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Kinematics JUNIOR EAMCET

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Distance and Displacement O A B 4m 3m 5m O to B: distance is 7m and displacement is 5m

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Distance and Displacement O A B C 4m 3m 5m O to C along OABC: distance is 11m and displacement is 3m

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Distance and Displacement O A B C 4m 3m 5m O to O along OABCO: distance is 14m and displacement is zero

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Arc of a Circle RR A B Arc length AB is R AB jaw length is

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Average Speed and Average Velocity u v tt v s u s Both averge speed and average velocity

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Average Speed and Average Velocity v s u s Both averge speed Both averge velocity

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Train Crossing a Bridge Post Train BridgeTrain u s Time of crossing the post is Time of crossing the bridge is u s L

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Average Speed and Averge Velocity AB R Average speed = Average velocity = Average acceleration = u u

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Direction of Acceleration uv uv a a Direction of acceleration is v – u Acceleration is positive if v > u Acceleration is negative if v < u

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Sign Application + g Downward direction is positive Freely falling body S, u, v, g, h are all positive Body projecte up Upward direction is positive a = – g u, v, h are all positive

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Body Projected From the Top of a Tower u Upward direction is positive a = – g s = – h u is positive s, g are negative

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Tossing in a Train If V t = V b : a = 0; the ball falls in the hand If V t > V b : a > 0 the ball falls bedhind If V t < V b : a < 0; the ball falls in front

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V n – V n-1 = a V n = u + an V n-1 = u + a(n – 1) S n – S n-1 = a

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u v s t Displacement = average velocity × time If u = 0

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A B a u A B a u A B a u A travels with acceleration a, B with uniform velocity u If they start simultaneously Time of meet is If A is ahead of B d d If B is ahead of A

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One Body Projected and the Other Falling Freely u = 0 u = u Time of meet is h Height of meet is h = h 1 + h 2

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u u h Time taken to reach the ground is t 1 when throuwn up Time taken to reach the ground is t 2 when thrown down t1t1 t2t2 Time of free fall is Initial velocity u = Height h =

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Water Drops from the Tap h (n – 1)t = t is time interval with which the drops are released Ratio of displacements of 4 th, 3 rd, 2 nd and 1 st drops is 1 : 3 : 5 : 7

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h height of the window u velocity at the top of the window t time of crossing the window

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If the velocity is reduced by after travelling a distnce x, then the total distance it can travel is

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If th front and back of the train cross a post with velocities u and v, the center will cross the same post with velocity ---- u v x

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Graphical Representation v t Uniform velocity a a O -a 10m/s 35 7 9 -5

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Graphical Representation v t Uniform velocity 10/3 a O -5 10m/s 35 7 9 -5 15 20 10 -5

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Graphical Representation v tO v max

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v t O P V – t Graph The slope gives acceleration Acceleration is positive if < 90 Acceleration is negative if > 90 Area represents displacement Displacement is positive if area is above x - axis

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Graphical Representation v t u constant a u = 0 a u -a O

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AB s u v Distance is 2s Displacement is zero Time for forward journey is Time for return journey is Average speed is

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u a t u a t d

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Equations of Motion X = a + bt + ct 2

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Problem Find the acceleration

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Problem

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