Simple Nature - Light and Matter

about the function x(t), given information about the acceleration.To go from acceleration to position, we need to integrate twice:∫ ∫x = a dt dt∫= (at + v o ) dt [v o is a constant of integration.]∫= at dt [v o is zero because she’s dropping from rest.]= 1 2 at2 + x o [x o is a constant of integration.]= 1 2 at2 [x o can be zero if we define it that way.]Note some of the good problem-solving habits demonstrated here.We solve the problem symbolically, and only plug in numbers atthe very end, once all the algebra and calculus are done. Oneshould also make a habit, after finding a symbolic result, of checkingwhether the dependence on the variables make sense. Agreater value of t in this expression would lead to a greater valuefor x; that makes sense, because if you want more time in theair, you’re going to have to jump from higher up. A greater accelerationalso leads to a greater height; this also makes sense,because the stronger gravity is, the more height you’ll need in orderto stay in the air for a given amount of time. Now we plug innumbers.x = 1 (9.8 m/s 2) (1.0 s) 22= 4.9 mNote that when we put in the numbers, we check that the unitswork out correctly, ( m/s 2) (s) 2 = m. We should also check thatthe result makes sense: 4.9 meters is pretty high, but not unreasonable.The notation dq in calculus represents an infinitesimally smallchange in the variable q. The corresponding notation for a finitechange in a variable is ∆q. For example, if q represents the valueof a certain stock on the stock market, and the value falls fromq o = 5 dollars initially to q f = 3 dollars finally, then ∆q = −2dollars. When we study linear functions, whose slopes are constant,the derivative is synonymous with the slope of the line, and dy/ dxis the same thing as ∆y/∆x, the rise over the run.Under conditions of constant acceleration, we can relate velocityand time,a = ∆v ,∆tor, as in the example 1, position and time,x = 1 2 at2 + v o t + x o .22 Chapter 0 Introduction and Review