Physics

1. Physics

2. Session Kinematics - 4

3. Y Y Y Y X X X X O O O O Session Opener Who is moving ? Who is at rest ? Everything is relative

4. Session Objective • Two dimensional motion withconstant acceleration • Relative displacement • Relative velocity • River and Boat problem • Rain and Man problem

5. 2 D Equations of motion

6. have different directions 2 D Equations of motion Both x and y components separately satisfy the equations Motion is curvilinear.

7. Relative Velocity Till now we have discussed motion w.r.t. only a single frame of reference. The position, velocity and acceleration of a particle depends on the frame chosen.

8. Consider position of a point P w.r.t. two frames of reference A & B : Differentiating the equation w.r.t. time : Where, is velocity of pt. P w.r.t A And so on … P Here B Important equation-(1) A Relative Velocity

9. Which one of the following represent(s) the displacement-time graph of two objects A and B moving with zero relative speed? (a) (b) (d) (c) Class Exercise - 1

10. Solution Displacement of A w.r.t. B remains constant with time. Hence answer is a, b

11. If the two frames A & B are fixed w.r.t to each other we get : & Note: To avoid non inertial frames we assume no relative acceleration between the frames A & B, hence on further differentiating the velocity equation we get : as Relative Velocity

12. 1) The particles move in same direction : 1 2 or Relative Velocity in 1D As there is 1 axis of motion there are two possible cases : From the equation we derived above we have : Here, g=ground

13. 1 2 Relative Velocity in 1D 2) The particles move in opposite directions : Thus relative velocity is higher when velocities are in opposite directions

14. VBA = 10 m/s at t = 0; aA= 2 m/s2; aB= 3 m / s2. Find the relative displacement of B with respect to A after t = 2 s. Class Exercise - 4 Solution:

15. An elevator whose floor to ceiling distance is 2.50 m, starts ascending with a constant acceleration of 1.25 m/s2. 1.0 s after the start, a bolt begins to fall from the ceiling of the elevator. Calculate (i) free fall time of the bolt, (ii) displacement and distance covered by the bolt during the free fall in the reference frame fixed to the shaft of the elevator. g = 10 ms–2. Class Exercise - 5

16. Solution Let us consider the line of motion of elevator and bolt as the Y-axis and the floor’s initial position (when the bolt starts falling) as origin. (i) At the moment when the bolt starts falling, speed of the elevator and the bolt = 0 + at = 0 + 1(1.25) = 1.25 m/s Let t’ be the time after which the bolt strikes the floor. The y-coordinate of the bolt and floor at time t’ are respectively (As the bolt is freely falling, its acceleration is –g)

17. Þ 2.5 + 1.25t’ – gt’2 = 0 + 1.25t’ + (1.25)t’2 Sb = vt’ + at’2 Solution As the bolt strikes the floor at time t’, yb = yf (b)

18. Solution The bolt goes up from P to Q and then goes down from Q to R. Þ Distance covered = 2(PQ) + QR

19. 1) Boat flows downstream (with the flow) 2) Boat flows upstream (against the flow) Note upstream flow will take place only if : River & boat problem in 1D Here, b=boat, r=river, g=ground

20. Relative Velocity in 2D The basic equation of relative velocity are the same. Here vectors are used for describing directional aspects of motion.

21. There are two reference frames XY with the origin O and X’Y’ with the origin O’. The velocity of an object as observed from O is m/s. The velocity of an object as observed from O’ is m/s. Find the velocity of O’ as observed from O. Class Exercise - 3 Solution:

22. Three persons are located at the vertices of an equilateral triangle whose sides equal a. They all start moving simultaneously with velocity v constant in magnitude with the first person heading continuously for the second, the second for the third and the third for the first. They will meet after time Class Exercise - 10

23. Solution For the motions of B and C xrel = a Hence answer is (a)

24. C B h A The motion of a boat or a man across a river • Motion through the shortest • distance from A to B: In the above case the man’s velocity w.r.t the river has two components 1)perpendicular to the flow 2)against the flow

25. Here, m=man r=river g=ground Velocity component: 1)Perpendicular to flow 2) against flow The motion of a boat or a man across a river Shortest Distance = AB From fig. we get velocity of man along AB.

26. The man will take least time to reach the other side of the river when the value of his velocity component perpendicular to the flow of river is maximum. From 1 & 2 it is clear that this happens when : h 1 2 The motion of a boat or a man across a river 2) Motion taking the shortest time to go across the river

27. Class Exercise - 7 The velocity of a river is 3 km/hr. The velocity of a man with respect to the river is 5 km/hr. Find the time taken by the man to reach the opposite bank if the path that he takes to swim across makes an angle of 60° with the horizontal. (Width of the river = 1.5 km) Solution: Vmr = 5 km/hr Vrg = 3 km/hr

28. In the above question, if the man reaches the opposite bank of the river at a distance x directly from the opposite point on the river, then x will be equal to • 1.6 km (b) 1.75 km • (c) 1.9 km (d) 2.1 km Class Exercise - 8 Solution: x = (Vrg + Vmr cos60°) × t x = 1.9 km Hence answer is (c).

29. Class Exercise - 9 Two ships are 10 km apart on a line running south to north. The one farther north is steaming west at 20 km/hr. The other is moving north at 20 km/hr. What is their distance of the closest approach and how long do they take to reach it? Solution: Suppose the two ships are at X and Y moving with velocities u and v respectively (each 20 km/hr) If A is an observer and B is moving towards X, then the closest approach of B to A is AN.

30. The velocity of Y relative to X = Solution (The perpendicular drawn from A to BX) \ q = 45º From DQPR, the ship Y will move along a direction QR relative to the ship X, where QR is at 45º to PQ, the north-south direction, when the relative velocity is considered, the ship X is at rest.

31. = PQ sin45º = (10 km) = 7.071 km Solution The distance of the closest approach PN The distance QN = PQ cos45º = 7.071 km

32. Man crossing a river - shortest path N W E S C B h A To cross along AB with respect to ground: If Vmr= 5 m/s as shown and Vrg= 4 m/s due east then what should be the velocity of man with respect to ground and the time taken to cross the river if d =30m?

33. Man crossing a river - shortest path Here, m=man r=river g=ground B C h A Shortest time For shortest time the velocity of man with respect to river should be parallel to AB.

34. Class Exercise - 6 A man standing on a road has to hold his umbrella at 30° vertical to keep the rain away. He throws the umbrella and starts running at 10 km/hr. He finds that raindrops are hitting his head vertically. Find the speed of rain-drops with respect to (i) the road, (ii) the moving man. Solution: Let v1 = Velocity of rain with respect to the ground, v2 = Velocity of the man running on the ground = 10 km/hr,

35. Solution v3 = Velocity of rain with respect to the moving man directed vertically. Resolving v1 along the X- and Y-axes, or v1 = 2v2 = 20 km/hr

36. Thank you