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Physics. Session Kinematics - 2 Session Opener REST ! or MOTION.

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Presentation on theme: "Physics. Session Kinematics - 2 Session Opener REST ! or MOTION."— Presentation transcript:

1 Physics

2 Session Kinematics - 2

3 Session Opener REST ! or MOTION

4 Session Objectives

5 Session Objective 1.Rest and motion 2.Distance and displacement 3.Uniform and non-uniform motion 4.Velocity 5.Acceleration 6.Equations of motion

6 Rest and Motion Body at rest : Position constant w.r.t. fixed point as time increases Fixed Point : Origin of a coordinate system Y O 90 0 x X r y P (position) (origin) 

7 What is motion ? Change in position of an object with time, with respect to a given co ordinate system. Motion Actual distance traveled : Curve P 0 P 1 P 2 P 3 P 4. Displacement : Straight line P 0 P 4 directed from P 0 to P 4 P 0 (t=0) P 1 (t=t 1 ) P 2 (t=t 2 ) P 3 (t=t 3 ) P 4 (t=t 4 ) x y

8 Distance and Displacement 1. Distance  Displacement. 1 2 x t o 2. Distance = Displacement (If direction remains same.) 3.Distance is always 0 or+ve. Displacement can be +ve,0 or –ve. 4. Distance always increases with motion.

9 Average Speed P 0 (t=0) P 1 (t=t 1 ) P 2 (t=t 2 ) P 3 (t=t 3 ) P 4 (t=t 4 ) x y

10 Courtesy:www.physicsclassroom.com

11 Instantaneous Speed When time is infinitesimal (=t  0), distance is infinitesimal(=s  0) Instantaneous speed = tt ss tt t ss PoPo P1P1 s Speed : Scalar Unit m/s Dimension LT -1

12 Average Velocity Average velocity is defined as displacement divided by time taken. Nature : vector Dimension : [LT -1 ] Unit : m/s Displacement and average velocity are in same direction

13 Instantaneous Velocity If t= (t 2 – t 1 )is extremely small (t  0) v is instantaneous velocity v is a vector. Unit of v : m/s

14 Uniform Motion (One dimension) Equal displacement (x) traveled in equal time interval (t) v is constant. x = x 0 + vt If x 0 = 0 at t = 0, x = vt x0x0 x1x1 x2x2 x3x3 x4x4 x 0 t1t1 t2t2 t3t3 t4t4 t

15 Class Exercise

16 Class Exercise - 3 Graph in the figure below shows the variation of displacement with respect to time for a particle in one-dimensional motion. Which of the following represents the velocity-time graph of the particle in motion? Options is in next slide

17 Class Exercise - 3 (a) (b) (c) (d)

18 Solution - 3 For t = 0 to 5 Hence answer is (c)

19 Class Exercise - 8 An object travels half the distance with v 1, with v 2 for half of the remaining time and with v 3 for the remaining half of the time. If the object never reverses the direction of motion, find the average velocity during the motion. Solution :

20 Average Acceleration Change in velocity divided by the time interval during which the change occurs. Nature : vector unit : m/s² Dimension: [LT -2 ]

21 Non uniform motion (constant acceleration) Instantaneous acceleration t v v u t1t1 t2t2 t=t 2 -t 1 For constant acceleration v = u + at Equation of motion (1)

22 Class Exercise

23 Class Exercise - 1 Which graph represents increasing acceleration? (a) A(b) B(c) C(d) None of these

24 Solution - 1 Hence answer is (b). Increasing a  Increasing slope of v-t curve. By observation, we find that the velocity is increasing at an increasing rate. So acceleration is increasing.

25 Class Exercise - 9 An object starts from rest. It accelerates at 2 m/s 2 till it reaches its maximum velocity. Then it retards at 4 m/s 2 and finally comes to rest. If the total time taken is 6s, find v max and the displacement of the object.

26 Solution - 9 V max = 2t 1 V max = 4t 2 s = Area of triangle  t 1 = 4 s, t 2 = 2 s Hence, V max = 8 m/s

27 Class Exercise - 2 A particle is thrown vertically upwards with velocity v. It returns to the ground in time T. Which of the following graphs correctly represents the motion? (a) (b) (d) (c)

28 Solution - 2 The acceleration is constant (= –g). So slope has to be negative throughout the motion and velocity varies between v and –v. Hence answer is (c).

29 Class Exercise - 5 If a particle has an initial velocity of and an acceleration of, its speed after 10 s is (a) 10 units(b) 7 units (c) units(d) 8.5 units

30 Solution - 5 Hence answer is (c).

31 One dimensional equations of motion Distance s = area under v-t graph = ½ (u+v)t But, s = v avg t Hence, v avg = (u+v)/2 t v v u t1t1 t2t2 t=t 2 -t 1 Using equation of motion (1) s = ut + ½ at 2 Equation of motion (2) Equation of motion (3)

32 One dimensional equations of motion v = u + at v² = u² + 2as Distance traveled in n th second

33 Class Exercise

34 Class Exercise - 6 A particle moves along the X-axis as x = u(t – 3) + a(t – 3) 2, then Which of the following are true? (a)initial velocity of particle is u at t = 0 (b) acceleration of particle is a (c) at t = 3 the particle was at origin (d) the particle may have negative velocity

35 Solution - 6 The observation of displacement has started at time t = 3 s, after the object has actually started. So if it represents the time for which the object has traveled and s be the displacement after the observation has started, then general form is

36 Class Exercise - 10 The magnitude of maximum acceleration, retardation of an object is ‘a’ m/s 2. What is the minimum time taken by the object to cover a displacement ‘s’ if it starts from rest and finally comes to rest? Solution : The minimum time would be when the acceleration is at maximum and deceleration is also maximum. Half the time accelerating at a and the rest of the time deceleration at ‘a’.

37 Uniform and Non-Uniform Motion Let us see a comparison of uniform and non-uniform motion Courtesy:www.physicsclassroom.com

38 Class Exercise

39 Class Exercise - 7 An object has one-dimensional motion. If V = 6t + 4t 3, then (a) what is the distance covered from t = 3 s to t = 5 s? (b) when is the acceleration < 0 for the first time? So acceleration is never negative. Solution :

40 Class Exercise - 4 A particle moves in a straight line, starting from rest. The acceleration of the particle is given by What is the distance traveled by the particle in the time interval 0 to  seconds.

41 Solution - 4 [Check that v(0) = 0]

42 Solution - 4  x = 2 – log( + 1)

43 Thank you


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