Class 9 solutions in science is a beneficial reference and guiding solution that helps students clear doubts instantly, in an effective way. 1. Neither of the statements are true. Fig 8.11 shows the distance-time graph of three objects A,B and C. Study the graph and answer the following questions: (a) Which of the three is travelling the fastest? You will find all the solutions in a detailed manner and without any mistake. Since a = 0.1 m.s-2, v = 12 m.s-1, u = 0 m.s-1, and t = 120s, the following value for s (distance) can be obtained. Therefore, the total acceleration of the bus is -1.112m.s-2. which will help you to revise the complete Syllabus and score more marks in your examinations. Joseph jogs from one end A to the other end B of a straight 300 m road in 2 minutes 30 seconds and then turns around and jogs 100 m back to point C in another 1 minute. NCERT for Class 9 Science Solutions are made available in both web and PDF format for ease of access. 4. Physics is one of the most important subjects in Class 9 CBSE Board examination. What distance will it cover in 10 s after start? The. It refers to the displacement of a given object over a time interval. Therefore, the net displacement will be equal to the diameter of the track, which is 200m. 1. Uniform and non-uniform motions of objects are explained through the graph and examples. Therefore, Abduls average speed for the entire trip is 24 kilometers per hour. Distance traveled to reach the school = distance traveled to reach home = d (say), therefore, average speed while going to school = total distance travelled/ total time taken = d/t1 = 20 kmph, Average speed while going home = total distance travelled/ total time taken = d/t2= 30 kmph, Now, the average speed for the entire trip is given by total distance travelled/ total time taken, = (d+d)/(t1+t2)kmph = (d+d)/(d/20+d/30)kmph. It refers to the distance moved by an object over a time interval. 1. All Chapter solutions are latest and up-to-date. Given, initial velocity (u) = 0 (the trolley begins from the rest position), Therefore, terminal velocity of the trolley (v) = 0 + (0.02 ms-2)(3s)= 0.06 ms-2, Therefore, the velocity of the trolley after 3 seconds will be 6 cm.s-2. Distance-time graph and velocity-time graph, which are considered as the important concepts for examination are explained in an easy way in NCERT Solutions. (b) Since the speed of the car does not change from the points (x=6) and (x=10), the car is said to be in uniform motion from the 6th to the 10th second. We provide its users access to a profuse supply of questions with their solutions. Refer NCERT Solutions for Class 9 for best scores in board and competitive exams. 2. (a) Find how far does the car travel in the first 4 seconds. If the initial position of the farmer is at the mid-point between two adjacent corners of the square, the net displacement of the farmer would be equal to the side of the square, which is 10m. The Lakhmir Singh Class 9 Physics Solutions for all chapters are provided here so that students can prepare for their examination more effectively. Later, he became a Professor at that college. Students who aspire to make a career in the medical must prepare for this subject efficiently to score good marks in the Class 9 Board examination. (c) an object moving in a certain direction with an acceleration in the perpendicular direction. Degree from Delhi University. It can be calculated as: (1/2)*4*6 = 12 meters. Our experts make the chapters interesting so that you enjoy your studying process and score well without stress. Terminal velocity (v) = 0 m/s (since the stone will reach a position of rest at the point of maximum height), Acceleration = 10 ms-2 in the direction opposite to the trajectory of the stone = -10 ms-2, As per the third motion equation, v2 – u2 = 2as, Therefore, the distance traveled by the stone (s) = (02 – 52)/ 2(10), As per the first motion equation, v = u + at, Therefore, time taken by the stone to reach a position of rest (maximum height) = (v – u) /a. Therefore, initial velocity (u) = 0 m/s, Acceleration is given by the equation a=(v-u)/t, Therefore, terminal velocity (v) = (at)+u. Therefore, the net distance covered by the athlete is 2200 meters and the total displacement of the athlete is 200m. A motorboat starting from rest on a lake accelerates in a straight line at a constant rate of 3.0 m s–2 for 8.0 s. How far does the boat travel during this time? The odometer measures the total distance travelled by automobile.