Introduction to linear kinematics


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Introduction to linear kinematics

  1. 1. An Introduction to Linear Kinematics LINEAR KINEMATICS
  2. 2. Kinematic Analysis  Linear Kinematics    Motion described in terms of (variables):      Distance, displacement, length (e.g. stride, stroke) Time, cadence (e.g. stride frequency, stroke frequency) Speed, velocity Acceleration Single point models   description of the motion of a body the appearance of a motion with respect to time e.g. Centre of mass (CM) during running/jumping Multi-segment models  e.g. Co-ordination of body segments during running/jumping
  3. 3. Distance & Displacement  Distance:    Displacement:    Length of path which a body covers during motion Units: metre (m), centimeter (cm), kilometer (km) The change in position of a body during motion Units: metre (m), centimeter (cm), kilometer (km) Distance is a scalar, and displacement is a vector variable
  4. 4. Speed and Velocity  Speed (scalar)   Length of path (distance) divided by change in time (∆t) Δp d v= = Δt Δt Average velocity (vector)  Change in position (∆p) divided by change in time (∆t)  Displacement (d) divided by change in time (∆t)  Vector equivalent of linear speed If displacement = 50 m If ∆t = 5 s v = 50 / 5 = 10 m·s-1
  5. 5. Velocity  Units of velocity   Current velocity m/s or m·s-1 Velocity is a vector   Magnitude and direction calculated using Pythagoras and trigonometry The velocity of a swimmer in a river is the vector sum of the velocities of swimmer and current. Swimmer’s velocity Resultant velocity
  6. 6. Exercise 2.2 : 6 1. A speedboat moving at 30.0 m s-1 approaches stationary buoy marker 100 m ahead. The pilot slows the boat with a constant acceleration of -3.50 m s-2 by reducing the throttle. a. How long does it take the boat to reach the buoy? b. What is the velocity of the boat when it reaches the buoy? No. 23,pg. 51,Physics for scientists and engineers with modern physics, Serway & Jewett,6th edition. ANS. : 4.53 s; 14.1 m s−1 2. An unmarked police car travelling a constant 95 km h -1 is passed by a speeder traveling 140 km h-1. Precisely 1.00 s after the speeder passes, the policemen steps on the accelerator; if the police car’s acceleration is 2.00 m s -2, how much time passes before the police car overtakes the speeder (assumed moving at constant speed)? No. 44, pg. 41,Physics for scientists and engineers with modern physics, Douglas C. Giancoli,3rd edition. ANS. : 14.4 s
  7. 7. Example 2.7 : 7 A ball is thrown from the top of a building is given an initial velocity of 10.0 m s−1 straight upward. The building is 30.0 m high and the ball just misses the edge of the roof on its way down, as shown in B figure 2.7. Calculate a. the maximum height of the stone from point A. b. the time taken from point A to C. u =10.0 m s−1 c. the time taken from point A to D. A d. the velocity of the ball when it reaches point D. C (Given g = 9.81 m s−2) 30.0 m Figure 2.7 D
  8. 8. 2.4. Projectile motion  A projectile motion consists of two components:  vertical component (y-comp.)   motion under constant acceleration, ay= −g horizontal component (x-comp.)   8 motion with constant velocity thus ax= 0 y The path followed by a projectile is called trajectory is shown in Figure 2.9. B v1y P Simulation 2.5 Figure 2.9 uy A θ1 v1x u θ ux v v1 Q sy=H v2y v2x θ2 v2 C t1 sx= R t2 x
  9. 9.  From the trigonometry identity, thus 9 sin 2θ = 2 sin θ cos θ 2 u R= sin 2θ g  The value of R maximum when θ = 45° and sin 2θ = 1 therefore 2 Rmax u = g Simulation 2.6
  10. 10. 10 2.4.5 Horizontal projectile  Figure 2.10 shows a ball bearing rolling off the end of a table with an initial velocity, u in the horizontal direction. u u vx vy h Figure 2.10  A v B x Horizontal component along path AB. velocity, u x = u = v x = constant displaceme nt, s x = x  Vertical component along path AB. initial velocity, u y = 0 displaceme nt, s y = − h Simulation 2.7
  11. 11. Example 2.9 : y u Figure 2.12 O 11 H θ = 60.0° P R v1y Figure 2.12 shows a ball thrown by superman with an initial speed, u = 200 m s-1 and makes an angle, θ = 60.0° to the horizontal. Determine a. the position of the ball, and the magnitude and direction of its velocity, when t = 2.0 s. v1x v1 Q v2y x v2x v2
  12. 12. 12 b. the time taken for the ball reaches the maximum height, H and calculate the value of H. c. the horizontal range, R d. the magnitude and direction of its velocity when the ball reaches the ground (point P). e. the position of the ball, and the magnitude and direction of its velocity at point Q if the ball was hit from a flat-topped hill with the time at point Q is 45.0 s. (Given g = 9.81 m s-2) Solution : The component of Initial velocity : u x = 200 cos 60.0 = 100 m s −1 u y = 200 sin 60.0 = 173 m s −1
  13. 13. Exercise 2.4 : 13 Use gravitational acceleration, g = 9.81 m s−2 1. A basketball player who is 2.00 m tall is standing on the floor 10.0 m from the basket, as in Figure 2.13. If he shoots the ball at a 40.0° angle above the horizontal, at what initial speed must he throw so that it goes through the hoop without striking the backboard? The basket height is 3.05 m. Figure 2.13 ANS. : 10.7 m s−1
  14. 14. Exercise 2.4 : 14 2. An apple is thrown at an angle of 30° above the horizontal from the top of a building 20 m high. Its initial speed is 40 m s−1. Calculate a. the time taken for the apple to strikes the ground, b. the distance from the foot of the building will it strikes the ground, c. the maximum height reached by the apple from the ground. ANS. : 4.90 s; 170 m; 40.4 m 3. A stone is thrown from the top of one building toward a tall building 50 m away. The initial velocity of the ball is 20 m s−1 at 40° above the horizontal. How far above or below its original level will the stone strike the opposite wall? ANS. : 10.3 m below the original level.
  15. 15. Acceleration  Acceleration = change in Velocity / time  (How fast you change how fast your going!)  Vector = Magnitude (size) + Direction  Units: mi/hr/sec; m/s/s; m/s2 v v ∆v v f − vi a= = t t a = acceleration ∆v = change in velocity (final – initial) t = time
  16. 16. Constant acceleration woman covers more distance every second in the same direction her change in speed is what’s constant!
  17. 17. Free Fall Speeds & Distances acceleration equation to solve for v : How FAST = Rearrange  f v v f − vi a= t v v f − vi at = v vf at + vi = v v f = vi + at How FAR = equation for free fall distances when falling from rest: v2 ∆x = 1 at 2 ∆x = change in position (displacement) a = acceleration t = time
  18. 18. Summary  Variables used to describe motion are either:  Scalar (magnitude only: e.g. time, distance and speed)  Vector (magnitude and direction: e.g. displacement, velocity and acceleration)  Displacement is the change in position of a body  Average velocity is the change in position divided by the change in time  Average acceleration is the change in velocity divided by the change in time
  19. 19. THE END… Linear motion 19
  20. 20. Recommended Reading      Enoka, R.M. (2002). Neuromechanics of Human Movement (3rd edition). Champaign, IL.: Human Kinetics. Pages 310 & 22-27. Hamill, J. & Knutzen, K.M. (2003). Biomechanical Basis of Human Movement (2nd edition). Philadelphia: Lippincott Williams & Wilkins. Pages 271289. werpoint/ch2-linear_motion