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NOTES 6 - Topic 2 - Mechanics

NOTES 6 - Topic 2 - Mechanics. 2.1.3 Equations for Uniformly Accelerated Motion 1. We start with the basic equation for acceleration: a av = ∆v/∆t (Eq. 1) 2............................................ a av = ∆v/∆t ... (add vector arrow where needed) an d ∆v = v f - v o ...

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NOTES 6 - Topic 2 - Mechanics

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  1. NOTES 6 - Topic 2 - Mechanics

  2. 2.1.3 Equations for Uniformly Accelerated Motion 1. We start with the basic equation for acceleration: aav = ∆v/∆t (Eq. 1) 2............................................ aav = ∆v/∆t ... (add vector arrow where needed) and ∆v = vf - vo... so ... aav = (vf - vo)/∆t; Therefore... vf = vo + aav ∆t(Eq. 2);

  3. 3................................... vav = (vo + vf) / 2 (Eq. 3) 4........................................ vav = ∆x / ∆t = (xf - xo) / ∆t... so xf = xo + vav ∆t (Eq. 4); 5...................................... if xf = xo + vav ∆t ... substitute (Eq. 3) for vav... xf = xo + (vo + vf / 2) ∆t; now substitute (Eq. 2) for vf... xf = xo + (vo + vo + a ∆t / 2) ∆t; simplify... xf= xo + vo ∆t + (1/2) a ∆t2 ∆x = vo ∆t + (1/2) a ∆t2 (Eq. 5)

  4. 6........................................ if xf = xo + vav ∆t; substitute Eq. 3 for vav... xf = xo + (vo + vf / 2) ∆t; if aav = ∆v / ∆t, then ∆t = ∆v / aav = (vf - vo) / aav; substitute into above equation; xf = xo + (vo + vf / 2) ((vf - vo) / aav); ...simplify to get... vf2 = vo2 + 2a∆x (Eq. 6)

  5. 2.1.4 Equations for Uniformly Accelerated Motion due to the Force of Gravity For every equation above, aav can be changed to g, the acceleration due to gravity at the surface of the earth ( g = -9.8 ms-2 ); three of the equations can be rewritten as follows for freely falling objects: Take Equation 2................. vf = vo + aav ∆t ...substitute g for a to get... vf = vo + g ∆t(Eq. 2g); Take Equation 5..............∆x = vo ∆t + (1/2) a ∆t2 ...substitute g for a to get... ∆x = vo ∆t + (1/2) g ∆t2 (Eq. 5g) Take Equation 6.................vf2 = vo2 + 2a∆x ...substitute g for a to get... vf2 = vo2 + 2g∆x (Eq. 6g)

  6. To use these equations with g, all ∆x’s are vertical distances and all v’s are vertical velocities!! A. Galileo, in the early 1600’s, was the first to determine that objects fall at the same constant acceleration, regardless of mass, if air resistance is neglected. B. This special acceleration, called the acceleration due to gravity, is represented by the symbol g, and has a value of -9.80 ms-2. The value actually varies according to latitude (the farther North, the greater the value of g) and altitude (the higher you are above sea level, the lower the value of g).

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