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Tips to study IX - Science - Moving Objects
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Tips to study IX - Science - Moving Objects

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Tips for IX - Science - Moving Objects by Ednexa

Tips for IX - Science - Moving Objects by Ednexa

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  • 1. 9011041155 / 9011031155 • Live Webinars (online lectures) with recordings. • Online Query Solving • Online MCQ tests with detailed solutions • Online Notes and Solved Exercises • Career Counselling 1
  • 2. 9011041155 / 9011031155 Moving Objects Motion A body is said to be in motion if it changes its position with respect to its surroundings and is said to be at rest if it does not change its position with respect to its surroundings. Distance It is the actual path followed by a body between the points in which it moves. Displacement Displacement is the shortest distance between the initial and final points of moment. 2
  • 3. 9011041155 / 9011031155 Speed It is related to distance while velocity is related to displacement Speed = Total distance travelled Total time taken Velocity = Displacement Time Velocity related to speed and direction Change in speed and Direction is same 1. Change in direction and Speed constant 2. Change in Bothe speed and direction Unit S.I. :- m/s C.G.S. – Cm/s 3
  • 4. 9011041155 / 9011031155 Motion The motion in which the object covers equal distance in equal interval of time is known as uniform motion. The motion in which the object covers unequal distance in equal interval of time is known as nonuniform motion Acceleration The rate of change of velocity is known as time A = Change in velocity / time = v - u /t Distance – Time Graphs Uniformly Accelerated 4 for Uniform And
  • 5. 9011041155 / 9011031155 Velocity time graph for uniform motion Velocity time graph for uniformly accelerated motion 5
  • 6. 9011041155 / 9011031155 Equations of motions by graphical method Motion of an object was studied by Newton and then it was summarized in a set of equations of motion. The equations analyse rectilinear motion of uniformly accelerated body. The position of such a body can be well predicted with the help of a set of three equations, called kinematical equations. A body is moving along a straight line with initial velocity ‘u’. After time interval ‘t’ it attains final velocity ‘v’ due to an acceleration ‘a’. Then the set of three equations is given as: v = u + at _________ (1) It represents velocity time relation. 1 2 s = ut + at _________ (2) 2 It describes position time relation. 6
  • 7. 9011041155 / 9011031155 v2 = u2 + 2as _________ (3) It represents relation between position and velocity. These equations can be derived by graphical method. (As shown in fig.) 1. Equation for velocity - time relation. An object having non zero initial velocity u starts from point P. It’s velocity goes on increasing with respective to time and becomes v when it reaches point Q. The change in velocity is at uniform rate. change in velocity a= time a= QP t ∴ QP = at _______ (a) 7
  • 8. 9011041155 / 9011031155 To find change in velocity, draw PR parallel to OS. QS = QR + RS = QR + PT Put QS = v, PT = u, and QR = at, we get. v = at + u v = u + at The velocity - time relation is obtained here. This is the first kinematical equation. 8
  • 9. 9011041155 / 9011031155 2. Equation for position - time relation. Let distance travelled by the object be ‘s’ in the time interval ‘t’ under uniform acceleration ‘a’. In figure, the distance travelled is given by area enclosed within TPQS. s = Area of quadrilateral TPQS. s = area of rectangle PRST + area of triangle PRQ. s = (TS × TP) + 1 (PR × QR) 2 But TP = u, TS = PR = t and QR = at s = ut + 1 a t2 2 This is the second kinematical equation. 9
  • 10. 9011041155 / 9011031155 3. Equation for position Velocity relation: The distance‘s’ travelled by object in time ‘t’ under uniform acceleration ‘a’ is Given by area enclosed within quadrilateral TPQS. s = Area of trapezium TPQS. s= 1 (QS + TP) × TS 2 Put TP = u, QS = v and OS = t ∴s= 1 (v + u) × t 2 __________(a) 10
  • 11. 9011041155 / 9011031155 From velocity time relation we have a= v−u v −u i.e. t = t a (v − u) 1 s = (v + u) × 2 a ∴ 2as = v2 - u2 v2 = u2 + 2as This is third kinematical equation 11
  • 12. 9011041155 / 9011031155 Answer the following Q.1Distinguish between i. Distance and displacement Ans: 12
  • 13. 9011041155 / 9011031155 ii. Speed and velocity Ans: 13
  • 14. 9011041155 / 9011031155 Q.2 Give reasons i. An object at rest can be considered to have uniform motion Ans: • An object is said to be in uniform motion when it has constant speed. • For an object at rest, its speed is always zero. • Hence, an object at rest can be considered to have uniform motion. ii. When a body falls freely to the ground, its motion has a uniform acceleration Ans: • When a body falls freely, the only force acting on it is earth’s gravitational force. • This force is practically uniform near the earth’s surface. • Hence, when a body falls freely to the ground, its motion has a uniform acceleration. 14
  • 15. 9011041155 / 9011031155 Solved Examples 1. A person swims 100 m in 45 second at the start. Then he covers 80 m in the next 45 second and in the last 10 second he covers 250 m. Calculate the average speed. Ans: Given: Total distance = 100 m + 80 m +25 m = 205 m Total time taken = 45 s + 45 s +10 s = 100 s Average speed =? total distance covered Average speed = total time taken = 205 100 = 2.05 m/s 15
  • 16. 9011041155 / 9011031155 2. An athlete runs on a circular track of length 400 m in 25 second and returns to the starting point. Calculate average speed and average velocity. Ans: Given: Total distance covered Total displacement = 400 m = 0 m/s as he returns to Starting point Total time taken = 25 second Average speed = ? Average velocity =? Average speed = total distance covered total time taken 400 = 25 = 16 m/s 16
  • 17. 9011041155 / 9011031155 total distance = total time taken Average velocity 0 = 25 = 0 m/s 17
  • 18. 9011041155 / 9011031155 3. A motorboat starts from rest and moves with uniform acceleration. If it reaches to a velocity of 15 m/s in 5 second, calculate acceleration and distance covered by it in given time. Ans: Given: Initial velocity = 0 m/s Final velocity = 15 m/s Time taken = 5 second Acceleration =? Distance covered = ? Acceleration = v−u t = 15 − 0 5 = 3 m/s2 18
  • 19. 9011041155 / 9011031155 By second kinematical equation, s = ut + 1 2 at 2 s =0+ 1 × 3 × 25 2 = 75 2 = 37.5 m • Ask Your Doubts • For inquiry and registration, call 9011041155 / 9011031155. 19