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Physics Session Kinematics - 2 Session Opener or ! REST MOTION Session Objectives Session Objective 1. Rest and motion 2. Distance and displacement 3. Uniform and non-uniform motion 4. Velocity 5. Acceleration 6. Equations of motion Rest and Motion Body at rest : Position constant w.r.t. fixed point as time increases Fixed Point : Origin of a coordinate system Y r ox oy 2 1 2 2 P (position) y r oy tan ox ox r cos oy r sin 900 O (origin) x X Motion What is motion ? Change in position of an object with time, with respect to a given co ordinate system. Actual distance traveled : Curve P0P1P2P3P4. Displacement : Straight line P0P4 directed from P0 to P4 y P4(t=t4) P3(t=t3) P2(t=t2) P1(t=t1) P0(t=0) x Distance and Displacement 1. Distance Displacement. 2. Distance = Displacement (If direction remains same.) 3.Distance is always 0 or+ve. Displacement can be +ve,0 or –ve. x 2 4. Distance always increases with motion. o 1 t Average Speed y P4(t=t4) P3(t=t3) P2(t=t2) P1(t=t1) P0(t=0) x actual dis tan ce cov ered Average speed time taken Curve lengthP0P1P2P3P4 time int erval (t 4 t 0 ) Courtesy:www.physicsclassroom.com Instantaneous Speed When time is infinitesimal (=t 0), distance is infinitesimal(=s 0) Instantaneous speed = Speed : Scalar Unit Lim t 0 s t m/s s Dimension LT-1 P1 s Po s t t t Average Velocity Average velocity is defined as displacement divided by time taken. r r r v av 2 1 t 2 t1 t Nature : vector vav Unit : m/s Dimension : [LT-1] r r2 r1 t t1 t2 Displacement and average velocity are in same direction Instantaneous Velocity If t= (t2 – t1)is extremely small (t 0) v lim t 0 r t v is instantaneous velocity v is a vector. Unit of v : m/s r r2 r2 r1 t1 t 2 t2 t Uniform Motion (One dimension) Equal displacement (x) traveled in equal time interval (t) x1 x0 x2 x1 x x0 ... v t1 0 t2 t1 t 0 v is constant. x = x0 + vt If x0= 0 at t = 0 , x = vt x0 x1 x2 x3 x4 x 0 t1 t2 t3 t4 t Class Exercise 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? x 0 5 10 t 15 20 Options is in next slide Class Exercise - 3 (a) v (d) 0 5 10 t 15 20 v (c) 0 5 t v (b) 0 5 10 t v 0 5 10 t 15 20 10 15 15 20 20 Solution - 3 For t = 0 to 5 v dx (ve and cons tan t) dt t 5 to 15 v dx ( 0) dt t 15 to 20 v dx (ve and constant) dt Hence answer is (c) Class Exercise - 8 An object travels half the distance with v1, with v2 for half of the remaining time and with v3 for the remaining half of the time. If the object never reverses the direction of motion, find the average velocity during the motion. Solution : 2v1 v2 v3 4 Average Acceleration Change in velocity divided by the time interval during which the change occurs. aav v2 v1 v t 2 t1 t Nature : vector aav unit : m/s² v v2 v1 Dimension: [LT-2] t1 t2 t Non uniform motion (constant acceleration) Instantaneous acceleration a lim t 0 v t For constant acceleration v (final velocity) u(initial velocity) a t (time taken) v = u + at Equation of motion (1) t=t2-t1 v v u t1 t2 t Class Exercise Class Exercise - 1 Which graph represents increasing acceleration? C A B v t (a) A (b) B (c) C (d) None of these Solution - 1 dv a dt 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. Hence answer is (b). Class Exercise - 9 An object starts from rest. It accelerates at 2 m/s2 till it reaches its maximum velocity. Then it retards at 4 m/s2 and finally comes to rest. If the total time taken is 6s, find vmax and the displacement of the object. Solution - 9 Vmax = 2t1 Vmax = 4t2 4t2 2t1 t2 1 t1 2 t1 t2 6 s = Area of triangle Vmax \ t1 = 4 s, t2 = 2 s Hence, Vmax = 8 m/s 1 s 6 8 24 m 2 O 6 t1 t2 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? (b) (a) v T 2 v t T 2 T –v (d) t (c) v v T 2 t –v T 2 T t 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). Class Exercise - 5 If a particle has an initial velocity of 3 i 4 j and an acceleration of 0.4 i 0.3 ,j its speed after 10 s is (a) 10 units (b) 7 units (c) 7 2 units (d) 8.5 units Solution - 5 v u at Hence, v 3 i 4 j 0.4 i 3 j 10 7 i7 j Speed v 7 2 m/s Hence answer is (c). One dimensional equations of motion Distance s = area under v-t graph = ½ (u+v)t Using equation of motion (1) s = ut + ½ at2 Equation of motion (2) But, s = vavgt Hence, vavg = (u+v)/2 As t v u , equation of motion (2) a v v t=t2-t1 u gives v2 u2 2 a.s Equation of motion (3) t1 t2 t One dimensional equations of motion v = u + at s ut 1 2 at 2 v² = u² + 2as Distance traveled in nth second 2n 1 Sn u a 2 Class Exercise 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 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 1 2 s ut at 2 compare with s u(t – 3) a(t – 3)2 Acceleration 2a, initial velocity at t 3s u Class Exercise - 10 The magnitude of maximum acceleration, retardation of an object is ‘a’ m/s2. 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’. 1 s t Hence a t s t 2 sec 2 a 2 Uniform and Non-Uniform Motion Let us see a comparison of uniform and non-uniform motion Courtesy:www.physicsclassroom.com Class Exercise Class Exercise - 7 An object has one-dimensional motion. If V = 6t + 4t3, 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? Solution : 5 (a) (6t 4t)3 dt 592 3 dv (b) Since is never zero dt So acceleration is never negative. Class Exercise - 4 A particle moves in a straight line, starting from rest. The acceleration of the particle is given by a sin t 1 t 1 2 What is the distance traveled by the particle in the time interval 0 to p seconds. Solution - 4 dv 1 sin t dt t 12 1 dt dv sint 2 t 1 0 0 t t t 1 v t cos t t 10 v t 2 cost 1 t 1 [Check that v(0) = 0] Solution - 4 Now v(t) dx t dt dx 1 2 cos t dt t 1 x p 0 0 dx 1 2 cos t dt t 1 x [2t sin t log t 1] x = 2p – log(p + 1) p 0 Thank you