Chapter 4 Applications of the definite integral to rates, velocities and ...

Math 103 Notes Chapter 4ButI =∫ T0v(t) dt =∫ T0(v 0 + at) dt =So, setting these two results equal means that( ) ∣ ∣∣∣v 0 t + a t2 T20x(T) − x 0 = v 0 T + a T 22 .=(v 0 T + a T 22But this is true for all final times, T, i.e. this holds for any time t so that).x(t) = x 0 + v 0 t + a t2 2 .This expression represents the position of a particle at time t given that it experienced a constantacceleration, a, after starting with initial velocity v 0 , from initial position x 0 at time t = 0.Non-constant acceleration and terminal velocityThe velocity of a sky-diver is described by a differential equationdvdt = g − kvi.e., a mathematical statement that relates changes in velocity to acceleration due to gravity, g,and drag forces due to friction with the atmosphere. A good approximation for such drag forcesis the term kv, proportional to the velocity, with k, a positive constant, representing a frictionalcoefficient. We will assume that initially the velocity is zero, i.e. v(0) = 0. In this example we showhow to determine the velocity at any time t. In this calculation, we will use the following resultfrom first term calculus material:The differential equation and initial conditiondydt = −ky, y(0) = y 0 (4.3)has a solutiony(t) = y 0 e −kt (4.4)The equation dv/dt = g − kv implies thata(t) = g − kv(t)where a(t) is the acceleration at time t. Taking a derivative of both sides of this equation leads todadt = −kdv dt ,which means thatdadt = −ka.v.2005.1 - January 5, 2009 4