{"id":31515,"date":"2022-01-25T21:54:43","date_gmt":"2022-01-25T16:24:43","guid":{"rendered":"https:\/\/infinitylearn.com\/surge\/?p=31515"},"modified":"2023-03-10T15:09:03","modified_gmt":"2023-03-10T09:39:03","slug":"extra-questions-class-9-science-chapter-8-motion","status":"publish","type":"post","link":"https:\/\/infinitylearn.com\/surge\/study-materials\/extra-questions\/class-9\/science-chapter-8-motion\/","title":{"rendered":"Extra Questions &#8211; Class 9 Science Chapter 8 Motion"},"content":{"rendered":"<div id=\"ez-toc-container\" class=\"ez-toc-v2_0_37 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\">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\" style=\"display: none;\"><label for=\"item\" aria-label=\"Table of Content\"><span style=\"display: flex;align-items: center;width: 35px;height: 30px;justify-content: center;\"><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><\/label><input type=\"checkbox\" id=\"item\"><\/a><\/span><\/div>\n<nav><ul class='ez-toc-list ez-toc-list-level-1' style='display:block'><li class='ez-toc-page-1 ez-toc-heading-level-2'><a class=\"ez-toc-link ez-toc-heading-1\" href=\"https:\/\/infinitylearn.com\/surge\/study-materials\/extra-questions\/class-9\/science-chapter-8-motion\/#Chapter_8_-_Motion_Extra_Questions\" title=\"Chapter 8 &#8211; Motion Extra Questions \">Chapter 8 &#8211; Motion Extra Questions <\/a><ul class='ez-toc-list-level-3'><li class='ez-toc-heading-level-3'><a class=\"ez-toc-link ez-toc-heading-2\" href=\"https:\/\/infinitylearn.com\/surge\/study-materials\/extra-questions\/class-9\/science-chapter-8-motion\/#Very_Short_Answer_Questions\" title=\"Very Short Answer Questions\">Very Short Answer Questions<\/a><ul class='ez-toc-list-level-4'><li class='ez-toc-heading-level-4'><a class=\"ez-toc-link ez-toc-heading-3\" href=\"https:\/\/infinitylearn.com\/surge\/study-materials\/extra-questions\/class-9\/science-chapter-8-motion\/#Short_Answer_Questions-I\" title=\"Short Answer Questions-I\">Short Answer Questions-I<\/a><ul class='ez-toc-list-level-5'><li class='ez-toc-heading-level-5'><a class=\"ez-toc-link ez-toc-heading-4\" href=\"https:\/\/infinitylearn.com\/surge\/study-materials\/extra-questions\/class-9\/science-chapter-8-motion\/#Short_Answer_Questions-II\" title=\"Short Answer Questions-II\">Short Answer Questions-II<\/a><\/li><\/ul><\/li><\/ul><\/li><li class='ez-toc-page-1 ez-toc-heading-level-3'><a class=\"ez-toc-link ez-toc-heading-5\" href=\"https:\/\/infinitylearn.com\/surge\/study-materials\/extra-questions\/class-9\/science-chapter-8-motion\/#Long_Answer_Questions\" title=\"Long Answer Questions\">Long Answer Questions<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-3'><a class=\"ez-toc-link ez-toc-heading-6\" href=\"https:\/\/infinitylearn.com\/surge\/study-materials\/extra-questions\/class-9\/science-chapter-8-motion\/#Numericals\" title=\"Numericals\">Numericals<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-3'><a class=\"ez-toc-link ez-toc-heading-7\" href=\"https:\/\/infinitylearn.com\/surge\/study-materials\/extra-questions\/class-9\/science-chapter-8-motion\/#HOTS_Higher_Order_Thinking_Skills\" title=\"HOTS (Higher Order Thinking Skills)\">HOTS (Higher Order Thinking Skills)<\/a><\/li><\/ul><\/li><\/ul><\/nav><\/div>\n<h2><span style=\"color: #0000ff;\"><strong>Chapter 8 &#8211; Motion Extra Questions <\/strong><\/span><\/h2>\n<p><span style=\"color: #800080;\"><strong>Get Extra Questions for Class 9 Science on Infinity Learn for free.<\/strong><\/span><\/p>\n<h3><span class=\"ez-toc-section\" id=\"Very_Short_Answer_Questions\"><\/span>Very Short Answer Questions<span class=\"ez-toc-section-end\"><\/span><\/h3>\n<p>Question 1.<br \/>\nThe phenomenon of motion was placed on a sound scientific footing by two scientists. Write their names.<br \/>\nAnswer:<br \/>\nGalileo Galilei and Isaac Newton.<\/p>\n<p>Question 2.<br \/>\nAre rest and motion absolute or relative terms?<br \/>\nAnswer:<br \/>\nThey are relative terms.<\/p>\n<p>Question 3.<br \/>\nSuppose a ball is thrown vertically upwards from a position P above the ground. It rises to the highest point Q and returns to the same point P. What is the net displacement and distance travelled by the ball?<br \/>\nAnswer:<br \/>\nDisplacement is zero. Distance is twice the distance between position P and Q.<\/p>\n<p>Question 4.<br \/>\nWhich speed is greater: 30 m\/s or 30 km\/h?<br \/>\nAnswer:<br \/>\n30 m\/s<\/p>\n<p>Question 5.<br \/>\nWhat do you mean by 2 m\/s<sup>2<\/sup>?<br \/>\nAnswer:<br \/>\nThe velocity of the body increases by 2 m\/s after every second.<\/p>\n<p>Question 6.<br \/>\nCan uniform linear motion be accelerated?<br \/>\nAnswer:<br \/>\nNo<\/p>\n<p>Question 7.<br \/>\nDefine one radian.<br \/>\nAnswer:<br \/>\nIt is the angle which is subtended at the centre by an arc having a length equal to the radius of the circle.<\/p>\n<p>Question 8.<br \/>\nWhat is the relation between linear velocity and angular velocity?<br \/>\nAnswer:<br \/>\nLinear velocity = Angular velocity \u00d7 Radius of circular path.<\/p>\n<p>Question 9.<br \/>\nGive an example when we infer the motion indirectly.<br \/>\nAnswer:<br \/>\nWe infer the motion of air by observing the movement of dust particles or leaves and branches of trees, or simply by feeling the blowing air on our face.<\/p>\n<p>Question 10.<br \/>\nWhat is essential to describe the position of an object?<br \/>\nAnswer:<br \/>\nWe need to specify a reference point called the origin.<\/p>\n<p>Question 11.<br \/>\nWhat is the simplest type of motion?<br \/>\nAnswer:<br \/>\nMotion in a straight line.<\/p>\n<p>Question 12.<br \/>\nWhat indicates the motion of the earth?<br \/>\nAnswer:<br \/>\nPhenomenon like day and night indicates the motion of the earth.<\/p>\n<p>Question 13.<br \/>\nIf the displacement of a body is zero, is it necessary that the distance coyered by it is also zero?<br \/>\nAnswer:<br \/>\nNo. When the body comes back to the same position after travelling a distance, its displacement is zero though it has travelled some distance.<\/p>\n<p>Question 14.<br \/>\nCan the displacement be greater than the distance travelled by an object?<br \/>\nAnswer:<br \/>\nNo, it is always either equal to or less than the distance travelled by the object.<\/p>\n<p>Question 15.<br \/>\nWhen do the distance and displacement of a moving object have the same magnitude?<br \/>\nAnswer:<br \/>\nThe magnitude of distance and displacement of a moving object are same when the object moves along the same straight line in the same fixed direction.<\/p>\n<p>Question 16.<br \/>\nDoes the speedometer of a car measure its average speed?<br \/>\nAnswer:<br \/>\nNo. It measures its instantaneous speed.<\/p>\n<p>Question 17.<br \/>\nA body is moving with a velocity of 10 m\/s. If the motion is uniform, what will be the velocity after 10 s?<br \/>\nAnswer:<br \/>\nAs the motion is uniform, the velocity remains 10 m\/s after 10 s.<\/p>\n<p>Question 18.<br \/>\nCan a body have constant speed but variable velocity?<br \/>\nAnswer:<br \/>\nYes, e.g. a body in uniform circular motion has constant speed but due to the change in the direction of motion, its velocity changes at every point.<\/p>\n<p>Question 19.<br \/>\nWhen is the acceleration taken as negative?<br \/>\nAnswer:<br \/>\nAcceleration is taken as negative if it is in the direction opposite to the direction of velocity.<\/p>\n<p>Question 20.<br \/>\nWhat is uniform acceleration?<br \/>\nAnswer:<br \/>\nThe acceleration of an object is said to be uniform if it travels in a straight line and its velocity increases or decreases by equal amounts in equal intervals of time. For example, the motion of a freely falling body.<\/p>\n<p>Question 21.<br \/>\nGive an example of non-uniform acceleration.<br \/>\nAnswer:<br \/>\nA car is travelling along a straight road increases its speed by unequal amounts in equal intervals of time.<\/p>\n<p>Question 22.<br \/>\nHow are the distances travelled by an object related to the time taken when an object travels equal distances in equal intervals of time?<br \/>\nAnswer:<br \/>\nIn this case, the distance travelled by the object is directly proportional to the time taken.<\/p>\n<p>Question 23.<br \/>\nWhat would be the acceleration of a body if its velocity-time graph is a line parallel to the time axis?<br \/>\nAnswer:<br \/>\nZero, as the body possesses uniform velocity.<\/p>\n<p>Question 24.<br \/>\nIs the motion of a body uniform or accelerated if it goes round the sun with constant speed in a circular orbit?<br \/>\nAnswer:<br \/>\nIt is accelerated, as its velocity changes due to changes in direction.<\/p>\n<h4><span class=\"ez-toc-section\" id=\"Short_Answer_Questions-I\"><\/span><span style=\"font-size: 14pt;\"><strong>Short Answer Questions-I<\/strong><\/span><span class=\"ez-toc-section-end\"><\/span><\/h4>\n<p>Question 1.<br \/>\nGive an example of a body that may appear to be moving for one person and stationary for the other.<br \/>\nAnswer:<br \/>\nThe passengers in a moving bus observe that the trees, buildings as well as people on the roadside appear to be moving backwards. Similarly, a person standing on the roadside observes that the bus (along with its passengers) is moving in a forwarding direction. But, at the same time, each passenger in a moving bus or train observes, his fellow passengers sitting and not moving. Thus, we can tell that motion is relative.<\/p>\n<p>Question 2.<br \/>\nHow can we describe the location of an object?<br \/>\nAnswer:<br \/>\nWe need to supply a reference point called the origin to define the position of an item.<br \/>\nAssume that a city library is 2 kilometres north of the railway station. The library&#8217;s location in relation to the railway station has been given; in this case, the railway station serves as the reference point.<\/p>\n<p>Question 3.<br \/>\nWhat do you mean by average speed? What are its units?<br \/>\nAnswer:<br \/>\nAverage speed is defined as the average distance travelled per unit time and is obtained by dividing the total distance travelled by the total time taken.<br \/>\nThe unit of average speed is the same as that of the speed, that is, ms<sup>-1<\/sup>.<\/p>\n<p>Question 4.<br \/>\nWhat is the difference between uniform velocity and non-uniform velocity?<br \/>\nAnswer:<br \/>\nUniform velocity: An object with uniform velocity covers equal distances in equal intervals of time in a specified direction, e.g., an object moving with the speed of 40 kmh<sup>-1<\/sup> towards the west has uniform velocity.<\/p>\n<p>Non-uniform velocity: When an object covers unequal distances in equal intervals of time in a specified direction, or if the direction of motion changes, it is said to be moving with a non-uniform or variable velocity, e.g., a revolving fan at a constant speed has variable velocity.<\/p>\n<p>Question 5.<br \/>\nWhat do you understand by instantaneous velocity?<br \/>\nAnswer:<br \/>\nInstantaneous velocity is the velocity of a body at any particular instant during its motion. For example, the instantaneous velocity of a motorcycle at a particular instant is 40 kmh<sup>-1<\/sup> if it is moving at 40 kmh<sup>-1<\/sup> at that particular instant. It is measured by the speedometers on the vehicles.<\/p>\n<p>Question 6.<br \/>\nWhat is negative acceleration?<br \/>\nAnswer:<br \/>\nWhen a body&#8217;s velocity declines with time, its end velocity is less than its starting velocity, resulting in a negative acceleration. Retardation or deceleration are terms used to describe negative acceleration. When brakes are applied to a moving truck, for example, the truck&#8217;s velocity gradually lowers. To put it another way, it is slowed down.<\/p>\n<p>Question 7.<br \/>\nHow will the equations of motion for an object moving with a uniform velocity change? [NCERT Exemplar]\nAnswer:<br \/>\nAcceleration a = 0, v = u<br \/>\nSo, the equations of motion will become<br \/>\ns = ut<br \/>\nv<sup>2<\/sup> \u2013 u<sup>2<\/sup> = 0<\/p>\n<p>Question 8.<br \/>\nExpress average velocity when the velocity of a body changes at a non-uniform rate and a uniform rate.<br \/>\nAnswer:<br \/>\nWhen the velocity of a body changes at a non-uniform rate, its average velocity is found by dividing the net displacement covered by the total time taken.<br \/>\ni.e., Average velocity = \\(\\frac{\\text { Net displacement }}{\\text { Total time taken }}\\)<br \/>\nIn case the velocity of a body changes at a uniform rate, then the average velocity is given by the arithmetic mean of initial velocity and final<br \/>\ni.e., Average velocity = \\(\\frac{\\text { Initial velocity + Final velocity }}{2}\\)<\/p>\n<p>Question 9.<br \/>\nA particle is moving in a circular path of radius r. What would be the displacement after half a circle?<br \/>\nAnswer:<br \/>\nDisplacement = AB<br \/>\n= Shortest distance between initial and final positions<br \/>\n= r + r = 2r<br \/>\n<img loading=\"lazy\" class=\"size-full wp-image-131045 aligncenter\" src=\"https:\/\/infinitylearn.com\/surge\/wp-content\/uploads\/2021\/12\/Motion-Class-9-Extra-Questions-Science-Chapter-8-1.png\" alt=\"Motion Class 9 Extra Questions Science Chapter 8 1\" width=\"194\" height=\"166\" \/><\/p>\n<h5><span class=\"ez-toc-section\" id=\"Short_Answer_Questions-II\"><\/span><span style=\"font-size: 14pt; color: #800080;\">Short Answer Questions-II<\/span><span class=\"ez-toc-section-end\"><\/span><\/h5>\n<p>Question 1.<br \/>\nDifferentiate between distance and displacement.<br \/>\nAnswer:<br \/>\n<img loading=\"lazy\" class=\"size-full wp-image-131046 aligncenter\" src=\"https:\/\/infinitylearn.com\/surge\/wp-content\/uploads\/2021\/12\/Motion-Class-9-Extra-Questions-Science-Chapter-8-2.png\" alt=\"Motion Class 9 Extra Questions Science Chapter 8 2\" width=\"731\" height=\"191\" \/><\/p>\n<p>Question 2.<br \/>\nWhat are the uses of a distance-time graph?<br \/>\nAnswer:<br \/>\nThe various uses of a distance-time graph are as follows:<\/p>\n<ul>\n<li>It tells us about the position of the body at any instant of time.<\/li>\n<li>From the graph, we can see the distance covered by the body during a particular interval of time.<\/li>\n<li>It also gives us information about the velocity of the body at any instant of time.<\/li>\n<\/ul>\n<p>Question 3.<br \/>\nDraw a velocity versus time graph of a stone thrown vertically upwards and then coming downwards after attaining the maximum height. [NCERT Exemplar]\nAnswer:<br \/>\nDuring upward motion, acceleration = -g and during downward motion acceleration = + g. Times of upward and downward motion is equal. Also initial and final velocities are equal to \\(\\left(g \\frac{T}{2}\\right)\\).<br \/>\n<img loading=\"lazy\" class=\"size-full wp-image-131048 aligncenter\" src=\"https:\/\/infinitylearn.com\/surge\/wp-content\/uploads\/2021\/12\/Motion-Class-9-Extra-Questions-Science-Chapter-8-3.png\" alt=\"Motion Class 9 Extra Questions Science Chapter 8 3\" width=\"224\" height=\"225\" \/><\/p>\n<p>Question 4.<br \/>\nThe data regarding the motion of two different objects P and Q is given in the following table. Examine them carefully and state whether the motion of the objects is uniform or non-uniform.<br \/>\n<img loading=\"lazy\" class=\"size-full wp-image-131049 aligncenter\" src=\"https:\/\/infinitylearn.com\/surge\/wp-content\/uploads\/2021\/12\/Motion-Class-9-Extra-Questions-Science-Chapter-8-4.png\" alt=\"Motion Class 9 Extra Questions Science Chapter 8 4\" width=\"732\" height=\"257\" \/><br \/>\nAnswer:<br \/>\nWe can see that the object P covers a distance of 10 m every fifteen minutes. In other words, it covers equal distance in equal intervals of time. So, the motion of object P is uniform. On the other hand, the object Q covers 7 m from 9:30 am to 9:45 am, 4 m from 9:45 am to 10:00 am, and so on. In other words, it covers an unequal distance in equal intervals of time. So, the motion of object Q is non-uniform.<\/p>\n<p>Question 5.<br \/>\nHow will you show that the slope of the displacement-time graph gives the velocity of the body?<br \/>\nAnswer:<br \/>\n<img loading=\"lazy\" class=\"size-full wp-image-131050 aligncenter\" src=\"https:\/\/infinitylearn.com\/surge\/wp-content\/uploads\/2021\/12\/Motion-Class-9-Extra-Questions-Science-Chapter-8-5.png\" alt=\"Motion Class 9 Extra Questions Science Chapter 8 5\" width=\"243\" height=\"241\" \/><br \/>\nThe adjoining figure shows the displacement-time graph for a body moving with uniform velocity. Clearly, it covers distance s<sub>1<\/sub> and s<sub>2<\/sub> at times t<sub>1<\/sub> and t<sub>2<\/sub> respectively.<br \/>\nSlope of line PQ = tan\u03b8 = \\(\\frac{Q R}{P R}\\)<br \/>\n= \\(\\frac{s_{2}-s_{1}}{t_{2}-t_{I}}=\\frac{\\text { Displacement }}{\\text { Time }}\\)<br \/>\nAs \\(\\frac{\\text { Displacement }}{\\text { Time }}\\) is velocity, so the slope of the distance-time graph gives velocity of the body.<\/p>\n<p>Question 6.<br \/>\nWhat are the characteristics of a distance-time graph for an object moving with a non-uniform speed?<br \/>\nAnswer:<br \/>\nThe characteristics of the distance-time graph for a non-uniform speed are:<br \/>\n<img loading=\"lazy\" class=\"size-full wp-image-131051 aligncenter\" src=\"https:\/\/infinitylearn.com\/surge\/wp-content\/uploads\/2021\/12\/Motion-Class-9-Extra-Questions-Science-Chapter-8-6.png\" alt=\"Motion Class 9 Extra Questions Science Chapter 8 6\" width=\"233\" height=\"234\" \/><\/p>\n<ul>\n<li>It is always a curve (parabola).<\/li>\n<li>The speed of the moving object at any point is given by the slope of the tangent to the curve at that point.<\/li>\n<\/ul>\n<p>Question 7.<br \/>\nGiven below is the velocity-time graph for the motion of the car. What does the nature of the graph show ? Also, find the acceleration of the car.<br \/>\n<img loading=\"lazy\" class=\"size-full wp-image-131052 aligncenter\" src=\"https:\/\/infinitylearn.com\/surge\/wp-content\/uploads\/2021\/12\/Motion-Class-9-Extra-Questions-Science-Chapter-8-7.png\" alt=\"Motion Class 9 Extra Questions Science Chapter 8 7\" width=\"425\" height=\"214\" \/><br \/>\nAnswer:<br \/>\nThe nature of the graph shows that velocity changes by equal amounts in equal intervals of time. For a uniformly accelerated motion, the velocity-time graph is always a straight line.<br \/>\nAs we know, acceleration is equal to the slope of the graph<br \/>\n<img loading=\"lazy\" class=\"alignnone size-full wp-image-131053\" src=\"https:\/\/infinitylearn.com\/surge\/wp-content\/uploads\/2021\/12\/Motion-Class-9-Extra-Questions-Science-Chapter-8-8.png\" alt=\"Motion Class 9 Extra Questions Science Chapter 8 8\" width=\"211\" height=\"185\" \/><\/p>\n<p>&nbsp;<\/p>\n<h3><span class=\"ez-toc-section\" id=\"Long_Answer_Questions\"><\/span>Long Answer Questions<span class=\"ez-toc-section-end\"><\/span><\/h3>\n<p>Question 1.<br \/>\nWith the help of a graph, derive the relation v = u + at.<br \/>\nAnswer:<br \/>\nConsider the velocity-time graph of an object that moves under uniform acceleration as shown in the figure (u \u2260 0).<br \/>\nFrom this graph, we can see that initial velocity of the object (at point A) is u and then it increases tow (at point B) in time t. The velocity changes at a uniform rate a. As shown in the figure, the lines BC and BE are drawn from point B on the time and the velocity axes respectively, so that the initial velocity is represented by OA, the final velocity is represented by BC and the time interval t is represented by OC. BD = BC \u2013 CD, represents the change in velocity in time interval t.<br \/>\n<img loading=\"lazy\" class=\"aligncenter wp-image-131054 size-full\" src=\"https:\/\/infinitylearn.com\/surge\/wp-content\/uploads\/2021\/12\/Motion-Class-9-Extra-Questions-Science-Chapter-8-9.png\" alt=\"Motion Class 9 Extra Questions Science Chapter 8 9\" width=\"279\" height=\"370\" \/><br \/>\nIf we draw AD parallel to OC, we observe that<br \/>\nBC = BD + DC = BD + OA<br \/>\nSubstituting, BC with v and OA with u,<br \/>\nwe get<br \/>\nv = BD + u<br \/>\nor BD = v \u2013 u \u2026(1)<br \/>\nThus, from the given velocity-time graph, the acceleration of the object is given by<br \/>\na = \\(\\frac{\\text { Change in velocity }}{\\text { Time taken }}\\)<br \/>\n= \\(\\frac{B D}{A D}=\\frac{B D}{O C}\\)<br \/>\nSubstituting, OC with t, we get<br \/>\na = \\(\\frac{\\mathrm{BD}}{t}\\) or BD= at \u2026..(2)<br \/>\nFrom equations (1) and (2),<br \/>\nwe have<br \/>\nv \u2013 u = at or v = u + at<\/p>\n<p>Question 2.<br \/>\nDeduce the following equations of motion:<br \/>\n(i) s = ut + \\(\\left(\\frac{1}{2}\\right)\\)at<sup>2<\/sup><br \/>\n(ii) v<sup>2<\/sup> = u<sup>2<\/sup> + 2as<br \/>\nAnswer:<br \/>\n(i) Consider a body which starts with initial velocity u and due to uniform acceleration a, its final velocity becomes v after time t. Then, its average velocity is given by<br \/>\n<img loading=\"lazy\" class=\"size-full wp-image-131055 aligncenter\" src=\"https:\/\/infinitylearn.com\/surge\/wp-content\/uploads\/2021\/12\/Motion-Class-9-Extra-Questions-Science-Chapter-8-10.png\" alt=\"Motion Class 9 Extra Questions Science Chapter 8 10\" width=\"720\" height=\"525\" \/><\/p>\n<p>Question 3.<br \/>\nObtain a relation for the distance travelled by an object moving with a uniform acceleration in the interval between 4<sup>th<\/sup> and 5<sup>th<\/sup> seconds. [NCERT Exemplar]\nAnswer:<br \/>\nUsing the equation of motion s = ut + \\(\\frac{1}{2}\\) at<sup>2<\/sup><br \/>\nDistance travelled in 5 seconds, s = u \u00d7 5 + \\(\\frac{1}{2}\\)a \u00d7 5 <sup>2<\/sup><br \/>\nor s = 5u + \\(\\frac{25}{2}\\)a<br \/>\nSimilarly, distance travelled in 4 seconds, s\u2019 = 4u + \\(\\frac{1}{4}\\)a<br \/>\nDistance travelled in the interval between 4<sup>th<\/sup> and 5<sup>th<\/sup> seconds<br \/>\n= (s \u2013 s\u2019) = (u + \\(\\frac{9}{2}\\) a)m<\/p>\n<p>Question 4.<br \/>\nTwo stones are thrown vertically upwards simultaneously with their initial velocities u] and w2 respectively. Prove that the heights reached by them would be in the ratio of \\(u_{1}^{2} : u_{2}^{2}\\).<br \/>\n(Assume upward acceleration is \u2014g and downward acceleration to be +g). [NCERT Exemplar]\nAnswer:<br \/>\n<img loading=\"lazy\" class=\"size-full wp-image-131056 aligncenter\" src=\"https:\/\/infinitylearn.com\/surge\/wp-content\/uploads\/2021\/12\/Motion-Class-9-Extra-Questions-Science-Chapter-8-11.png\" alt=\"Motion Class 9 Extra Questions Science Chapter 8 11\" width=\"445\" height=\"359\" \/><\/p>\n<p>Question 5.<br \/>\nThe driver of train A travelling at a speed of 54 kmh<sup>-1<\/sup> applies brakes and retards the train uniformly. The train stops in 5 seconds. Another train B is travelling on the parallel with a speed of 36 kmh<sup>-1<\/sup>. Its driver applies the brakes and the train retards uniformly; train B stops in 10 seconds. Plot speed-time graphs for both the trains on the same axis. Which of the trains travelled farther after the brakes were applied?<br \/>\nAnswer:<br \/>\nFor train A, the initial velocity,<br \/>\nu = 54 kmh<sup>-1<\/sup> = 54 \u00d7 \\(\\frac{5}{18}\\) = 15 ms<sup>\u2013 1<\/sup><br \/>\nFinal velocity, v = 0 and time, t = 5 s<br \/>\nFor train B, u = 36 kmh<sup>\u2013 1<\/sup> = 36 \u00d7 \\(\\frac{5}{18}\\) = 10 ms<sup>\u2013 1<\/sup><br \/>\nv = 0; t = 10 s<br \/>\nSpeed-time graphs for trains A and B are shown in the figure.<br \/>\n<img loading=\"lazy\" class=\"size-full wp-image-131057 aligncenter\" src=\"https:\/\/infinitylearn.com\/surge\/wp-content\/uploads\/2021\/12\/Motion-Class-9-Extra-Questions-Science-Chapter-8-12.png\" alt=\"Motion Class 9 Extra Questions Science Chapter 8 12\" width=\"407\" height=\"290\" \/><br \/>\nDistance travelled by train A<br \/>\n= Area under straight line graph RS<br \/>\n= Area of \u2206ORS<br \/>\n= \\(\\frac{1}{2}\\) \u00d7 OR \u00d7 OS = \\(\\frac{1}{2}\\) \u00d7 15 ms<sup>\u2013 1<\/sup> \u00d7 5 s = 37.5 m<br \/>\nDistance travelled by train B = Area under PQ = Area of \u2206OPQ<br \/>\n= \\(\\frac{1}{2}\\) \u00d7 OP \u00d7 OQ = \\(\\frac{1}{2}\\) \u00d7 10 ms<sup>-1<\/sup> \u00d7 10 s = 50 m<br \/>\nThus, train B travelled farther after the brakes were applied.<\/p>\n<p>&nbsp;<\/p>\n<h3><span class=\"ez-toc-section\" id=\"Numericals\"><\/span><strong>Numericals<\/strong><span class=\"ez-toc-section-end\"><\/span><\/h3>\n<p>Question 1.<br \/>\nStudy the speed-time graph of a body given here and answer the following questions:<br \/>\n<img loading=\"lazy\" class=\"size-full wp-image-131058 aligncenter\" src=\"https:\/\/infinitylearn.com\/surge\/wp-content\/uploads\/2021\/12\/Motion-Class-9-Extra-Questions-Science-Chapter-8-13.png\" alt=\"Motion Class 9 Extra Questions Science Chapter 8 13\" width=\"379\" height=\"216\" \/><br \/>\n(a) What type of motion is represented by OA?<br \/>\n(b) What type of motion is represented by AB?<br \/>\n(c) What type of motion is represented by BC?<br \/>\n(d) Find out the acceleration of the body.<br \/>\n(e) Calculate the retardation of the body.<br \/>\n(f) Find out the distance travelled by the body from A to B.<br \/>\nAnswer:<br \/>\n(a) OA is a straight line graph between speed and time, and it is sloping upward from O to A. Therefore, the graph line OA represents uniform acceleration.<br \/>\n(b) AB is a straight line graph between speed and time, which is parallel to the time axis (x-axis).<br \/>\nSo, AB represents uniform speed. There is no acceleration from A to B.<br \/>\n(c) BC is a straight line graph between speed and time which is sloping downwards from B to C. Therefore, BC represents uniform retardation or negative acceleration.<br \/>\n(d) Acceleration of the body as we see from graph line OA represents it. So, the slope of velocity-time graph OA will give the acceleration of the body. Thus,<br \/>\nAcceleration = Slope of line OA = AD\/OD<br \/>\nWe have, AD = 6 m\/s, and OD = 4 s<br \/>\nSo, acceleration = \\(\\frac{6 \\mathrm{m} \/ \\mathrm{s}}{4 \\mathrm{s}}\\) = 1.5 m\/s<br \/>\n(e) The slope of line graph BC represents the retardation of the body.<br \/>\nSo, retardation = Slope of line BC = BE\/EC<br \/>\nWe have, BE = 6m\/s, EC = 16 \u2013 10 = 6s<br \/>\nRetardation = \\(\\frac{6 \\mathrm{m} \/ \\mathrm{s}}{4 \\mathrm{s}}\\) = 1m\/s<sup>2<\/sup><br \/>\n(f) Distance travelled from A to B = Area under the line AB and the time axis<br \/>\n= Area of rectangle DABE = DA \u00d7 DE<br \/>\nHere, DA = 6 m\/s and DE = 10 \u2013 4 = 6 s<br \/>\nDistance travelled from A to B = 6 \u00d7 6 = 36m<\/p>\n<p>Question 2.<br \/>\nThe graph given below shows the positions of a body at different times. Calculate the speed of the body as it moves from<br \/>\n(i) A to B<br \/>\n(ii) B to C and<br \/>\n(iii) C to D.<br \/>\n<img loading=\"lazy\" class=\"size-full wp-image-131059 aligncenter\" src=\"https:\/\/infinitylearn.com\/surge\/wp-content\/uploads\/2021\/12\/Motion-Class-9-Extra-Questions-Science-Chapter-8-14.png\" alt=\"Motion Class 9 Extra Questions Science Chapter 8 14\" width=\"303\" height=\"250\" \/><br \/>\nAnswer:<br \/>\n(i) The distance-time graph represents the line AB which shows the speed of the body. So,<br \/>\nSpeed = \\(\\frac{\\text { Distance }}{\\text { Time }}=\\frac{3 \\mathrm{cm}}{(5-2) \\mathrm{s}}\\) = 1 cm\/s<br \/>\n(ii) The distance-time graph shows that the body is at rest between graph line B to C, it means no movement. So speed is zero i.e.,<br \/>\nSpeed = \\(\\frac{\\text { Distance }}{\\text { Time }}=\\frac{0}{(7-5) \\mathrm{s}}=\\frac{0}{2 \\mathrm{s}}\\) = 0<br \/>\n(iii) The distance-time graph represents the line CD which shows the speed of the body. So,<br \/>\nSpeed = \\(\\frac{\\text { Distance }}{\\text { Time }}=\\frac{(7-3) \\mathrm{cm}}{(9-7) \\mathrm{s}}=\\frac{4 \\mathrm{cm}}{2 \\mathrm{s}}\\) = 2 cm\/s<\/p>\n<p>Question 3.<br \/>\nDistance travelled by train and time taken by it is shown in the following table, (i) Plot distance-time graph. (ii) What is the average speed of the train? (iii) When is the train travelling at the highest speed? (iv) At what distance does the train slow down? (v) Calculate the speed of the train between 10:40 AM to 11:00 AM.<br \/>\n<img loading=\"lazy\" class=\"size-full wp-image-131060 aligncenter\" src=\"https:\/\/infinitylearn.com\/surge\/wp-content\/uploads\/2021\/12\/Motion-Class-9-Extra-Questions-Science-Chapter-8-15.png\" alt=\"Motion Class 9 Extra Questions Science Chapter 8 15\" width=\"729\" height=\"248\" \/><br \/>\nAnswer:<br \/>\n<img loading=\"lazy\" class=\"size-full wp-image-131061 aligncenter\" src=\"https:\/\/infinitylearn.com\/surge\/wp-content\/uploads\/2021\/12\/Motion-Class-9-Extra-Questions-Science-Chapter-8-16.png\" alt=\"Motion Class 9 Extra Questions Science Chapter 8 16\" width=\"430\" height=\"381\" \/><br \/>\n<img loading=\"lazy\" class=\"size-full wp-image-131062 aligncenter\" src=\"https:\/\/infinitylearn.com\/surge\/wp-content\/uploads\/2021\/12\/Motion-Class-9-Extra-Questions-Science-Chapter-8-17.png\" alt=\"Motion Class 9 Extra Questions Science Chapter 8 17\" width=\"545\" height=\"202\" \/><br \/>\n(iii) We know, speed = slope of the distance-time graph. The greater the slope, the greater is the speed.<br \/>\nFrom the graph, it is clear that the slope of the distance-time graph is maximum between 10:00 AM to 10:30 AM, so the train was travelling at the highest speed during this interval of time.<br \/>\n(iv) The part CD of the graph has a minimum slope, so the train had a minimum speed between 11:00 AM and 11:15 AM. Thus, the train had slowed down between 40 km and 42 km.<br \/>\n(v) Speed between 10:40 AM to 11:00 AM = \\(\\frac{\\text { Distance }}{\\text { Time }}=\\frac{(40-28) \\mathrm{km}}{20 \\mathrm{min}}=\\frac{12 \\mathrm{km}}{\\frac{20}{60} \\mathrm{h}}\\) = 36 kmh<sup>-1<\/sup><\/p>\n<p>Question 4.<br \/>\nThe velocity-time graph shows the motion of a cyclist. Find (i) its acceleration (ii) its velocity and (iii) the distance covered by the cyclist in 15 seconds. [NCERT Exemplar]\n<img loading=\"lazy\" class=\"size-full wp-image-131063 aligncenter\" src=\"https:\/\/infinitylearn.com\/surge\/wp-content\/uploads\/2021\/12\/Motion-Class-9-Extra-Questions-Science-Chapter-8-18.png\" alt=\"Motion Class 9 Extra Questions Science Chapter 8 18\" width=\"257\" height=\"251\" \/><br \/>\nAnswer:<br \/>\n(i) Since velocity is not changing acceleration is equal to zero. [a = \\(\\frac{\\Delta v}{\\Delta t}\\) = 0]\n(ii) Reading the graph, velocity = 20 ms<sup>-1<\/sup> (constant)<br \/>\n(iii) Distance covered in 15 seconds = Area of ABNO<br \/>\n= v \u00d7 t<br \/>\n= 20 \u00d7 15 = 300 m<\/p>\n<p>Question 5.<br \/>\nA body starts to slide over a horizontal surface with an initial velocity of 0.5 m\/s. Due to friction, its velocity decreases at the rate of 0.05 m\/s<sup>2<\/sup>. How much time will it take for the body to stop?<br \/>\nAnswer:<br \/>\nInitial velocity, u = 0.5 m\/s<br \/>\nFinal velocity, v= 0<br \/>\nAcceleration, a = -0.05 m\/s<sup>2<\/sup><br \/>\nNow, from the first equation of the motion,<br \/>\nv = u + at<br \/>\nor, 0 = 0.5 + (- 0.05t) \u21d2 0.5 = 0.05t,<br \/>\n\u2234 t = 0.5\/0.05 = 10 s<br \/>\nThus, the body will take 10 s to stop.<\/p>\n<p>Question 6.<br \/>\nA particle moves in a circle with O as centre and AO = OB = 5 cm, as radius, as shown in the figure. It starts from A. Calculate:<br \/>\n(a) the distance covered, and<br \/>\n(b) the displacement, when it reaches B.<br \/>\n<img loading=\"lazy\" class=\"size-full wp-image-131064 aligncenter\" src=\"https:\/\/infinitylearn.com\/surge\/wp-content\/uploads\/2021\/12\/Motion-Class-9-Extra-Questions-Science-Chapter-8-19.png\" alt=\"Motion Class 9 Extra Questions Science Chapter 8 19\" width=\"185\" height=\"163\" \/><br \/>\nAnswer:<br \/>\n(a) Distance covered = \u03c0 \u00d7 OA = \u03c0 \u00d7 5 = 5\u03c0 cm<br \/>\n(b) Displacement = 2 \u00d7 OB<br \/>\n= 2 \u00d7 5 = 10 cm along AB<\/p>\n<p>Question 7.<br \/>\nA body travels along a circular path of a radius of 70 m. After travelling half a revolution in 20 s, find the<br \/>\n(i) average velocity,<br \/>\n(ii) average speed.<br \/>\nAnswer:<br \/>\n<img loading=\"lazy\" class=\"size-full wp-image-131065 aligncenter\" src=\"https:\/\/infinitylearn.com\/surge\/wp-content\/uploads\/2021\/12\/Motion-Class-9-Extra-Questions-Science-Chapter-8-20.png\" alt=\"Motion Class 9 Extra Questions Science Chapter 8 20\" width=\"486\" height=\"94\" \/><br \/>\n<img loading=\"lazy\" class=\"size-full wp-image-131066 aligncenter\" src=\"https:\/\/infinitylearn.com\/surge\/wp-content\/uploads\/2021\/12\/Motion-Class-9-Extra-Questions-Science-Chapter-8-21.png\" alt=\"Motion Class 9 Extra Questions Science Chapter 8 21\" width=\"612\" height=\"89\" \/><\/p>\n<p>Question 8.<br \/>\nA cheetah is the fastest land animal and can achieve a peak velocity of 100 km\/h up to distances less than 500 m. If a cheetah spots its prey at a distance of 100 m, what is the minimum time it will take to get its prey, if the average velocity attained by it is 90 km\/h?<br \/>\nAnswer:<br \/>\nAverage velocity = 90 km\/h = \\(\\frac{90 \\mathrm{km}}{1 \\mathrm{h}}\\)<br \/>\n= \\(\\frac{90 \\times 1000 \\mathrm{m}}{60 \\times 60 \\mathrm{s}}\\) = 25 ms<sup>-1<\/sup><br \/>\nAlso, Average velocity = \\(\\frac{\\text { Displacement }}{\\text { Time taken }}\\)<br \/>\n\u2235 Cheetah moves in a straight line displacement is equal to 100 m.<br \/>\nTherefore, time taken = \\(\\frac{100}{25}\\) = 4 s<\/p>\n<p>Question 9.<br \/>\nThe brakes applied to a car produce an acceleration of 6 ms<sup>-2<\/sup> in the opposite direction to the motion. If the car takes 2s to stop after the application of brakes, calculate the distance it travels during this time.<br \/>\nAnswer:<br \/>\nWe have been given<br \/>\n<img loading=\"lazy\" class=\"size-full wp-image-131067 aligncenter\" src=\"https:\/\/infinitylearn.com\/surge\/wp-content\/uploads\/2021\/12\/Motion-Class-9-Extra-Questions-Science-Chapter-8-22.png\" alt=\"Motion Class 9 Extra Questions Science Chapter 8 22\" width=\"427\" height=\"212\" \/><br \/>\nThus, the car will move 12 m before it stops after the application of brakes.<\/p>\n<p>Question 10.<br \/>\nA car starts from rest and moves along the x-axis with constant acceleration 5 ms<sup>-2<\/sup> for 8 seconds. If it then continues with constant velocity, what distance will the car cover in 12 seconds since it started from the rest? [NCERT Exemplar]\nAnswer:<br \/>\nInitial velocity ,u = 0<br \/>\nUsing s = ut + \\(\\frac{1}{2}\\)at<sup>2<\/sup><br \/>\nThe distance travelled in first 8s,<br \/>\ns<sub>1<\/sub> = 0 + \\(\\frac{1}{2}\\) \u00d7 5 \u00d7 8<sup>2<\/sup> = 160 m<br \/>\nAt this point the velocity, v = u + at<br \/>\n= 0 + 5 \u00d7 8 = 40 ms<sup>-1<\/sup><br \/>\nSo, the distance covered in last four seconds<br \/>\ns<sub>2<\/sub> = 40 \u00d7 4 = 160 m<br \/>\nHence, total distance, s = s<sub>1<\/sub> + s<sub>2<\/sub><br \/>\n= 160 m + 160 m = 320 m<\/p>\n<p>Question 11.<br \/>\nA motorcyclist drives from A to B with a uniform speed of 30 kmh<sup>-1<\/sup> and returns back with a speed of 20 kmh<sup>-1<\/sup>. Find its average speed. [NCERT Exemplar]\nAnswer:<br \/>\n<img loading=\"lazy\" class=\"size-full wp-image-131068 aligncenter\" src=\"https:\/\/infinitylearn.com\/surge\/wp-content\/uploads\/2021\/12\/Motion-Class-9-Extra-Questions-Science-Chapter-8-23.png\" alt=\"Motion Class 9 Extra Questions Science Chapter 8 23\" width=\"519\" height=\"153\" \/><\/p>\n<p>Question 12.<br \/>\nAn object is dropped from rest at a height of 150 m and simultaneously another object is dropped from rest at a height 100 m. What is the difference in their heights after 2 s if both the objects drop with same accelerations? How does the difference in heights vary with time? [NCERT Exemplar]\nAnswer:<br \/>\nInitial difference in height = (150 \u2013 100) m = 50 m<br \/>\nDistance travelled by first body in 2 s = h<sub>1<\/sub> = 0 + \\(\\frac{1}{2}\\) g(2)<sup>2<\/sup> = 2g<br \/>\nDistance travelled by another body in 2 s = h<sub>2<\/sub> = 0 + \\(\\frac{1}{2}\\) g(2)<sup>2<\/sup> = 2g<br \/>\nAfter 2 s, height at which the first body will be = h<sub>1<\/sub>\u2018 = 150 \u2013 2g<br \/>\nAfter 2 s, height at which the second body will be = h\u2019<sub>2<\/sub> = 100 \u2013 2g<br \/>\nThus, after 2 s, difference in height = 150 \u2013 2g \u2013 (100 \u2013 2g) = 50 m = initial difference in height Thus, difference in height does not vary with time.<\/p>\n<p>Question 13.<br \/>\nAn object starting from rest travels 20 m in first 2s and 160 m in next 4s. What will be the velocity after 7s from the start? [NCERT Exemplar]\nAnswer:<br \/>\ns<sub>1<\/sub> = ut + \\(\\frac{1}{2}\\) at<sup>2<\/sup> or 20 = 0 + \\(\\frac{1}{2}\\)a(2)<sup>2<\/sup> or a = 10ms<sup>-2<\/sup><br \/>\nv = u + at = 0 + (10 \u00d7 2) = 20 ms<sup>-1<\/sup><br \/>\ns<sub>2<\/sub> = 160 = vt\u2019 + \\(\\frac{1}{2}\\)a'(t\u2019)<sup>2<\/sup> = (20 \u00d7 4) + (\\(\\frac{1}{2}\\)a\u2019 \u00d7 16) \u21d2 a\u2019 = 10ms<sup>-2<\/sup><br \/>\nSince accelerations is the same, we have v\u2019 = 0 + (10 \u00d7 7) =70 ms<sup>-1<\/sup><\/p>\n<p>Question 14.<br \/>\nAn electron moving with a velocity of 5 \u00d7 10<sup>4<\/sup> ms<sup>-1<\/sup> enters into a uniform electric field and acquires a uniform acceleration of 104 ms<sup>-2<\/sup> in the direction of its initial motion.<br \/>\n(i) Calculate the time in which the electron would acquire a velocity double of its initial velocity.<br \/>\n(ii) How much distance the electron would cover at this time? [NCERT Exemplar]\nAnswer:<br \/>\nGiven initial velocity, u = 5 \u00d7 10<sup>4<\/sup> ms<sup>-1<\/sup><br \/>\nand acceleration, a = 10<sup>4<\/sup>ms<sup>-2<\/sup><br \/>\n(i) Final velocity = v = 2u = 2 \u00d7 5 \u00d7 10<sup>4<\/sup>ms<sup>-1<\/sup> = 10 \u00d7 10<sup>4<\/sup>ms<sup>-1<\/sup><br \/>\nTo find t, use v = u + at<br \/>\n<img loading=\"lazy\" class=\"size-full wp-image-131069 aligncenter\" src=\"https:\/\/infinitylearn.com\/surge\/wp-content\/uploads\/2021\/12\/Motion-Class-9-Extra-Questions-Science-Chapter-8-24.png\" alt=\"Motion Class 9 Extra Questions Science Chapter 8 24\" width=\"622\" height=\"180\" \/><\/p>\n<h3><\/h3>\n<h3><span class=\"ez-toc-section\" id=\"HOTS_Higher_Order_Thinking_Skills\"><\/span>HOTS (Higher Order Thinking Skills)<span class=\"ez-toc-section-end\"><\/span><\/h3>\n<p>Question 1.<br \/>\nFour cars A, B, C and D are moving on a levelled road. Their distance versus time graphs are shown in figure. Which car is the slowest?<br \/>\n<img loading=\"lazy\" class=\"size-full wp-image-131070 aligncenter\" src=\"https:\/\/infinitylearn.com\/surge\/wp-content\/uploads\/2021\/12\/Motion-Class-9-Extra-Questions-Science-Chapter-8-25.png\" alt=\"Motion Class 9 Extra Questions Science Chapter 8 25\" width=\"306\" height=\"299\" \/><br \/>\nAnswer:<br \/>\nSpeed = Slope of the distance-time graph. The smaller the slope, the smaller is the speed.<br \/>\nFrom the figure, the slope is minimum for car D. So, D is the slowest car.<\/p>\n<p>Question 2.<br \/>\nA girl walks along a straight path to drop a letter in the letterbox and comes back to her initial position. Her displacement-time graph is shown in figure. Plot a velocity-time graph for the same. [NCERT Exemplar]\n<img loading=\"lazy\" class=\"size-full wp-image-131071 aligncenter\" src=\"https:\/\/infinitylearn.com\/surge\/wp-content\/uploads\/2021\/12\/Motion-Class-9-Extra-Questions-Science-Chapter-8-26.png\" alt=\"Motion Class 9 Extra Questions Science Chapter 8 26\" width=\"206\" height=\"201\" \/><br \/>\nAnswer:<br \/>\nVelocity from 0 to 50 s is v<sub>1<\/sub> = \\(\\frac{\\Delta s}{\\Delta t}=\\frac{100-0}{50-0}\\) = 2 ms<sup>-1<\/sup><br \/>\nAt 50 \u2013 0<br \/>\nVelocity from 50 s to 100 s, v<sub>2<\/sub> = \\(\\frac{\\Delta s}{\\Delta t}=\\frac{0-100}{100-50}\\) = \u2013 2 ms<sup>-1<\/sup><br \/>\nAccordingly the velocity-time graph is shown in figure below.<br \/>\n<img loading=\"lazy\" class=\"size-full wp-image-131072 aligncenter\" src=\"https:\/\/infinitylearn.com\/surge\/wp-content\/uploads\/2021\/12\/Motion-Class-9-Extra-Questions-Science-Chapter-8-27.png\" alt=\"Motion Class 9 Extra Questions Science Chapter 8 27\" width=\"257\" height=\"252\" \/><\/p>\n<p>Question 3.<br \/>\nSuppose a squirrel is moving at a steady speed from the base of a tree towards some nuts. It then stays in the same position for a while, eating the nuts, before returning to the tree at the same speed. A graph can be plotted with distance on the x-axis and the time on y-axis.<br \/>\n<img loading=\"lazy\" class=\"size-full wp-image-131073 aligncenter\" src=\"https:\/\/infinitylearn.com\/surge\/wp-content\/uploads\/2021\/12\/Motion-Class-9-Extra-Questions-Science-Chapter-8-29.png\" alt=\"Motion Class 9 Extra Questions Science Chapter 8 29\" width=\"610\" height=\"265\" \/><br \/>\nObserve the graph carefully and answer the following questions.<br \/>\n(i) Which part of the graph shows the squirrel moving away from the tree?<br \/>\n(ii) Name the point on the graph which is 6 m away from the base of the tree.<br \/>\n(iii) Which part of the graph shows that the squirrel is not moving?<br \/>\n(iv) Which part of the graph shows that the squirrel is returning to the tree?<br \/>\n(v) Calculate the speed of the squirrel from the graph during its journey.<br \/>\nAnswer:<br \/>\n(i) Part AB<br \/>\n(ii) Part B<br \/>\n(iii) Part BC<br \/>\n(iv) Part CD<br \/>\n(v) Total distance travelled = 6m + 6m=12m<br \/>\nTime = 11 s.<br \/>\nSpeed = \\(\\frac{\\text { Total distance }}{\\text { Time }}=\\frac{12}{11}\\) = 1.09m\/s.<\/p>\n<p>Question 4.<br \/>\nThe table given below shows distance (in cm) travelled by bodies A, B and C. Read this data carefully and answer the following questions.<br \/>\n<img loading=\"lazy\" class=\"size-full wp-image-131074 aligncenter\" src=\"https:\/\/infinitylearn.com\/surge\/wp-content\/uploads\/2021\/12\/Motion-Class-9-Extra-Questions-Science-Chapter-8-28.png\" alt=\"Motion Class 9 Extra Questions Science Chapter 8 28\" width=\"706\" height=\"170\" \/><br \/>\n(i) Which of the bodies is moving with<br \/>\n(a) constant speed?<br \/>\n(b) constant acceleration?<br \/>\n(c) non-uniform acceleration?<br \/>\n(ii) Which of the bodies covers<br \/>\n(a) maximum distance in 3rd second?<br \/>\n(b) minimum distance in 3rd second?<br \/>\nAnswer:<br \/>\n(i) (a) Body A (b) Body C (c) Body B<br \/>\n(ii) (a) Body C. Total distance travelled = 100 \u2013 60 = 40 cm<br \/>\n(b) Body B. Total distance travelled = 24 \u2013 36 = (-) 12 cm<br \/>\nThe negative sign implies decceleration.<\/p>\n","protected":false},"excerpt":{"rendered":"<p>Chapter 8 &#8211; Motion Extra Questions Get Extra Questions for Class 9 Science on Infinity Learn for free. Very Short [&hellip;]<\/p>\n","protected":false},"author":7,"featured_media":0,"comment_status":"closed","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"_yoast_wpseo_focuskw":"Chapter 8 Motion Extra Questions","_yoast_wpseo_title":"Extra Questions - Class 9 Science Chapter 8 Motion","_yoast_wpseo_metadesc":"class 9 science chapter 8 extra questions and answers. 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