{"id":31530,"date":"2026-05-20T16:46:52","date_gmt":"2026-05-20T11:16:52","guid":{"rendered":"https:\/\/mycbseguide.com\/blog\/?p=31530"},"modified":"2026-05-20T16:48:16","modified_gmt":"2026-05-20T11:18:16","slug":"describing-motion-around-us-ncert-solutions-class-9-science-exlporation","status":"publish","type":"post","link":"https:\/\/mycbseguide.com\/blog\/describing-motion-around-us-ncert-solutions-class-9-science-exlporation\/","title":{"rendered":"Describing Motion Around Us &#8211; NCERT Solutions Class 9 Science (Exlporation)"},"content":{"rendered":"\n<p>Describing Motion Around Us &#8211; NCERT Solutions Class 9 Science Exploration includes all the questions with solutions given in NCERT Class 9 <a href=\"https:\/\/ncert.nic.in\/textbook.php?iesc1=0-13\">Science Exploration<\/a> textbook.<\/p>\n\n\n\n<h2 class=\"wp-block-heading\">NCERT Solutions Class 9<\/h2>\n\n\n<a class=\"mks_button mks_button_small rounded\" href=\"https:\/\/mycbseguide.com\/blog\/category\/ncert-solutions\/ncert-solutions-class-9\/ncert-solutions-class-9-english-kaveri\/\" target=\"_self\" style=\"color: #FFFFFF; background-color: #0066bf;\" >English Kaveri<\/a>\n\n\n<a class=\"mks_button mks_button_small rounded\" href=\"https:\/\/mycbseguide.com\/blog\/category\/ncert-solutions\/ncert-solutions-class-9\/ncert-solutions-class-9-hindi-ganga\/\" target=\"_self\" style=\"color: #FFFFFF; background-color: #0066bf;\" >Hindi Ganga<\/a>\n\n\n<a class=\"mks_button mks_button_small rounded\" href=\"https:\/\/mycbseguide.com\/blog\/category\/ncert-solutions\/ncert-solutions-class-9\/ncert-solutions-class-9-sanskrit-sharada\/\" target=\"_self\" style=\"color: #FFFFFF; background-color: #0066bf;\" >Sanskrit Sharada<\/a>\n\n\n<a class=\"mks_button mks_button_small rounded\" href=\"https:\/\/mycbseguide.com\/blog\/category\/ncert-solutions\/ncert-solutions-class-9\/ncert-solutions-class-9-maths-ganita-manjari\/\" target=\"_self\" style=\"color: #FFFFFF; background-color: #0066bf;\" >Maths Ganita Manjari<\/a>\n\n\n<a class=\"mks_button mks_button_small rounded\" href=\"https:\/\/mycbseguide.com\/blog\/category\/ncert-solutions\/ncert-solutions-class-9\/ncert-solutions-class-9-science-exploration\/\" target=\"_self\" style=\"color: #FFFFFF; background-color: #0066bf;\" >Science Exploration<\/a>\n\n\n<a class=\"mks_button mks_button_small rounded\" href=\"https:\/\/mycbseguide.com\/blog\/category\/ncert-solutions\/ncert-solutions-class-9\/ncert-solutions-class-9-social-understanding-society\/\" target=\"_self\" style=\"color: #FFFFFF; background-color: #0066bf;\" >Social Understanding Society<\/a>\n\n\n\n<h2 class=\"wp-block-heading\">Describing Motion Around Us \u2013 NCERT Solutions<\/h2>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p><strong>Q.1: How much distance should we maintain from the truck ahead to avoid a collision if it suddenly applies the brakes?<\/strong><\/p>\n\n\n\n<p>Solution: This involves stopping distance and relative motion using kinematic equations.<\/p>\n\n\n\n<p>The <strong>time-gap rule (3-4 seconds)<\/strong> is essentially a practical version of:<\/p>\n\n\n\n<p>Safe&nbsp;distance {tex}\\propto{\/tex} speed<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p><strong>Q.2: Does this distance depend upon the speed with which we are moving?<\/strong><\/p>\n\n\n\n<p>Solution: Yes. The safe distance <strong>depends on speed<\/strong> &#8211; higher speed \u21d2 <strong>larger stopping distance<\/strong> (distance increases rapidly, \u221d speed<sup>2<\/sup>).<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p><strong>Q.3: As shown in fig, a ball is thrown vertically upwards from O. It moves up straight till B and then falls back to O. Can this be considered a motion in a straight line?<\/strong><br><img decoding=\"async\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/question_images\/1776934767-zquhve.jpg\"><\/p>\n\n\n\n<p>Solution: Yes, because the ball moves along a single vertical straight path (up and down along the same line).<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p><strong>Q.4: In the example of an athlete running back and forth on a straight track, when will the displacement of the athlete be zero? What will be the total distance travelled in that case?<\/strong><br><img decoding=\"async\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/question_images\/1776935412-7h853m.jpg\"><\/p>\n\n\n\n<p>Solution:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li><strong>Displacement = 0<\/strong> when the athlete returns to the <strong>starting point (A)<\/strong>.<\/li>\n\n\n\n<li><strong>Total distance travelled<\/strong> = length of path covered = <strong>A \u2192 B \u2192 A = 2 \u00d7 AB<\/strong>.<\/li>\n<\/ul>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p><strong>Q.5: Fuel used up in a vehicle depends on which of the following? Justify your answer.<\/strong><\/p>\n\n\n\n<ol start=\"1\" class=\"wp-block-list\">\n<li>Total distance travelled<\/li>\n\n\n\n<li>Displacement<\/li>\n<\/ol>\n\n\n\n<p>Solution: Fuel used depends on total distance travelled, not displacement.<\/p>\n\n\n\n<p><strong>Justification:<\/strong><br>Fuel consumption is related to the actual path covered by the vehicle. Even if displacement is zero (returning to the start), fuel is still used because distance travelled is not zero. Hence, fuel depends on distance, not displacement.<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p><strong>Q.6: A ball rolls down an inclined track as shown in the figure 1.<\/strong><br><img decoding=\"async\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/question_images\/1777264218-3ufaqf.jpg\"><br>Is its motion, a straight line motion? Assuming the starting point of the ball (O) to be the origin, can its motion from O to D be depicted using a horizontal line as shown in the figure 2?<br><img decoding=\"async\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/question_images\/1777264116-aqfwnd.jpg\"><br>Are the values of total distance travelled and magnitude of displacement from O equal or different at positions A, B, C and D?<\/p>\n\n\n\n<p>Solution: The motion of the ball on the inclined track is not a straight line motion, because it follows a curved path along the track.<\/p>\n\n\n\n<p>Yes, its motion from O to D can be represented by a horizontal straight line if we only consider the distance covered along the track (path length), ignoring the actual shape of the path.<\/p>\n\n\n\n<p>At positions A, B, C, and D:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>The total distance travelled is always greater than the magnitude of displacement, except at the starting point.<\/li>\n\n\n\n<li>This is because distance depends on the actual path, while displacement is the shortest straight-line distance between two points.<\/li>\n<\/ul>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p><strong>Q.7: During a family road trip, you drive 200 km north in three hours. Afterwards, you drive 200 km south in two hours. Find the average speed and average velocity for your entire trip.<\/strong><\/p>\n\n\n\n<p>Solution: Total distance = 200 + 200 = 400 km<\/p>\n\n\n\n<p>Total time = 3 + 2 = 5 h<\/p>\n\n\n\n<p>Average speed = 400 \/ 5 = 80 km\/h<\/p>\n\n\n\n<p>Displacement = 0 (back to starting point)<\/p>\n\n\n\n<p>Average velocity = 0 km\/h<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p><strong>Q.8: Under what condition(s) is the<\/strong><\/p>\n\n\n\n<ol start=\"1\" class=\"wp-block-list\">\n<li>magnitude of average velocity of an object equal to its average speed?<\/li>\n\n\n\n<li>magnitude of average velocity of an object zero while its average speed is not zero?<\/li>\n<\/ol>\n\n\n\n<p>Solution: The magnitude of average velocity equals average speed when the object moves in a straight line without changing direction, making displacement equal to distance. When an object returns to its starting point, its displacement becomes zero, so average velocity is zero. However, since it has covered some distance, its average speed remains non-zero.<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p><strong>Q.9: Consider two postmen. They start walking towards each other from a distance of 210 yojanas (Yojana is a unit of distance used in ancient India). One travels 9 yojanas per day and the other covers 5 yojanas per day. Can you determine in how many days they will meet each other?<\/strong><\/p>\n\n\n\n<p>Solution: Total of distance covered by each postman in one day {tex}=9{\/tex} yojanas +5 yojanas {tex}=14{\/tex} yojanas.<br>To meet with each other, the postmen need to cover 210 yojanas together.<br>Time taken by them to cover 210 yojanas together {tex}=\\frac{210}{14}=15{\/tex} days.<br>So, both postmen will meet each other after 15 days.<br>(In 15 days, first postman will cover 135 yojanas and the second postman will cover 75 yojanas).<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p><strong>Q.10: Sarang takes 50 seconds to swim from one end to the other end and back in the swimming pool shown in figure.\u00a0Find his average speed and average velocity within the time interval of 50 s.<\/strong><br><img decoding=\"async\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/question_images\/1776936276-g6d4zy.jpg\"><\/p>\n\n\n\n<p>Solution: Total distance travelled by Sarang in\u00a050\u00a0s\u00a0= 50\u00a0m<br>displacement of Sarang in\u00a050 s = 0 m<br>{tex}\\text {average speed }=\\frac{\\text { total distance travelled }}{\\text { time interval }}=\\frac{50 {~m}}{50 {~s}}=1 {~m} {~s}^{-1}{\/tex}<br>{tex}\\text {average velocity }=\\frac{\\text { displacement }}{\\text { time interval }}=\\frac{0 {~m}}{50 {~s}}=0 {~m} {~s}^{-1}{\/tex}<br>During the 50\u2009s time interval, the average speed of Sarang is approximately 1 m s<sup>-1<\/sup> while his average velocity is 0 m s\u200a<sup>-1<\/sup>.<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p><strong>Q.11: A bus is moving on a long straight highway with a velocity of 36 km h<sup>-1<\/sup>. The driver presses the accelerator for a time interval of 10 s and velocity of the bus increases to 54 km h<sup>-1<\/sup>. For some time, the bus moves at a constant velocity. Then, the driver notices an obstacle on the road ahead and presses the brake. The bus comes to a stop in a time interval of 5 s. Find the average acceleration in the two time intervals, (i) when the accelerator was pressed, and (ii) when the brakes were pressed.<\/strong><br><img decoding=\"async\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/question_images\/1776937736-v4622d.jpg\"><\/p>\n\n\n\n<p>Solution:<\/p>\n\n\n\n<ol start=\"1\" class=\"wp-block-list\">\n<li>When the driver presses the accelerator<br>{tex} u=36 {~km} {~h}^{-1}=36 \\times \\frac{1000 {~m}}{60 \\times 60 {~s}}=10 {~m} {~s}^{-1}, {\/tex}\u00a0{tex}v=54 {~km} {~h}^{-1}=15 {~m} {~s}^{-1}, t=10 {~s}, a=?{\/tex}<br>Using Eq. {tex}a=\\frac{v-u}{t_2-t_1}{\/tex}, we obtain the average acceleration<br>{tex} a=\\frac{15 {~m} {~s}^{-1}-10 {~m} {~s}^{-1}}{10 {~s}}=\\frac{5 {~m} {~s}^{-1}}{10 {~s}}=0.5 {~m} {~s}^{-2}{\/tex}<br>Since the magnitude of velocity of the bus is increasing, the acceleration is acting in the direction of velocity.<\/li>\n\n\n\n<li>When the driver presses the brake<br>{tex} u=54 {~km} {~h}^{-1}=15 {~m} {~s}^{-1}, v=0 {~m} {~s}^{-1}, t=5 {~s}, a=?{\/tex}<br>Using Eq. {tex}a=\\frac{v-u}{t_2-t_1}{\/tex} again, we obtain<br>{tex} a=\\frac{0 {~m} {~s}^{-1}-15 {~m} {~s}^{-1}}{5 {~s}}=\\frac{-15 {~m} {~s}^{-1}}{5 {~s}}=-3 {~m} {~s}^{-2}{\/tex}<br>The minus sign indicates that the acceleration is acting opposite to the direction of velocity (since the magnitude of velocity of the bus is decreasing).<\/li>\n<\/ol>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p><strong>Q.12: As we learnt earlier, when an object is dropped from a height, it takes a straight vertical path downwards before touching the ground. While coming down, the velocity of the object increases as shown in figure at different instants.<\/strong><br><img decoding=\"async\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/question_images\/1776939656-5vg64c.jpg\"><br>Find the magnitude of the average acceleration of the object in every successive interval of a second. Is the average acceleration constant across all intervals? What is the direction of this average acceleration?<\/p>\n\n\n\n<p>Solution: The magnitude of the average acceleration in every successive interval is<br>average acceleration between 0 s and {tex}1 {~s}=\\frac{(9.8-0) {m} {s}^{-1}}{(1-0) {s}}=9.8 {~m} {~s}^{-2}{\/tex}<br>average acceleration between 1 s and {tex}2 {~s}=\\frac{(19.6-9.8) {m} {s}^{-1}}{(2-1) {s}}=9.8 {~m} {~s}^{-2}{\/tex}<br>average acceleration between 2 s and {tex}3 {~s}=\\frac{(29.4-19.6) {m} {s}^{-1}}{(3-2) {s}}=9.8 {~m} {~s}^{-2}{\/tex}<br>average acceleration between 3 s and {tex}4 {~s}=\\frac{(39.2-29.4) {m} {s}^{-1}}{(4-3) {s}}=9.8 {~m} {~s}^{-2}{\/tex}<br>We note that the average acceleration is constant and equal to 9.8 m s<sup>-2<\/sup>. As the velocity is increasing in the direction of motion, the acceleration is in the direction of motion. This acceleration is called the acceleration due to gravitational force by the Earth and is denoted by g.<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p><strong>Q.13: For a vehicle starting from rest and speeding up, the data for position and time are given in table. Plot the position-time graph corresponding to it.<\/strong><\/p>\n\n\n\n<p><strong>Positions of vehicle at different instants of time<\/strong><\/p>\n\n\n\n<figure class=\"wp-block-table\"><table class=\"has-fixed-layout\"><tbody><tr><td><strong>Time<\/strong><\/td><td><strong>Position<\/strong><\/td><\/tr><tr><td>0 s<\/td><td>0 m<\/td><\/tr><tr><td>2 s<\/td><td>1 m<\/td><\/tr><tr><td>4 s<\/td><td>4 m<\/td><\/tr><tr><td>6 s<\/td><td>9 m<\/td><\/tr><tr><td>8 s<\/td><td>16 m<\/td><\/tr><tr><td>10 s<\/td><td>25 m<\/td><\/tr><tr><td>12 s<\/td><td>36 m<\/td><\/tr><\/tbody><\/table><\/figure>\n\n\n\n<p>Solution: Choosing the scale to be<br>x-axis: 5 divisions = 2 s<br>y-axis: 5 divisions = 5 m<br><img decoding=\"async\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/question_images\/1777265894-6n34jf.jpg\"><\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p><strong>Q.14: What does the graph shown in figure indicate about the nature of motion of the vehicle?<\/strong><br><img decoding=\"async\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/question_images\/1776940222-wtvx7y.jpg\"><\/p>\n\n\n\n<p>Solution: The position of the vehicle is 40 m from the origin and is not changing with time. Thus, the vehicle is at rest at 40 m from the origin. A straight line parallel to the time axis on a position-time graph represents a stationary object (In this case, the position-time graph is not the distance-time graph).<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p><strong>Q.15: The position-time graphs of two objects A and B are given in the figure. Which object of the magnitude of average velocity is higher?<\/strong><br><img decoding=\"async\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/question_images\/1777266218-kspm8m.jpg\"><\/p>\n\n\n\n<p>Solution: By making lines parallel to axes as shown in figure b, it is found that the displacement of object B is more than object A for the same time interval. That is, the slope of line for B is steeper than the slope for line A. Thus, the velocity of B is higher than that of A.<br><img decoding=\"async\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/question_images\/1777266269-ymrpcr.jpg\"><\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p><strong>Q.16: Suppose a car is moving on a highway and brakes are applied, which cause an acceleration of -4 m s<sup>-2<\/sup>. How much will be the distance travelled by the car before coming to a stop, if the car was moving with a velocity of (i) 54 km h<sup>-1<\/sup>, and (ii) 108 km h<sup>-1<\/sup> when the brakes were applied?<\/strong><\/p>\n\n\n\n<p>Solution: Given: {tex}a=-4 {~m} {s}^{-2}, v=0 {~m} {s}^{-1}{\/tex}<br>Suppose the initial velocity is {tex}u{\/tex} and the distance travelled is {tex}s{\/tex}.<\/p>\n\n\n\n<ol start=\"1\" class=\"wp-block-list\">\n<li>{tex}u=54 {~km} {h}^{-1}=15 {~m} {s}^{-1}{\/tex}<\/li>\n\n\n\n<li>{tex}u=108 {~km} {h}^{-1}=30 {~m} {s}^{-1}{\/tex}<\/li>\n<\/ol>\n\n\n\n<p>Using Eq.&nbsp;{tex}v^2=u^2+2 a s{\/tex}<br>{tex}\\left(0 {~m} {~s}^{-1}\\right)^2=u^2+2 \\times\\left(-4 {~m} {~s}^{-2}\\right) \\times s{\/tex}<br>{tex}0=u^2-8 \\times s \\Rightarrow s=\\frac{u^2}{8}{\/tex}<\/p>\n\n\n\n<p>Substituting the value of {tex}u{\/tex}, we obtain (i) 28.1 m, and (ii) 112.5 m.<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p><strong>Q.17: 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?<\/strong><\/p>\n\n\n\n<p>Solution: Total distance = 250 + 250 + 250 + 250 = 1000 m<br>Displacement = 0 m (came back to starting point)<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p><strong>Q.18: 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:<\/strong><\/p>\n\n\n\n<ol start=\"1\" class=\"wp-block-list\">\n<li>the total vertical distance travelled, and<\/li>\n\n\n\n<li>their displacement from the starting point.<\/li>\n<\/ol>\n\n\n\n<p>Solution: <strong>Given:<\/strong> Each floor = 3 m<\/p>\n\n\n\n<p>Ground \u2192 4th floor = 4 \u00d7 3 = 12 m<br>4th \u2192 2nd floor = 2 \u00d7 3 = 6 m<\/p>\n\n\n\n<ol start=\"1\" class=\"wp-block-list\">\n<li>Total distance = 12 + 6 = 18 m<\/li>\n\n\n\n<li>Displacement = final &#8211; initial = 2nd floor \u2212 ground = 6 m upward<\/li>\n<\/ol>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p><strong>Q.19: 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?<\/strong><\/p>\n\n\n\n<p>Solution: Yes, it is possible. Even if the speed (speedometer reading) is constant, the scooter can still be accelerating if its direction is changing. For example, when moving in a circular path, velocity changes continuously due to change in direction, so acceleration exists even with constant speed.<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p><strong>Q.20: A car starts from rest and its velocity reaches 24 m s<sup>-1<\/sup> in 6 s. Find the average acceleration and the distance travelled in these 6 s.<\/strong><\/p>\n\n\n\n<p>Solution: <strong>Given:<\/strong><\/p>\n\n\n\n<p>{tex} {u}=0 {~m} \/ {s}, {v}=24 {~m} \/ {s}, {t}=6 {~s}{\/tex}<\/p>\n\n\n\n<ol start=\"1\" class=\"wp-block-list\">\n<li>Average acceleration<br>{tex} {a}=({v}-{u}) \/ {t}=(24-0) \/ 6=4 {~m} \/ {s}^2{\/tex}<\/li>\n\n\n\n<li>Distance travelled<br>{tex}s=u t+1 \/ 2 a t^2{\/tex}<br>{tex}s=0+1 \/ 2 \\times 4 \\times 6^2=2 \\times 36=72 {~m}{\/tex}<\/li>\n<\/ol>\n\n\n\n<p><strong>Acceleration {tex}=4 {~m} \/ {s}^2{\/tex}<br>Distance {tex}=72 {~m}{\/tex}<\/strong><\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p><strong>Q.21: A motorbike moving with initial velocity 28 m s<sup>-1<\/sup> and constant acceleration stops after travelling 98 m. Find the acceleration of the motorbike and the time taken to come to a stop.<\/strong><\/p>\n\n\n\n<p>Solution: <strong>Given:<\/strong><\/p>\n\n\n\n<p>{tex}u=28 {~m} \/ {s}{\/tex}<br>{tex}v=0 {~m} \/ {s}{\/tex}<br>{tex}s=98 {~m}{\/tex}<\/p>\n\n\n\n<ol start=\"1\" class=\"wp-block-list\">\n<li><strong>Acceleration<\/strong> Using: {tex}v^2=u^2+2 a s{\/tex}<br>{tex}0=28^2+2 a(98){\/tex}<br>{tex}0=784+196 a{\/tex}<br>{tex}196 a=-784{\/tex}<br>{tex}a=-4 {~m} \/ {s}^2{\/tex}<\/li>\n\n\n\n<li><strong>Time<\/strong> Using:<br>{tex}{v}={u}+{at}{\/tex} {tex}0=28+(-4) t{\/tex}<br>{tex}4 t=28{\/tex}<br>t = 7s<\/li>\n<\/ol>\n\n\n\n<p>Acceleration = -4 m\/s<sup>2<\/sup> (retardation)<\/p>\n\n\n\n<p>Time taken = 7 s<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p><strong>Q.22: 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.<\/strong><br><img decoding=\"async\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/question_images\/1776942219-fxypyg.jpg\"><\/p>\n\n\n\n<p>Solution: Velocity is the slope of a position\u2013time graph. In the given graph, objects A and B have different slopes at all points.<\/p>\n\n\n\n<p>Since their slopes are never equal, their velocities are never equal at any time.<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p><strong>Q.23: A graph in the figure 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).<\/strong><br><img decoding=\"async\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/question_images\/1776942567-ghpqba.jpg\"><\/p>\n\n\n\n<p>Options:<br>(1) The average velocity of both over the 10 s time interval is equal since they have the same initial and final positions. \u2705<br>(2) The average speeds of both over the 10 s time interval are equal since both cover equal distance in equal time.<br>(3) 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.<br>(4) 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<p>Explanation: From the position-time graph: both A and B start and end at the same position, so displacement is same \u2192 average velocity same. But distance covered is different (B is curved, so longer path).<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p><strong>Q.24: A truck driver driving at the speed of 54 km h<sup>-1<\/sup> notices a road sign with a speed limit of 40 km h<sup>-1<\/sup>\u00a0(figure) for trucks. He slows down to 36 km h<sup>-1<\/sup> in 36 s. What was the distance travelled by him during this time? Assume the acceleration to be constant while slowing down.<\/strong><br><img decoding=\"async\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/question_images\/1776942928-zcu7t7.jpg\"><\/p>\n\n\n\n<p>Solution: Convert speeds to {tex}{m} \/ {s}{\/tex}:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Initial speed, {tex}u=54 {~km} \/ {h}=15 {~m} \/ {s}{\/tex}<\/li>\n\n\n\n<li>Final speed, {tex}v=36 {~km} \/ {h}=10 {~m} \/ {s}{\/tex}<\/li>\n\n\n\n<li>Time, {tex}t=36 {~s}{\/tex}<\/li>\n<\/ul>\n\n\n\n<p>For constant acceleration, distance travelled:<\/p>\n\n\n\n<p>{tex}s=\\frac{u+v}{2} \\times t{\/tex}<br>{tex}s=\\frac{15+10}{2} \\times 36=\\frac{25}{2} \\times 36=12.5 \\times 36=450 {~m}{\/tex}<\/p>\n\n\n\n<p><strong>Distance travelled {tex}=450 {~m}{\/tex}<\/strong><\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p><strong>Q.25: A car starts from rest and accelerates uniformly to 20 m s<sup>-1<\/sup> in 5 seconds. It then travels at 20 m s<sup>-1<\/sup> for 10 seconds and finally applies the brake (with uniform acceleration) to stop in 6 seconds. Find the total distance travelled.<\/strong><\/p>\n\n\n\n<p>Solution:<\/p>\n\n\n\n<ol start=\"1\" class=\"wp-block-list\">\n<li>Acceleration phase (0 to {tex}20 {~m} \/ {s}{\/tex} in 5 s):<br>{tex} s_1=\\frac{u+v}{2} t=\\frac{0+20}{2} \\times 5=10 \\times 5=50 {~m}{\/tex}<\/li>\n\n\n\n<li>Constant speed phase ( {tex}20 {~m} \/ {s}{\/tex} for 10 s):<br>{tex} s_2=v t=20 \\times 10=200 {~m}{\/tex}<\/li>\n\n\n\n<li>Braking phase (20 to {tex}0 {~m} \/ {s}{\/tex} in 6 s):<br>{tex} s_3=\\frac{20+0}{2} \\times 6=10 \\times 6=60 {~m}{\/tex}<\/li>\n<\/ol>\n\n\n\n<p><strong>Total distance:<\/strong><\/p>\n\n\n\n<p>{tex} s=50+200+60=310 {~m}{\/tex}<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p><strong>Q.26: A bus is travelling at 36 km h<sup>-1<\/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<sup>-2<\/sup>. Will the bus be able to stop before reaching the obstacle?<\/strong><\/p>\n\n\n\n<p>Solution: <strong>Step 1: Convert speed<\/strong><\/p>\n\n\n\n<p>{tex} 36 {~km} \/ {h}=10 {~m} \/ {s}{\/tex}<\/p>\n\n\n\n<p><strong>Step 2: Reaction distance<\/strong><br>During reaction time (0.5 s), speed is constant:<\/p>\n\n\n\n<p>{tex} s_1=v t=10 \\times 0.5=5 {~m}{\/tex}<\/p>\n\n\n\n<p><strong>Step 3: Braking distance<\/strong><br>Initial speed {tex}u=10 {~m} \/ {s}{\/tex}, final speed {tex}v=0{\/tex}, acceleration {tex}a=-2.5 {~m} \/ {s}^2{\/tex}<br>Use:<br>{tex}v^2=u^2+2 a s{\/tex}<br>{tex}0=100+2(-2.5) s{\/tex}<br>{tex}0=100-5 s{\/tex}<br>{tex}s_2=20 {~m}{\/tex}<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p><strong>Q.27: 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.<\/strong><\/p>\n\n\n\n<p>Solution: An object kept on Earth cannot be considered truly at rest because motion is relative.<\/p>\n\n\n\n<p>Even though the student says \u201cthe Earth moves around the Sun,\u201d the object shares the same motion as the Earth. So, relative to the Earth\u2019s surface, it is at rest. However, relative to the Sun (or space), it is continuously moving along with the Earth\u2019s orbital motion.<\/p>\n\n\n\n<p><strong>Conclusion:<\/strong><br>An object on Earth is at rest only in the Earth\u2019s frame of reference, but not in an absolute sense.<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p><strong>Q.28: The velocity-time graph from 0 s to 120 s for a cyclist is shown in the figure.<\/strong><br><img decoding=\"async\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/question_images\/1776944619-kyhefk.jpg\"><br>Shade the areas (in different colours) representing the displacement of the cyclist<\/p>\n\n\n\n<ol start=\"1\" class=\"wp-block-list\">\n<li>while cyclist is moving with constant velocity.<\/li>\n\n\n\n<li>when the velocity of cyclist is decreasing.<\/li>\n<\/ol>\n\n\n\n<p>Also, calculate the displacement and average acceleration in the 120 s time interval.<\/p>\n\n\n\n<p>Solution: Displacement is found from the area under a velocity-time graph. The cyclist moves with constant velocity for the first 40 s, so the area is a rectangle: {tex}20 \\times 40=800 {~m}{\/tex}. From 40 s to 120 s, velocity decreases uniformly to zero, forming a triangle with area {tex}\\frac{1}{2} \\times 80 \\times 20=800 {~m}{\/tex}. Total displacement {tex}=1600 {~m}{\/tex}. Average acceleration {tex}=(0- 20) \/ 120=-0.167 {~m} \/ {s}^2{\/tex}.<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p><strong>Q.29: 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\u00a0depicts her velocity versus time. Estimate the running distance based on the graph.<\/strong><br><img decoding=\"async\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/question_images\/1776944801-tks82p.jpg\"><\/p>\n\n\n\n<p>Solution: To estimate distance, we calculate the area under the velocity\u2013time graph (since displacement = area under v\u2013t graph).<\/p>\n\n\n\n<p>From the graph:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>First part (constant velocity): rectangle area<\/li>\n\n\n\n<li>Second part (changing velocity): triangle area<\/li>\n<\/ul>\n\n\n\n<p>Adding both areas gives the total running distance.<\/p>\n\n\n\n<p>Estimated distance \u2248 900 m (from graph area).<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p><strong>Q.30: On entering a state highway, a car continues to move with a constant velocity of 6 m s<sup>-1<\/sup> for 2 minutes and then accelerates with a constant acceleration 1 m s<sup>-2<\/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.<\/strong><\/p>\n\n\n\n<p>Solution: <strong>Given:<\/strong><\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Constant velocity {tex}=6 {~m} \/ {s}{\/tex} for {tex}2 {~min}=120 {~s}{\/tex}<\/li>\n\n\n\n<li>Acceleration {tex}=1 {~m} \/ {s}^2{\/tex} for 6 s<\/li>\n\n\n\n<li>Initial velocity for acceleration phase {tex}=6 {~m} \/ {s}{\/tex}<\/li>\n<\/ul>\n\n\n\n<ol start=\"1\" class=\"wp-block-list\">\n<li><strong>(constant velocity)<\/strong><br>{tex} s_1=v t=6 \\times 120=720 {~m}{\/tex}<\/li>\n\n\n\n<li><strong>(uniform acceleration)<\/strong><br>Final velocity:<br>{tex} v=u+a t=6+(1 \\times 6)=12 {~m} \/ {s}{\/tex}<br>Displacement:<br>{tex} s_2=\\frac{u+v}{2} \\times t=\\frac{6+12}{2} \\times 6=9 \\times 6=54 {~m}{\/tex}<\/li>\n\n\n\n<li><strong>Total displacement<\/strong> {tex} s=720+54=774 {~m}{\/tex}<\/li>\n<\/ol>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p><strong>Q.31: Two cars A and B start moving with a constant acceleration from rest in a straight line. Car A attains a velocity of 5 m s<sup>-1<\/sup> in 5 s. Car B attains a velocity of 3 m s<sup>-1<\/sup> in 10 s. Plot the velocity-time graphs for both the cars in the same graph. Using the graph, calculate the displacement mentioned in the two time intervals (Hint: Calculate the acceleration in both cases. Then calculate their velocities at five instants of time to plot the graph).<\/strong><\/p>\n\n\n\n<p>Solution: <strong>Step 1: Find accelerations<\/strong><br>Car A:\u00a0{tex} a_A=\\frac{5}{5}=1 {~m} \/ {s}^2{\/tex}<br>Car B:\u00a0{tex} a_B=\\frac{3}{10}=0.3 {~m} \/ {s}^2{\/tex}<br><strong>Step 2: Velocities at 5 s intervals<\/strong><br>{tex}\\operatorname{CarA}({v}={at}):{\/tex}<\/p>\n\n\n\n<p>{tex} t=0,1,2,3,4,5 \\rightarrow v=0,1,2,3,4,5{\/tex}<\/p>\n\n\n\n<p>Car B ({tex}{v}=0.3 {t}{\/tex}):<\/p>\n\n\n\n<p>{tex} t=0,2,4,6,8,10 \\rightarrow v=0,0.6,1.2,1.8,2.4,3{\/tex}<\/p>\n\n\n\n<p>(Plot both straight-line graphs on same axes.)<\/p>\n\n\n\n<p><strong>Step 3: Displacement (area under v\u2013t graph)<\/strong><br>Since both are triangles:<br>Car A:&nbsp;{tex}s_A=12.5 \\mathrm{~m}{\/tex}<br>Car B:&nbsp;{tex}s_B=15 \\mathrm{~m}{\/tex}<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p><strong>Q.32: Rohan studies science from 6 PM to 7:30 PM at home. Consider the tip of the minute\u2019s hand of the wall clock.<\/strong><br><img decoding=\"async\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/question_images\/1776945716-n7kkup.jpg\"><br>During the given time interval, what is its:<\/p>\n\n\n\n<ol start=\"1\" class=\"wp-block-list\">\n<li>distance travelled,<\/li>\n\n\n\n<li>displacement,<\/li>\n\n\n\n<li>speed, and<\/li>\n\n\n\n<li>velocity.<\/li>\n<\/ol>\n\n\n\n<p>The length of the minute\u2019s hand is 7 cm.<\/p>\n\n\n\n<p>Solution: Time interval = 6:00 PM to 7:30 PM = 90 min<br>Radius of minute hand = 7 cm<\/p>\n\n\n\n<p>The minute hand completes 1 rotation in 60 min \u2192 in 90 min it completes 1.5 rotations.<\/p>\n\n\n\n<ol start=\"1\" class=\"wp-block-list\">\n<li><strong>Distance travelled<\/strong> {tex} \\text { Distance }=1.5 \\times 2 \\pi r={\/tex}\u00a0{tex}1.5 \\times 2 \\pi \\times 7=21 \\pi {~cm}{\/tex}<\/li>\n\n\n\n<li><strong>Displacement<\/strong> At 6:00 \u2192 tip at 12<br>At 7:30 \u2192 tip at 6 (opposite point)<br>So displacement = diameter: {tex} =2 r=14 {~cm}{\/tex}<\/li>\n\n\n\n<li><strong>Speed<\/strong> {tex} \\text { Speed }=\\frac{21 \\pi}{90}=\\frac{7 \\pi}{30} {~cm} \/ {min}{\/tex}<\/li>\n\n\n\n<li><strong>Velocity<\/strong><br>Velocity {tex}=\\frac{14}{90}=\\frac{7}{45} {~cm} \/ {min}{\/tex}<br>(direction from 12 to 6 in a straight line)<\/li>\n<\/ol>\n","protected":false},"excerpt":{"rendered":"<p>Describing Motion Around Us &#8211; NCERT Solutions Class 9 Science Exploration includes all the questions with solutions given in NCERT Class 9 Science Exploration textbook. NCERT Solutions Class 9 Describing Motion Around Us \u2013 NCERT Solutions Q.1: How much distance should we maintain from the truck ahead to avoid a collision if it suddenly applies &#8230; <a title=\"Describing Motion Around Us &#8211; NCERT Solutions Class 9 Science (Exlporation)\" class=\"read-more\" href=\"https:\/\/mycbseguide.com\/blog\/describing-motion-around-us-ncert-solutions-class-9-science-exlporation\/\" aria-label=\"More on Describing Motion Around Us &#8211; NCERT Solutions Class 9 Science (Exlporation)\">Read more<\/a><\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"closed","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[281,2071,2072],"tags":[216],"class_list":["post-31530","post","type-post","status-publish","format-standard","hentry","category-ncert-solutions","category-ncert-solutions-class-9","category-ncert-solutions-class-9-science-exploration","tag-ncert-solutions"],"yoast_head":"<!-- This site is optimized with the Yoast SEO plugin v26.0 - https:\/\/yoast.com\/wordpress\/plugins\/seo\/ -->\n<title>Describing Motion Around Us - NCERT Solutions Class 9 Science (Exlporation) | myCBSEguide<\/title>\n<meta name=\"description\" content=\"Describing Motion Around Us - NCERT Solutions Class 9 Science Exploration includes all the questions with solutions given in NCERT book.\" \/>\n<meta name=\"robots\" content=\"index, follow, max-snippet:-1, max-image-preview:large, max-video-preview:-1\" \/>\n<link rel=\"canonical\" href=\"https:\/\/mycbseguide.com\/blog\/describing-motion-around-us-ncert-solutions-class-9-science-exlporation\/\" \/>\n<meta property=\"og:locale\" content=\"en_US\" \/>\n<meta property=\"og:type\" content=\"article\" \/>\n<meta property=\"og:title\" content=\"Describing Motion Around Us - NCERT Solutions Class 9 Science (Exlporation) | myCBSEguide\" \/>\n<meta property=\"og:description\" content=\"Describing Motion Around Us - NCERT Solutions Class 9 Science Exploration includes all the questions with solutions given in NCERT book.\" \/>\n<meta property=\"og:url\" content=\"https:\/\/mycbseguide.com\/blog\/describing-motion-around-us-ncert-solutions-class-9-science-exlporation\/\" \/>\n<meta property=\"og:site_name\" content=\"myCBSEguide\" \/>\n<meta property=\"article:publisher\" content=\"https:\/\/www.facebook.com\/mycbseguide\/\" \/>\n<meta property=\"article:published_time\" content=\"2026-05-20T11:16:52+00:00\" \/>\n<meta property=\"article:modified_time\" content=\"2026-05-20T11:18:16+00:00\" \/>\n<meta property=\"og:image\" content=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/question_images\/1776934767-zquhve.jpg\" \/>\n<meta name=\"author\" content=\"myCBSEguide\" \/>\n<meta name=\"twitter:card\" content=\"summary_large_image\" \/>\n<meta name=\"twitter:creator\" content=\"@mycbseguide\" \/>\n<meta name=\"twitter:site\" content=\"@mycbseguide\" \/>\n<meta name=\"twitter:label1\" content=\"Written by\" \/>\n\t<meta name=\"twitter:data1\" content=\"myCBSEguide\" \/>\n\t<meta name=\"twitter:label2\" content=\"Est. reading time\" \/>\n\t<meta name=\"twitter:data2\" content=\"19 minutes\" \/>\n<script type=\"application\/ld+json\" class=\"yoast-schema-graph\">{\"@context\":\"https:\/\/schema.org\",\"@graph\":[{\"@type\":\"Article\",\"@id\":\"https:\/\/mycbseguide.com\/blog\/describing-motion-around-us-ncert-solutions-class-9-science-exlporation\/#article\",\"isPartOf\":{\"@id\":\"https:\/\/mycbseguide.com\/blog\/describing-motion-around-us-ncert-solutions-class-9-science-exlporation\/\"},\"author\":{\"name\":\"myCBSEguide\",\"@id\":\"https:\/\/mycbseguide.com\/blog\/#\/schema\/person\/10b8c7820ff29025ab8524da7c025f65\"},\"headline\":\"Describing Motion Around Us &#8211; NCERT Solutions Class 9 Science (Exlporation)\",\"datePublished\":\"2026-05-20T11:16:52+00:00\",\"dateModified\":\"2026-05-20T11:18:16+00:00\",\"mainEntityOfPage\":{\"@id\":\"https:\/\/mycbseguide.com\/blog\/describing-motion-around-us-ncert-solutions-class-9-science-exlporation\/\"},\"wordCount\":3566,\"publisher\":{\"@id\":\"https:\/\/mycbseguide.com\/blog\/#organization\"},\"image\":{\"@id\":\"https:\/\/mycbseguide.com\/blog\/describing-motion-around-us-ncert-solutions-class-9-science-exlporation\/#primaryimage\"},\"thumbnailUrl\":\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/question_images\/1776934767-zquhve.jpg\",\"keywords\":[\"NCERT Solutions\"],\"articleSection\":[\"NCERT Solutions\",\"NCERT Solutions Class 9\",\"NCERT Solutions Class 9 Science Exploration\"],\"inLanguage\":\"en-US\"},{\"@type\":\"WebPage\",\"@id\":\"https:\/\/mycbseguide.com\/blog\/describing-motion-around-us-ncert-solutions-class-9-science-exlporation\/\",\"url\":\"https:\/\/mycbseguide.com\/blog\/describing-motion-around-us-ncert-solutions-class-9-science-exlporation\/\",\"name\":\"Describing Motion Around Us - NCERT Solutions Class 9 Science (Exlporation) | myCBSEguide\",\"isPartOf\":{\"@id\":\"https:\/\/mycbseguide.com\/blog\/#website\"},\"primaryImageOfPage\":{\"@id\":\"https:\/\/mycbseguide.com\/blog\/describing-motion-around-us-ncert-solutions-class-9-science-exlporation\/#primaryimage\"},\"image\":{\"@id\":\"https:\/\/mycbseguide.com\/blog\/describing-motion-around-us-ncert-solutions-class-9-science-exlporation\/#primaryimage\"},\"thumbnailUrl\":\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/question_images\/1776934767-zquhve.jpg\",\"datePublished\":\"2026-05-20T11:16:52+00:00\",\"dateModified\":\"2026-05-20T11:18:16+00:00\",\"description\":\"Describing Motion Around Us - NCERT Solutions Class 9 Science Exploration includes all the questions with solutions given in NCERT book.\",\"breadcrumb\":{\"@id\":\"https:\/\/mycbseguide.com\/blog\/describing-motion-around-us-ncert-solutions-class-9-science-exlporation\/#breadcrumb\"},\"inLanguage\":\"en-US\",\"potentialAction\":[{\"@type\":\"ReadAction\",\"target\":[\"https:\/\/mycbseguide.com\/blog\/describing-motion-around-us-ncert-solutions-class-9-science-exlporation\/\"]}]},{\"@type\":\"ImageObject\",\"inLanguage\":\"en-US\",\"@id\":\"https:\/\/mycbseguide.com\/blog\/describing-motion-around-us-ncert-solutions-class-9-science-exlporation\/#primaryimage\",\"url\":\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/question_images\/1776934767-zquhve.jpg\",\"contentUrl\":\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/question_images\/1776934767-zquhve.jpg\"},{\"@type\":\"BreadcrumbList\",\"@id\":\"https:\/\/mycbseguide.com\/blog\/describing-motion-around-us-ncert-solutions-class-9-science-exlporation\/#breadcrumb\",\"itemListElement\":[{\"@type\":\"ListItem\",\"position\":1,\"name\":\"Home\",\"item\":\"https:\/\/mycbseguide.com\/blog\/\"},{\"@type\":\"ListItem\",\"position\":2,\"name\":\"NCERT Solutions\",\"item\":\"https:\/\/mycbseguide.com\/blog\/category\/ncert-solutions\/\"},{\"@type\":\"ListItem\",\"position\":3,\"name\":\"NCERT Solutions Class 9\",\"item\":\"https:\/\/mycbseguide.com\/blog\/category\/ncert-solutions\/ncert-solutions-class-9\/\"},{\"@type\":\"ListItem\",\"position\":4,\"name\":\"NCERT Solutions Class 9 Science Exploration\",\"item\":\"https:\/\/mycbseguide.com\/blog\/category\/ncert-solutions\/ncert-solutions-class-9\/ncert-solutions-class-9-science-exploration\/\"},{\"@type\":\"ListItem\",\"position\":5,\"name\":\"Describing Motion Around Us &#8211; NCERT Solutions Class 9 Science (Exlporation)\"}]},{\"@type\":\"WebSite\",\"@id\":\"https:\/\/mycbseguide.com\/blog\/#website\",\"url\":\"https:\/\/mycbseguide.com\/blog\/\",\"name\":\"myCBSEguide\",\"description\":\"\",\"publisher\":{\"@id\":\"https:\/\/mycbseguide.com\/blog\/#organization\"},\"potentialAction\":[{\"@type\":\"SearchAction\",\"target\":{\"@type\":\"EntryPoint\",\"urlTemplate\":\"https:\/\/mycbseguide.com\/blog\/?s={search_term_string}\"},\"query-input\":{\"@type\":\"PropertyValueSpecification\",\"valueRequired\":true,\"valueName\":\"search_term_string\"}}],\"inLanguage\":\"en-US\"},{\"@type\":\"Organization\",\"@id\":\"https:\/\/mycbseguide.com\/blog\/#organization\",\"name\":\"myCBSEguide\",\"url\":\"https:\/\/mycbseguide.com\/blog\/\",\"logo\":{\"@type\":\"ImageObject\",\"inLanguage\":\"en-US\",\"@id\":\"https:\/\/mycbseguide.com\/blog\/#\/schema\/logo\/image\/\",\"url\":\"https:\/\/mycbseguide.com\/blog\/wp-content\/uploads\/2016\/04\/books_square.png\",\"contentUrl\":\"https:\/\/mycbseguide.com\/blog\/wp-content\/uploads\/2016\/04\/books_square.png\",\"width\":180,\"height\":180,\"caption\":\"myCBSEguide\"},\"image\":{\"@id\":\"https:\/\/mycbseguide.com\/blog\/#\/schema\/logo\/image\/\"},\"sameAs\":[\"https:\/\/www.facebook.com\/mycbseguide\/\",\"https:\/\/x.com\/mycbseguide\",\"https:\/\/www.linkedin.com\/company\/mycbseguide\/\",\"http:\/\/in.pinterest.com\/mycbseguide\/\",\"https:\/\/www.youtube.com\/channel\/UCxuqSnnygFzwJG0pwogCNEQ\"]},{\"@type\":\"Person\",\"@id\":\"https:\/\/mycbseguide.com\/blog\/#\/schema\/person\/10b8c7820ff29025ab8524da7c025f65\",\"name\":\"myCBSEguide\",\"sameAs\":[\"http:\/\/mycbseguide.com\"]}]}<\/script>\n<!-- \/ Yoast SEO plugin. -->","yoast_head_json":{"title":"Describing Motion Around Us - NCERT Solutions Class 9 Science (Exlporation) | myCBSEguide","description":"Describing Motion Around Us - NCERT Solutions Class 9 Science Exploration includes all the questions with solutions given in NCERT book.","robots":{"index":"index","follow":"follow","max-snippet":"max-snippet:-1","max-image-preview":"max-image-preview:large","max-video-preview":"max-video-preview:-1"},"canonical":"https:\/\/mycbseguide.com\/blog\/describing-motion-around-us-ncert-solutions-class-9-science-exlporation\/","og_locale":"en_US","og_type":"article","og_title":"Describing Motion Around Us - NCERT Solutions Class 9 Science (Exlporation) | myCBSEguide","og_description":"Describing Motion Around Us - NCERT Solutions Class 9 Science Exploration includes all the questions with solutions given in NCERT book.","og_url":"https:\/\/mycbseguide.com\/blog\/describing-motion-around-us-ncert-solutions-class-9-science-exlporation\/","og_site_name":"myCBSEguide","article_publisher":"https:\/\/www.facebook.com\/mycbseguide\/","article_published_time":"2026-05-20T11:16:52+00:00","article_modified_time":"2026-05-20T11:18:16+00:00","og_image":[{"url":"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/question_images\/1776934767-zquhve.jpg","type":"","width":"","height":""}],"author":"myCBSEguide","twitter_card":"summary_large_image","twitter_creator":"@mycbseguide","twitter_site":"@mycbseguide","twitter_misc":{"Written by":"myCBSEguide","Est. reading time":"19 minutes"},"schema":{"@context":"https:\/\/schema.org","@graph":[{"@type":"Article","@id":"https:\/\/mycbseguide.com\/blog\/describing-motion-around-us-ncert-solutions-class-9-science-exlporation\/#article","isPartOf":{"@id":"https:\/\/mycbseguide.com\/blog\/describing-motion-around-us-ncert-solutions-class-9-science-exlporation\/"},"author":{"name":"myCBSEguide","@id":"https:\/\/mycbseguide.com\/blog\/#\/schema\/person\/10b8c7820ff29025ab8524da7c025f65"},"headline":"Describing Motion Around Us &#8211; NCERT Solutions Class 9 Science (Exlporation)","datePublished":"2026-05-20T11:16:52+00:00","dateModified":"2026-05-20T11:18:16+00:00","mainEntityOfPage":{"@id":"https:\/\/mycbseguide.com\/blog\/describing-motion-around-us-ncert-solutions-class-9-science-exlporation\/"},"wordCount":3566,"publisher":{"@id":"https:\/\/mycbseguide.com\/blog\/#organization"},"image":{"@id":"https:\/\/mycbseguide.com\/blog\/describing-motion-around-us-ncert-solutions-class-9-science-exlporation\/#primaryimage"},"thumbnailUrl":"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/question_images\/1776934767-zquhve.jpg","keywords":["NCERT Solutions"],"articleSection":["NCERT Solutions","NCERT Solutions Class 9","NCERT Solutions Class 9 Science Exploration"],"inLanguage":"en-US"},{"@type":"WebPage","@id":"https:\/\/mycbseguide.com\/blog\/describing-motion-around-us-ncert-solutions-class-9-science-exlporation\/","url":"https:\/\/mycbseguide.com\/blog\/describing-motion-around-us-ncert-solutions-class-9-science-exlporation\/","name":"Describing Motion Around Us - NCERT Solutions Class 9 Science (Exlporation) | myCBSEguide","isPartOf":{"@id":"https:\/\/mycbseguide.com\/blog\/#website"},"primaryImageOfPage":{"@id":"https:\/\/mycbseguide.com\/blog\/describing-motion-around-us-ncert-solutions-class-9-science-exlporation\/#primaryimage"},"image":{"@id":"https:\/\/mycbseguide.com\/blog\/describing-motion-around-us-ncert-solutions-class-9-science-exlporation\/#primaryimage"},"thumbnailUrl":"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/question_images\/1776934767-zquhve.jpg","datePublished":"2026-05-20T11:16:52+00:00","dateModified":"2026-05-20T11:18:16+00:00","description":"Describing Motion Around Us - NCERT Solutions Class 9 Science Exploration includes all the questions with solutions given in NCERT book.","breadcrumb":{"@id":"https:\/\/mycbseguide.com\/blog\/describing-motion-around-us-ncert-solutions-class-9-science-exlporation\/#breadcrumb"},"inLanguage":"en-US","potentialAction":[{"@type":"ReadAction","target":["https:\/\/mycbseguide.com\/blog\/describing-motion-around-us-ncert-solutions-class-9-science-exlporation\/"]}]},{"@type":"ImageObject","inLanguage":"en-US","@id":"https:\/\/mycbseguide.com\/blog\/describing-motion-around-us-ncert-solutions-class-9-science-exlporation\/#primaryimage","url":"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/question_images\/1776934767-zquhve.jpg","contentUrl":"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/question_images\/1776934767-zquhve.jpg"},{"@type":"BreadcrumbList","@id":"https:\/\/mycbseguide.com\/blog\/describing-motion-around-us-ncert-solutions-class-9-science-exlporation\/#breadcrumb","itemListElement":[{"@type":"ListItem","position":1,"name":"Home","item":"https:\/\/mycbseguide.com\/blog\/"},{"@type":"ListItem","position":2,"name":"NCERT Solutions","item":"https:\/\/mycbseguide.com\/blog\/category\/ncert-solutions\/"},{"@type":"ListItem","position":3,"name":"NCERT Solutions Class 9","item":"https:\/\/mycbseguide.com\/blog\/category\/ncert-solutions\/ncert-solutions-class-9\/"},{"@type":"ListItem","position":4,"name":"NCERT Solutions Class 9 Science Exploration","item":"https:\/\/mycbseguide.com\/blog\/category\/ncert-solutions\/ncert-solutions-class-9\/ncert-solutions-class-9-science-exploration\/"},{"@type":"ListItem","position":5,"name":"Describing Motion Around Us &#8211; NCERT Solutions Class 9 Science (Exlporation)"}]},{"@type":"WebSite","@id":"https:\/\/mycbseguide.com\/blog\/#website","url":"https:\/\/mycbseguide.com\/blog\/","name":"myCBSEguide","description":"","publisher":{"@id":"https:\/\/mycbseguide.com\/blog\/#organization"},"potentialAction":[{"@type":"SearchAction","target":{"@type":"EntryPoint","urlTemplate":"https:\/\/mycbseguide.com\/blog\/?s={search_term_string}"},"query-input":{"@type":"PropertyValueSpecification","valueRequired":true,"valueName":"search_term_string"}}],"inLanguage":"en-US"},{"@type":"Organization","@id":"https:\/\/mycbseguide.com\/blog\/#organization","name":"myCBSEguide","url":"https:\/\/mycbseguide.com\/blog\/","logo":{"@type":"ImageObject","inLanguage":"en-US","@id":"https:\/\/mycbseguide.com\/blog\/#\/schema\/logo\/image\/","url":"https:\/\/mycbseguide.com\/blog\/wp-content\/uploads\/2016\/04\/books_square.png","contentUrl":"https:\/\/mycbseguide.com\/blog\/wp-content\/uploads\/2016\/04\/books_square.png","width":180,"height":180,"caption":"myCBSEguide"},"image":{"@id":"https:\/\/mycbseguide.com\/blog\/#\/schema\/logo\/image\/"},"sameAs":["https:\/\/www.facebook.com\/mycbseguide\/","https:\/\/x.com\/mycbseguide","https:\/\/www.linkedin.com\/company\/mycbseguide\/","http:\/\/in.pinterest.com\/mycbseguide\/","https:\/\/www.youtube.com\/channel\/UCxuqSnnygFzwJG0pwogCNEQ"]},{"@type":"Person","@id":"https:\/\/mycbseguide.com\/blog\/#\/schema\/person\/10b8c7820ff29025ab8524da7c025f65","name":"myCBSEguide","sameAs":["http:\/\/mycbseguide.com"]}]}},"_links":{"self":[{"href":"https:\/\/mycbseguide.com\/blog\/wp-json\/wp\/v2\/posts\/31530","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/mycbseguide.com\/blog\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/mycbseguide.com\/blog\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/mycbseguide.com\/blog\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/mycbseguide.com\/blog\/wp-json\/wp\/v2\/comments?post=31530"}],"version-history":[{"count":1,"href":"https:\/\/mycbseguide.com\/blog\/wp-json\/wp\/v2\/posts\/31530\/revisions"}],"predecessor-version":[{"id":31531,"href":"https:\/\/mycbseguide.com\/blog\/wp-json\/wp\/v2\/posts\/31530\/revisions\/31531"}],"wp:attachment":[{"href":"https:\/\/mycbseguide.com\/blog\/wp-json\/wp\/v2\/media?parent=31530"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/mycbseguide.com\/blog\/wp-json\/wp\/v2\/categories?post=31530"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/mycbseguide.com\/blog\/wp-json\/wp\/v2\/tags?post=31530"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}