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Motion Graphs

Motion Graphs. 1 Dimensional Motion. Position-time graph. The object in motion would have positive and constant velocity. Position-time graph. The object in motion would have a positive but changing velocity (acceleration). Both graphs depict an object with constant and negative slope

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Motion Graphs

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  1. Motion Graphs 1 Dimensional Motion

  2. Position-time graph • The object in motion would have positive and constant velocity

  3. Position-time graph • The object in motion would have a positive but changing velocity (acceleration)

  4. Both graphs depict an object with constant and negative slope • The object on the right as a greater velocity

  5. As the slope goes, so goes the velocity!!

  6. Negative acceleration • Object is moving in the negative direction and is speeding up

  7. Velocity-Time Graph • the slope of the line on a velocity versus time graph is equal to the acceleration of the object

  8. Slope if a velocity-time graph • Since the slope of a velocity-time graph determines the object acceleration, the objects acceleration is calculated with the following equation:

  9. Slope of velocity-time graph • Example

  10. Distance vs. Displacement • Distance is a scalar quantity which refers to "how much ground an object has covered" during its motion. (12m)

  11. Distance vs. Displacement • Displacement is a vector quantity which refers to "how far out of place an object is"; it is the object's overall change in position. (0m)

  12. Finding the displacement • The shaded area is representative of the displacement during from 0 seconds to 6 seconds. This area takes on the shape of a rectangle can be calculated using the appropriate equation.

  13. Finding the displacement • Area of a rectangle A= b x h A= (6s) x (30m/s) A = 180m

  14. Finding the displacement • The shaded area is representative of the displacement during from 0 seconds to 4 seconds. This area takes on the shape of a triangle can be calculated using the appropriate equation.

  15. Finding the displacement • Area of a triangle A = ½ x b x h A = (½) x (4s) x (40m/s) A = 80 m

  16. Finding the displacement • The shaded area is representative of the displacement during from 2 seconds to 5 seconds. This area takes on the shape of a trapezoid can be calculated using the appropriate equation.

  17. Finding the displacement • Area of a Trapezoid A = (½) x (b) x ( h1 + h2) A = (½ )x (2s) x (10m/s +30m/s) A = 40 m

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