For t < 0, an object of mass m experiences no force and moves in the positive x
ID: 2172998 • Letter: F
Question
For t < 0, an object of mass m experiences no force and moves in the positive x direction with a constant speed vi. Beginning at t = 0, when the object passes position x = 0, it experiences a net resistive force proportional to the square of its speed: Fnet = ?mkv^2 i, where k is a constant. The speed of the object after t = 0 is given byv = vi/(1 + kvit).
(a) Find the position x of the object as a function of time. (Use the following as necessary: k, m, t, and vi.)
x(t) =
(b) Find the object's velocity as a function of position. (Use the following as necessary: k, m, t, vi, and x.)
v(x) =
Explanation / Answer
Some basic calculus first: If y = ln(x) (where 'ln' means natural logarithm) dy/dx = 1/x So if dy/dx = 1/x, then y =ln(x) + constant If y = 1/(1+ax) then dy/dx = (1/a)ln(1+ax) So if dy/dx = 1/(1 + ax), then y =ln(1 + ax) + constant _________________________________ Since you are given dx/dt = vi/(1 + k.vi.t) use this equation: Integrating gives: x = vi(1/(vi.k))ln( 1 + k.v.i.t) + C (see above) =(1/k)ln( 1 + k.v.i.t) + C Since x = 0 when t = 0 0 = (1/k) ln(1 ) + C C = 0 (because ln(1) = 0) So the answer to part (a) is: x = (1/k)ln(1 + k.v.i.t) (equation 1) ____________________________ To find v as a function of position, start with equation 1 and rearrange to get ‘t’ as the subject: x = (1/k)ln( 1 + k.v.i.t) k.x = ln(1 + k.v.i.t) Remember if y=ln(x), x = e^y e^(k.x) = 1+ k.vi.t k.vi.t = e^(k.x) - 1 t = (e^(k.x) -1) / (k.vi) (equation 2) We know v = dx/dt = vi/(1 + k.vi.t). Eliminate t using equation 2. v = vi / (1+ (k.vi((e^(k.x) -1) / (k.vi))) = vi / (1+ e^(k.x) -1)) = vi / e^(k.x) = vi.e^-(k.x) which is the answer to part b) I think that’s OK but there’s always the risk of an algebraic error. (And there might be a simpler solution but I don’t know what it is.)