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KINEMATICS

KINEMATICS. The study of motion. Translational. Rotational. Vibrational. DEFINITIONS. Distance, Δ d – change in position. Displacement, Δ d – Change in position in a particular direction. NOTE ON DELTA NOTATION : Δ INDICATES THE CHANGE IN A QUANTITY.

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KINEMATICS

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  1. KINEMATICS • The study of motion Translational Rotational Vibrational

  2. DEFINITIONS • Distance, Δd – change in position. • Displacement, Δd – Change in position in • a particular direction. NOTE ON DELTA NOTATION : Δ INDICATES THE CHANGE IN A QUANTITY.

  3. Velocity, v – Speed in a particular direction. DEFINITIONS • Speed, v – time rate of change of position.

  4. DEFINITIONS Acceleration, a – Time rate of change of a velocity.

  5. KINEMATIC GRAPHICINTERPRETATION Go to v vs t graph Go To : Next slide 16 12 8 4 0 v a vave d (m) Draw tan vinst Δd Δd 0 5 10 15 20 25t(s)

  6. Velocity is the slope of a position vs. time curve. Since slope = rise/run, Just as ,

  7. QUALITATIVE ANALYSISAt what time(s) is the velocity : • Positive? • Negative? • Zero? • Increasing? • Decreasing? • Constant? {Back to graph}

  8. QUALITATIVE ANALYSISAt what time(s) is the acceleration : • Positive? • Negative? • Zero? Hints • When is the velocity increasing? • When is the velocity decreasing? • When is the velocity constant? {Back to graph}

  9. QUANTITATIVE ANALYSISFind the average velocity betweent = 5.0 and 18.0 s Click here for a HINT Go to graph Click here for the SOLUTION

  10. QUANTITATIVE ANALYSISFind the instantaneous velocity at 20.0s Go to graph Click here for HINT 1 : Draw a tangent to the curve at 20.0 s. Pick 2 convenient points to determine its slope. ( try 13 s and 23s) Click here for HINT 2 : Click here for the SOLUTION

  11. What is the total displacement between 11.0 and 18.0 s? Click here for a HINT : Note that this is a graph of Position vs. Time Go to graph Click here for the SOLUTION :

  12. Δd = d1 – d0forward + d1 – d0 backward Δd = 18.0 m – 14.0m + 12.0m – 18.0m What is the distance traveled between 11.0 and 18.0 s? Click here for a HINT : In the time period given, the body moves forwards and backwards. Go to graph Click here for the SOLUTION : Δd = Δdforward + Δdbackward Δd = 4.0m + 6.0m = 10.0 m

  13. I Δd Δdarea II III Go to d vs t graph KINEMATIC GRAPHICINTERPRETATION Next Slide Go To : +2 +1 0 -1 -2 a1 a2 Show v (m/s) On this graph Δd(hint) Δdsoln) 0 5 10 15 20 25t(s)

  14. Acceleration is the slope of a velocity vs. time curve. Since slope = rise/run, Just as ,

  15. At what time(s) is the acceleration: • Positive? • Negative? • Zero? • Increasing? • Decreasing? • Constant? {Back to graph}

  16. Find the average accelerationbetween 8.0 and 25.0 s. Go to v vs t graph Click here for a HINT : Click here for the Solution :

  17. From , If v is constant, area is rectangular : If a is constant, area is trapezoidal : Displacement is the area between the curve and the horizontal origin on a v vs. t graph.

  18. Find the displacement between 11.0 and 19.0s. Go to v vs t graph Click here for a HINT : Above the origin, the displacement is positive. Below the origin, the displacement is negative.

  19. Δdtotal = Δdabove + Δdbelow Δdtotal = + + Go to v vs t graph Click here for the SOLUTION : {Total area = Area I + Area II + Area III} Δdtotal = ½(+2.0m/s+0)3.0s + ½(0+(-2.0m/s))3.0s +(-2m/s)2.0s Δdtotal = +3.0m + (-3.0m) + (-4.0m) = -4.0m

  20. What distance is traveled between 11.0 and 19.0 s? Go to v vs t graph Click here for a HINT : Add the absolute values of the areas under the curve.

  21. Δdtotal = Δdabove + Δdbelow Δdtotal= + + Go to v vs t graph DISTANCE Click here for the SOLUTION : {AREA II} {AREA III} {AREA I} Δdtotal= |½(+2.0m/s+0)3.0s| + |½(0+(-2.0m/s))3.0s +(-2m/s)2.0s| Δdtotal = 3.0m + 3.0m + 4.0m = 10.0m

  22. SIMPLE KINEMATICS PROBLEMS {1} A car moving south along a level road increases its velocity uniformly from 16 m/s to 32 m/s in 10.0 s. (a) What is the car’s acceleration? (b) What is it’s average velocity?

  23. V0 V1 SOLVING PROCEDURE • List given and identify the unknown. Let south ≡ + V0 = + 16 m/s V1 = + 32 m/s Δt = 10.0 s a = ? Vave = ? Draw and label a diagram.

  24. SOLVING PROCEDURE • Determine relevant equations relating the known and the unknown.

  25. Is already a working equation! Derive (if needed) working equations for the unknown in terms of the known.

  26. PLUG IN THE KNOWN

  27. Do the dimensional analysis. ARE THE UNITS CONSISTANT WITH WHAT YOU ARE SOLVING FOR?

  28. Do the math. And tag on the units.

  29. Acceleration due to gravity.[Bodies in freefall] • At the earth’s surface, all falling bodies accelerate at the same rate in a vacuum. g = 9.80 m/s2

  30. SAMPLE PROBLEM • A ball is dropped from a bridge and strikes the water after 8.0 s. a) At what speed does the ball strike the water? b) How high is the bridge? ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

  31. g V0 = 0 Δd v1 Click here for the Solution : Click here for the Given: a) V0 = 0 m/s Δt = 8.0s g = -9.80 m/s2 Click here for the Diagram :

  32. Click here for the Solution: b)

  33. Move on to KINEMATIC DERIVATIONS

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