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June 6 -10, 2005. Solving Remedial Challenges KTeam and SCANS Institute The Little Red School House Approach:

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Presentation on theme: "June 6 -10, 2005. Solving Remedial Challenges KTeam and SCANS Institute The Little Red School House Approach:"— Presentation transcript:

1 June 6 -10, 2005

2 Solving Remedial Challenges KTeam and SCANS Institute The Little Red School House Approach:

3 Measuring Acceleration Due to Gravity

4 Three basic linear (straight-line) equations of the form y = mx + b, having variables of distance, time, velocity and acceleration, are the following: v f = v o + at v f 2 = v o 2 + 2 ad d = v o t + ½ at 2 where v o is initial velocity, v f is final velocity, d is distance, a is acceleration, and t is time.

5 Three basic linear (straight-line) equations… v f = v o + at v f 2 = v o 2 + 2 ad d = v o t + ½ at 2 Given any three of the five variables, the other two can be determined when v o is zero.

6 Three basic linear (straight-line) equations… v f = v o + at v f 2 = v o 2 + 2 ad d = v o t + ½ at 2 These equations can be used to demonstrate that an object, falling freely under the influence of earth’s gravity, undergoes constant acceleration.

7 1/ v f = v o + at 2/ v f 2 = v o 2 + 2 ad 3/ d = v o t + ½ at 2 When an object is dropped from a distance, or height h, acceleration “a” becomes “g” for gravitational acceleration during time t. If initial velocity is zero, Equation 3 becomes h = ½ gt 2

8 3/ d = v o t + ½ at 2 h = ½ gt 2 Using y = mx + b and graphing h versus t 2 give a straight line with slope equal to ½ g, so g equals 2 x slope

9 Time (sec) Time Squared Distance (cm) 0.530.2809139.5 0.480.2304118.2 0.440.193699.1 000

10 h = ½ g t^2 Slope = ½ g g = 2 x slope

11

12 Dropped Ball “Measured” Data "Measured" DistanceTimeTime^2Calculated Distance (m) 0.050.100.010.049 0.450.300.090.441 1.240.500.251.226 2.390.700.492.403 3.990.900.813.973 4.911.00 4.905 5.851.101.215.935 8.11.301.698.289 19.42.004.0019.620

13 “Measured” Video Frame Data

14 “Measured” d vs Time Squared Slope =LINEST(D3:D11, F3:F11) = 4.84 Slope = 1/2 g g = 2 x 4.84 = 9.68 Correlation Cofficient = CORREL(D3:D11, F3:F11) = 0.99997

15 Dropped Ball “Measured” Data "Measured" DistanceTimeTime^2Calculated Distance (m) 0.050.100.010.049 0.450.300.090.441 1.240.500.251.226 2.390.700.492.403 3.990.900.813.973 4.911.00 4.905 5.851.101.215.935 8.11.301.698.289 19.42.004.0019.620

16 “Measured” d vs Time


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