# describing motion class 9 solutions

NCERT Solution for Class 9 science - motion 107 , Question 4. Obtain a relation for the distance travelled by an object moving with a uniform acceleration in the interval between 4th and 5th seconds. NCERT Solution for Class 9 science - motion 113 , Question 9. At the most, it can become zero. NCERT Solutions Class 9 Science Chapter 8 Motion – Here are all the NCERT solutions for Class 9 Science Chapter 8. : 1), What can you say about the motion of an object whose distance. (i) Part AB Answer: An object that has moved through a distance can have zero displacement. Answer: 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? How are the distances travelled by an object related to the time taken when an object travels equal distances in equal intervals of time? Answer: NCERT Solution for Class 9 science - motion 113 , Question 7. or s = 5u + $$\frac{25}{2}$$a A body is said to have uniform acceleration if it travels in a straight path in such a way that its velocity changes at a uniform rate, i.e., the velocity of a body increases or decreases by equal amounts in an equal interval of time. Shade the area on the graph that represents the distance travelled by the car during the period. Linear velocity = Angular velocity × Radius of circular path. Examine them carefully and state whether the motion of the objects is uniform or non-uniform. It is the angle which is subtended at the centre by an arc having a length equal to the radius of the circle. Initially, object C is 4 blocks away from the origin. Radius of the circular orbit, r= 42250 km, Time taken to revolve around the earth, t= 24 h. Hence, the speed of the artificial satellite is 3.069 km/s. BD = BC – CD, represents the change in velocity in time interval t. Question 6. Consider the following situation. Displacement is the shortest measurable distance between the initial and the final position of an object. Answer: What was the distance of the spaceship from the ground station? NCERT Solution for Class 9 science - motion 112 , Question 5. Measuring the Rate of Motion - Exercise No. (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. In this interval of time, he moves at the opposite end of the initial position. It tells us about the position of the body at any instant of time. But, at the same time, each passenger in a moving bus or train observes, his fellow passengers sitting and not moving. There is no acceleration from A to B. We have been given Question 6. A body starts to slide over a horizontal surface with an initial velocity of 0.5 m/s. What are Joseph’s average speeds and velocities in jogging (a) from A to B and (b) from A to C? The velocity of the body increases by 2 m/s after every second. Can the displacement be greater than the distance travelled by an object? The characteristics of distance-time graph for a non-uniform speed are: Question 7. Total distance = Distance travelled to reach school = d, Total distance = Distance travelled while returning from school = d, Total distance covered in the trip = d + d = 2d, Total time taken, t = Time taken to go to school + Time taken to return to school, Average Speed=2dd20+d30=23+260Average Speed=1205=24  km/h. An artificial satellite is moving in a circular orbit of radius 42250 km. we have Therefore, the graph line OA represents uniform acceleration. Another train B is travelling on the parallel with a speed of 36 kmh-1.

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