Kinematics

1. Kinematics JUNIOR EAMCET

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

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

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

5. Arc of a Circle B A R  R Arc length AB is R AB jaw length is

6. Average Speed and Average Velocity v u t t u v s s Both averge speed and average velocity

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

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

9. Average Speed and Averge Velocity Average speed = u B A Average velocity = R u Average acceleration =

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

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

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

13. Tossing in a Train If Vt = Vb : a = 0; the ball falls in the hand If Vt > Vb : a > 0 the ball falls bedhind If Vt < Vb : a < 0; the ball falls in front

14. Vn – Vn-1 = a Vn = u + an Vn-1 = u + a(n – 1) Sn – Sn-1 = a

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

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

20. One Body Projected and the Other Falling Freely Time of meet is u = 0 h Height of meet is u = u h = h1 + h2

21. Time taken to reach the ground is t1 when throuwn up Time taken to reach the ground is t2 when thrown down t1 u u t2 h Time of free fall is Initial velocity u = Height h =

22. Water Drops from the Tap (n – 1)t = ‘t’ is time interval with which the drops are released h Ratio of displacements of 4th, 3rd, 2nd and 1st drops is 1 : 3 : 5 : 7

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

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

25. 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 x v

26. Graphical Representation v Uniform velocity 10m/s a -a 9 O t 7 3 5 a -5

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

28. Graphical Representation v vmax O t

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

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

31. u s A B Distance is 2s Displacement is zero v Time for forward journey is Average speed is Time for return journey is

32. u t a u t a d

34. Equations of Motion X = a + bt + ct2

35. Problem Find the acceleration

36. Problem