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motion

  1. 1. © 1John Parkinson
  2. 2. Constant Velocity Distance travelled - s Time taken - t Velocity - v s v = t v s/t Velocity = Speed in a Specified Direction© 2John Parkinson
  3. 3. N 100 m in 4 seconds Distance travelled = ? 100 m Displacement = ? 100 m to the East Speed = ? Speed = 100/4 = 25 m s-1 Velocity = ? Velocity = 25 m s-1 to the East© 3John Parkinson
  4. 4. DISPLACEMENT – TIME GRAPHS Constant velocity Displacement - s What will the graph look like? Δs GRADIENT = ? Δt Time - t ∆s VELOCITY v= ∆t© 4John Parkinson
  5. 5. Displacement - s What about this graph? A body at rest Time - t Displacement - s And this graph? Δs Δt The gradient Time - t is increasing …….? The body must be accelerating ……..?© 5John Parkinson
  6. 6. VELOCITY – TIME GRAPHS Velocity - v This body has a constant or uniform ………? acceleration Δv The gradient = the acceleration ? Δt ∆v a= Time - t ∆t Velocity - v 1 = …… ? Uniform acceleration 2 = …… ? Constant velocity 2 1 3 3 = …… ? Uniform retardation [deceleration] A Area under the graph = A Time - t DISTANCE TRAVELLED = …….. ?© 6John Parkinson
  7. 7. QUESTION The graph represents the motion of a Velocity – v/ms-1 tube train between two stations Find 30 1. The acceleration 2. The maximum velocity 3. The retardation 20 50 80 Time – t/s 4. The distance travelled 1. The acceleration = the initial gradient = 30÷20 = 1.5 m s-2 2. The maximum velocity is read from the graph = 30 m s-1 3. The retardation = the final gradient = -30 ÷ [80-50] = -1.0 m s-2 4. The distance travelled = the area under the graph =½ x 20 x 30 + 30 x 30 + ½ x 30 x 30 = 1650 m© 7John Parkinson
  8. 8. What will the distance – time, velocity - time and acceleration time graphs look like for this bouncing ball? Displacement - s s2 s1 s1 s2 Time - t Velocity - v Time - t© 8John Parkinson
  9. 9. Velocity - v Time - t Acceleration - a 9.81ms-2 Time - t© 9John Parkinson
  10. 10. Velocity What might this graph represent? Terminal Velocity Time Can you draw an acceleration time graph for this motion? Acceleration 9.81 m s-1 Time© 10John Parkinson
  11. 11. EQUATIONS OF MOTION ACCELERATION EQUATIONS OF UNIFORM For Constant Velocity DISTANCE VELOCITY = TIME s v= t If Velocity is not constant , this equation just gives the average velocity for the journey DISTANCE = VELOCITY x TIME s =v t© 11John Parkinson
  12. 12. EQUATIONS OF MOTION For UNIFORM ACCELERATION SYMBOLS • a = ACCELERATION • u = INITIAL VELOCITY • v = FINAL VELOCITY • s = DISTANCE TRAVELLED • t = TIME TAKEN© 12John Parkinson
  13. 13. For UNIFORM ACCELERATION 1. Distance travelled = average velocity times the time taken Velocity u+v v s= t 2 u Time t 2. Acceleration = the change in velocity per second v-u Rearranging a= t v = u + at© 13John Parkinson
  14. 14. 3. Substituting equation [2] into equation [1] u+v v = u + at s= t 2  u + u + at   2ut + at  2 s=  t =    2   2  1 s = ut + 2 at2 v −u 4. Rearrange equation 1. to make t the subject t = a Now substitute this in equation 3 and rearrange to give : v2 = u2 + 2as© 14John Parkinson
  15. 15. USING THE EQUATIONS OF MOTION 1. v =u +at 1. Write down the symbols of the quantities that you know 2. v =u +2as 2 2 1 3. s = ut + at 2 2 2. Write down the symbol of the quantity that you require 1 4. s = ( u +v )t 2 3. Select the equation that contains all of the symbols e.g. in 1. and 2. aboveA stone is released from a height of 20 m above the ground. Neglecting air resistanceand using the acceleration due to gravity as 9.81 ms-2, find the velocity with which thestone will hit the ground . This must be equation 2 as it is the u = 0 from rest only one with “v”, “u”, “a” and “s” in it s = 20 m v2 = u2 + 2as a = 9.81 ms-2 v =? v2 = 02 + 2 x 9.81 x 20 v= 392 = 19.8 m s-1© 15John Parkinson

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