoding=\"async\" style=\"height: 22px; width: 71px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/science\/ch08\/image071.png\" \/> applies the brakes and accelerates uniformly in the opposite direction. The car stops in 5 s. Another driver going at <img decoding=\"async\" style=\"height: 22px; width: 64px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/science\/ch08\/image072.png\" \/>in another car applies his brakes slowly and stops in 10 s. On the same graph paper, plot the speed versus time graphs for the two cars. Which of the two cars travelled farther after the brakes were applied?<\/strong><span class=\"ez-toc-section-end\"><\/span><\/h6>\n<p style=\"text-align: justify;\"><strong>Ans. <\/strong><\/p>\n<p style=\"text-align: justify;\"><img decoding=\"async\" id=\"Picture 33\" style=\"height: 190px; width: 289px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/science\/ch08\/image073.jpg\" alt=\"cbse-science-ix-chapter-8-motion-5.PNG\" \/><\/p>\n<p style=\"text-align: justify;\">As given in the figure below AB (in red line) and CD(in red line) are the Speed-time graph for given two cars with initial speeds <img decoding=\"async\" style=\"height: 22px; width: 147px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/science\/ch08\/image074.png\" \/>respectively.<\/p>\n<p style=\"text-align: justify;\">Distance Travelled by first car before coming to rest =Area of <img decoding=\"async\" style=\"height: 19px; width: 47px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/science\/ch08\/image075.png\" \/><\/p>\n<p style=\"text-align: justify;\"><img decoding=\"async\" style=\"height: 26px; width: 144px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/science\/ch08\/image076.png\" \/><\/p>\n<p style=\"text-align: justify;\"><img decoding=\"async\" style=\"height: 26px; width: 183px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/science\/ch08\/image077.png\" \/><\/p>\n<p style=\"text-align: justify;\">=<img decoding=\"async\" style=\"height: 26px; width: 203px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/science\/ch08\/image078.png\" \/><\/p>\n<p style=\"text-align: justify;\"><img decoding=\"async\" style=\"height: 26px; width: 183px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/science\/ch08\/image079.png\" \/><\/p>\n<p style=\"text-align: justify;\">= 325\/9 m<\/p>\n<p style=\"text-align: justify;\">= 36.11 m<\/p>\n<p style=\"text-align: justify;\">Distance Travelled by second car before coming to rest =Area of <img decoding=\"async\" style=\"height: 19px; width: 49px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/science\/ch08\/image080.png\" \/><\/p>\n<p style=\"text-align: justify;\"><img decoding=\"async\" style=\"height: 26px; width: 145px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/science\/ch08\/image081.png\" \/><\/p>\n<p style=\"text-align: justify;\"><img decoding=\"async\" style=\"height: 26px; width: 176px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/science\/ch08\/image082.png\" \/><\/p>\n<p style=\"text-align: justify;\">=<img decoding=\"async\" style=\"height: 26px; width: 189px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/science\/ch08\/image083.png\" \/>m<\/p>\n<p style=\"text-align: justify;\"><img decoding=\"async\" style=\"height: 26px; width: 175px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/science\/ch08\/image084.png\" \/><\/p>\n<p style=\"text-align: justify;\"><img decoding=\"async\" style=\"height: 26px; width: 105px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/science\/ch08\/image085.png\" \/><\/p>\n<p style=\"text-align: justify;\">= 25\/6 m<\/p>\n<p style=\"text-align: justify;\">= 4.16 m<\/p>\n<p style=\"text-align: justify;\">\u2234 Clearly the first car will travel farther (36.11 m) than the first car(4.16 m).<\/p>\n<hr \/>\n<p style=\"text-align: center;\">NCERT Solutions for Class 9 Science Motion part 2<\/p>\n<h6 style=\"text-align: justify;\"><span class=\"ez-toc-section\" id=\"6_Fig_811_shows_the_distance-time_graph_of_three_objects_A_B_and_C_Study_the_graph_and_answer_the_following_questions\"><\/span><strong>6. Fig 8.11 shows the distance-time graph of three objects A, B and C. Study the graph and answer the following questions: <\/strong><span class=\"ez-toc-section-end\"><\/span><\/h6>\n<p style=\"text-align: justify;\"><img decoding=\"async\" id=\"Picture 2\" style=\"height: 201px; width: 264px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/science\/ch08\/image086.png\" \/><\/p>\n<p style=\"text-align: justify;\"><strong>(a) Which of the three is travelling the fastest?<\/strong><\/p>\n<p style=\"text-align: justify;\"><strong>(b) Are all three ever at the same point on the road? <\/strong><\/p>\n<p style=\"text-align: justify;\"><strong>(c) How far has C travelled when B passes A?<\/strong><\/p>\n<p style=\"text-align: justify;\"><strong>(d) How far has B travelled by the time it passes C?<\/strong><\/p>\n<p style=\"text-align: justify;\"><strong>Ans. <\/strong><\/p>\n<p style=\"text-align: justify;\"><img decoding=\"async\" id=\"Picture 15\" style=\"height: 205px; width: 355px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/science\/ch08\/image087.jpg\" alt=\"cl9SciCh8ncertP6.jpg\" \/><\/p>\n<p style=\"text-align: justify;\"><strong>(a)<\/strong> It is clear from graph that B covers more distance in less time. Therefore, B is the fastest.<\/p>\n<p style=\"text-align: justify;\"><strong>(b)<\/strong> All of them never come at the same point at the same time.<\/p>\n<p style=\"text-align: justify;\"><strong>(c)<\/strong> According to graph; each small division shows about 0.57 km.<\/p>\n<p style=\"text-align: justify;\">A is passing B at point S which is in line with point P (on the distance axis) and shows about 9.14 km<\/p>\n<p style=\"text-align: justify;\">Thus, at this point C travels about<\/p>\n<p style=\"text-align: justify;\"><img decoding=\"async\" style=\"height: 26px; width: 139px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/science\/ch08\/image088.png\" \/>km = 9.14 km \u2013 2.1375 km = 7.0025 km <img decoding=\"async\" style=\"height: 19px; width: 53px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/science\/ch08\/image089.png\" \/><\/p>\n<p style=\"text-align: justify;\">Thus, when A passes B, C travels about 7 km.<\/p>\n<p style=\"text-align: justify;\"><strong>(d)<\/strong> B passes C at point Q at the distance axis which is <img decoding=\"async\" style=\"height: 19px; width: 145px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/science\/ch08\/image090.png\" \/><img decoding=\"async\" style=\"height: 19px; width: 68px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/science\/ch08\/image091.png\" \/><\/p>\n<p style=\"text-align: justify;\">Therefore, B travelled about 5.28 km when passes to C.<\/p>\n<hr \/>\n<p style=\"text-align: center;\">NCERT Solutions for Class 9 Science Motion part 2<\/p>\n<h6 style=\"text-align: justify;\"><span class=\"ez-toc-section\" id=\"7_A_ball_is_gently_dropped_from_a_height_of_20_m_If_its_velocity_increases_uniformly_at_the_rate_of_with_what_velocity_will_it_strike_the_ground_After_what_time_will_it_strike_the_ground\"><\/span><strong>7. A ball is gently dropped from a height of 20 m. If its velocity increases uniformly at the rate of<img decoding=\"async\" style=\"height: 22px; width: 60px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/science\/ch08\/image092.png\" \/>, with what velocity will it strike the ground? After what time will it strike the ground?<\/strong><span class=\"ez-toc-section-end\"><\/span><\/h6>\n<p style=\"text-align: justify;\"><strong>Ans. <\/strong>Let us assume, the final velocity with which ball will strike the ground be &#8216;v&#8217; and time it takes to strike the ground be &#8216;t&#8217;<\/p>\n<p style=\"text-align: justify;\">Initial Velocity of ball u=0<\/p>\n<p style=\"text-align: justify;\">Distance or height of fall s=20m<\/p>\n<p style=\"text-align: justify;\">Downward acceleration a<img decoding=\"async\" style=\"height: 22px; width: 70px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/science\/ch08\/image093.png\" \/><\/p>\n<p style=\"text-align: justify;\">As we know, <img decoding=\"async\" style=\"height: 22px; width: 92px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/science\/ch08\/image094.png\" \/><\/p>\n<p style=\"text-align: justify;\">Or, 2as = <img decoding=\"async\" style=\"height: 22px; width: 47px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/science\/ch08\/image095.png\" \/><\/p>\n<p style=\"text-align: justify;\"><img decoding=\"async\" style=\"height: 22px; width: 96px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/science\/ch08\/image096.png\" \/><\/p>\n<p style=\"text-align: justify;\"><img decoding=\"async\" style=\"height: 19px; width: 119px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/science\/ch08\/image097.png\" \/><\/p>\n<p style=\"text-align: justify;\"><img decoding=\"async\" style=\"height: 22px; width: 82px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/science\/ch08\/image098.png\" \/><\/p>\n<p style=\"text-align: justify;\"><img decoding=\"async\" style=\"height: 14px; width: 16px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/science\/ch08\/image099.png\" \/>Final velocity of ball, v<img decoding=\"async\" style=\"height: 22px; width: 71px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/science\/ch08\/image100.png\" \/><\/p>\n<p style=\"text-align: justify;\">t= (v-u)\/a<\/p>\n<p style=\"text-align: justify;\"><img decoding=\"async\" style=\"height: 14px; width: 16px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/science\/ch08\/image099.png\" \/>Time taken by the ball to strike= (20-0)\/10<\/p>\n<p style=\"text-align: justify;\">= 20\/10<\/p>\n<p style=\"text-align: justify;\">= 2 seconds<\/p>\n<hr \/>\n<p style=\"text-align: center;\">NCERT Solutions for Class 9 Science Motion part 2<\/p>\n<h6 style=\"text-align: justify;\"><span class=\"ez-toc-section\" id=\"8_The_speed-time_graph_for_a_car_is_shown_is_Fig_812\"><\/span><strong>8. The speed-time graph for a car is shown is Fig. 8.12.<\/strong><span class=\"ez-toc-section-end\"><\/span><\/h6>\n<p style=\"text-align: justify;\"><strong><img decoding=\"async\" id=\"Picture 18\" style=\"height: 147px; width: 366px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/science\/ch08\/image101.jpg\" alt=\"cl9ch8fig7.jpg\" \/><\/strong><\/p>\n<p style=\"text-align: justify;\"><strong><em>Fig. 8.12<\/em><\/strong><\/p>\n<p style=\"text-align: justify;\"><strong>(a) Find how far does the car travel in the first 4 seconds. Shade the area on the graph that represents the distance travelled by the car during the period.<\/strong><\/p>\n<p style=\"text-align: justify;\"><strong>(b) Which part of the graph represents uniform motion of the car?<\/strong><\/p>\n<p style=\"text-align: justify;\"><strong>Ans. <\/strong><\/p>\n<p style=\"text-align: justify;\"><img decoding=\"async\" id=\"Picture 46\" style=\"height: 167px; width: 291px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/science\/ch08\/image102.png\" alt=\"cbse-class-ix-science--chapter-8-q8.PNG\" \/><\/p>\n<p style=\"text-align: justify;\"><strong>(a)<\/strong> Distance travelled by car in the 4 second<\/p>\n<p style=\"text-align: justify;\">The area under the slope of the speed \u2013 time graph gives the distance travelled by an object.<\/p>\n<p style=\"text-align: justify;\">In the given graph<\/p>\n<p style=\"text-align: justify;\">56 full squares and 12 half squares come under the area slope for the time of 4 second.<\/p>\n<p style=\"text-align: justify;\">Total number of squares = 56 + 12\/2 = 62 squares<\/p>\n<p style=\"text-align: justify;\">The total area of the squares will give the distance travelled by the car in 4 second. on the time axis,<\/p>\n<p style=\"text-align: justify;\">5 squares = 2seconds, therefore 1 square = 2\/5 seconds<\/p>\n<p style=\"text-align: justify;\">on speed axis there are 3 squares = 2m\/s<\/p>\n<p style=\"text-align: justify;\">therefore, area of one square = <img decoding=\"async\" style=\"height: 19px; width: 105px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/science\/ch08\/image103.png\" \/>= 4\/15 m<\/p>\n<p style=\"text-align: justify;\">so area of 62 squares= <img decoding=\"async\" style=\"height: 19px; width: 77px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/science\/ch08\/image104.png\" \/> = 248\/15 m = 16.53 m<\/p>\n<p style=\"text-align: justify;\">Hence the car travels 16.53m in the first 4 seconds.<\/p>\n<p style=\"text-align: justify;\"><strong>(b)<\/strong> The straight line part of graph, from point A to point B represents a uniform motion of car.<\/p>\n<hr \/>\n<p style=\"text-align: center;\">NCERT Solutions for Class 9 Science Motion part 2<\/p>\n<h6 style=\"text-align: justify;\"><span class=\"ez-toc-section\" id=\"9_State_which_of_the_following_situations_are_possible_and_give_an_example_for_each_of_these\"><\/span><strong>9. State which of the following situations are possible and give an example for each of these:<\/strong><span class=\"ez-toc-section-end\"><\/span><\/h6>\n<p style=\"text-align: justify;\"><strong>(a) an object with a constant acceleration but with zero velocity<\/strong><\/p>\n<p style=\"text-align: justify;\"><strong>(b) an object moving in a certain direction with an acceleration in the perpendicular direction.<\/strong><\/p>\n<p style=\"text-align: justify;\"><strong>Ans. (a)<\/strong> An object with a constant acceleration can still have the zero velocity. For example an object which is at rest on the surface of earth will have zero velocity but still being acted upon by the gravitational force of earth with an acceleration of 9.81 ms<sup>-2<\/sup> towards the center of earth. Hence when an object starts falling freely can have contant acceleration but with zero velocity.<\/p>\n<p style=\"text-align: justify;\"><strong>(b)<\/strong> When an athlete moves with a velocity of constant magnitude along the circular path, the only change in his velocity is due to the change in the direction of motion. Here, the motion of the athlete moving along a circular path is, therefore, an example of an accelerated motion where acceleration is always perpendicular to direction of motion of an object at a given instance. Hence it is possible when an object moves on a circular path.<\/p>\n<hr \/>\n<p style=\"text-align: center;\">NCERT Solutions for Class 9 Science Motion part 2<\/p>\n<h6 style=\"text-align: justify;\"><span class=\"ez-toc-section\" id=\"10_An_artificial_satellite_is_moving_in_a_circular_orbit_of_radius_42250_km_Calculate_its_speed_if_it_takes_24_hours_to_revolve_around_the_earth\"><\/span><strong>10. An artificial satellite is moving in a circular orbit of radius 42250 km. Calculate its speed if it takes 24 hours to revolve around the earth.<\/strong><span class=\"ez-toc-section-end\"><\/span><\/h6>\n<p style=\"text-align: justify;\"><strong>Ans. <\/strong>Let us assume an artificial satellite, which is moving in a circular orbit of radius 42250 km covers a distance &#8216;s&#8217; as it revolve around earth with speed &#8216;v&#8217; in given time &#8216;t&#8217; of 24 hours.<\/p>\n<p style=\"text-align: justify;\">= 42250 km<\/p>\n<p style=\"text-align: justify;\">Radius of circular orbit r<img decoding=\"async\" style=\"height: 19px; width: 132px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/science\/ch08\/image105.png\" \/><\/p>\n<p style=\"text-align: justify;\">Time taken by artificial satellite t=24 hours<\/p>\n<p style=\"text-align: justify;\"><img decoding=\"async\" style=\"height: 19px; width: 128px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/science\/ch08\/image106.png\" \/><\/p>\n<p style=\"text-align: justify;\">Distance covered by satellite s =Circumference of circular orbit<\/p>\n<p style=\"text-align: justify;\"><img decoding=\"async\" style=\"height: 19px; width: 48px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/science\/ch08\/image107.png\" \/><\/p>\n<p style=\"text-align: justify;\"><img decoding=\"async\" style=\"height: 14px; width: 14px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/science\/ch08\/image108.png\" \/> Speed of satellite v<img decoding=\"async\" style=\"height: 26px; width: 76px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/science\/ch08\/image109.png\" \/><\/p>\n<p style=\"text-align: justify;\"><img decoding=\"async\" style=\"height: 52px; width: 198px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/science\/ch08\/image110.png\" \/><\/p>\n<p style=\"text-align: justify;\"><img decoding=\"async\" style=\"height: 49px; width: 167px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/science\/ch08\/image111.png\" \/>]\n<p style=\"text-align: justify;\">=<img decoding=\"async\" style=\"height: 22px; width: 95px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/science\/ch08\/image112.png\" \/><\/p>\n<p style=\"text-align: justify;\">= 3.073 km\/s<\/p>\n<h2><span class=\"ez-toc-section\" id=\"NCERT_Solutions_for_Class_9_Science\"><\/span>NCERT Solutions for Class 9 Science<span class=\"ez-toc-section-end\"><\/span><\/h2>\n<p>NCERT Solutions Class 9 Science PDF (Download) Free from myCBSEguide app and myCBSEguide website. Ncert solution class 9 Science includes text book solutions .NCERT Solutions for CBSE Class 9 Science have total 15 chapters. 9 Science NCERT Solutions in PDF for free Download on our website. Ncert Science class 9 solutions PDF and Science ncert class 9 PDF solutions with latest modifications and as per the latest CBSE syllabus are only available in myCBSEguide.<\/p>\n<h2><span class=\"ez-toc-section\" id=\"CBSE_app_for_Class_9\"><\/span>CBSE app for Class 9<span class=\"ez-toc-section-end\"><\/span><\/h2>\n<p>To download\u00a0NCERT Solutions for class 9 Science, Computer Science, Home Science,Hindi ,English, Maths Social Science do check myCBSEguide app or website. myCBSEguide provides sample papers with solution, test papers for chapter-wise practice, NCERT solutions, NCERT Exemplar solutions, quick revision notes for ready reference, CBSE guess papers and CBSE important question papers. Sample Paper all are made available through\u00a0<a href=\"https:\/\/play.google.com\/store\/apps\/details?id=in.techchefs.MyCBSEGuide&amp;referrer=utm_source%3Dmycbse_bottom%26utm_medium%3Dtext%26utm_campaign%3Dmycbseads\"><strong>the best app for CBSE students<\/strong><\/a>\u00a0and myCBSEguide website.<\/p>\n","protected":false},"excerpt":{"rendered":"<p>NCERT Solutions for Class 9 Science Motion part 2 Class 9 Science book solutions are available in PDF format for free download. These ncert book chapter wise questions and answers are very helpful for CBSE exam. CBSE recommends NCERT books and most of the questions in CBSE exam are asked from NCERT text books. 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