ACCELERATION


 * ACCELERATION- RATE OF CHANGE **

ESSENTIAL QUESTION ESSENTIAL QUESTION: How can one determine the ** changes in motion **each second of an object experiencing ** acceleration? **


 * USE THE FOLLOWING VOCABULARY IN YOUR CORNEL NOTES: [[file:CORNEL NOTESCHANGE IN SPEED.pdf]][[file:C NOTES ACC.ppt]] **
 * Uniform acceleration
 * Average acceleration
 * Constant velocity
 * Negative Acceleration
 * Positive Acceleration
 * m/s 2
 * initial velocity, final velocity
 * initial time, final time


 * CLICK THE DEFINITION FOR MORE INFORMATION **




 * ACCELERATION FORMULAS AND CALCULATIONS:**

IT WOULD BE VERY HELPFUL TO COPY THESE EQUATIONS ONTO A NOTECARD AND KEEP THEM HANDY IN THE FUTURE:

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CHECK OUT THIS VIDEO FROM THE KAHN ACADEMY

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Run the following PHET SIMULATION -Set it to graph acceleration -Move the object and contrast constant speed, constant acceleration, negative acceleration, and positive acceleration -If time permits take some time to review speed and displacement graphs. media type="custom" key="20838572"

SAMPLE PROBLEMS

An airplane accelerates down a runway at 3.20 m/s2 for 32.8 s until is finally lifts off the ground. Determine the distance traveled before takeoff. AFTER YOU HAVE TRIED SCROLL TO THE BOTTOM TO SEE THE SOLUTION

A car traveling at 22.4 m/s skids to a stop in 2.55 s. Determine the skidding distance of the car (assume uniform acc)

d = ?? || d = vi*t + 0.5*a*t2 d = (0 m/s)*(32.8 s)+ 0.5*(3.20 m/s2)*(32.8 s)2 d = 1720 m
 * Given: || a = +3.2 m/s2 || t = 32.8 s || vi = 0 m/s ||  || Find:

d = ?? || d = (vi + vf)/2 *t d = (22.4 m/s + 0 m/s)/2 *2.55 s d = (11.2 m/s)*2.55 s  d = 28.6 m
 * Given: || vi = 22.4 m/s || vf = 0 m/s || t = 2.55 s ||  || Find: