Question

1. (10 pts) A particle moving along a straight line decelerates according to a -kv. This represents a drag-induced deceleration. Determine: a) velocity v as a function of time t b) position s as a function of time t c) velocity v as a function of position s At time t-0, the initial velocity is vo and position is s 0

Answer 1

a = -kv^2 a = dv/dt dv/dt = -kv^2 - integral dv/v^2 = integral k dt 1/v = kt v = k^2/t v = dS/dt ds/t = k Prime/t integral dS = integral k prime/t middot dt s = k prime ln t v middot dv/ds = -kv^2 - integral dv/v = integral_0^s k dS ln (v/v_0) = k prime S