the motion of a body falling from rest in a resistive medium is dv/dt= A-Bv a. f

Question

the motion of a body falling from rest in a resistive medium is dv/dt= A-Bv
a. find its initial acceleration
b. find the velocity at which the acceleration becomes zero ( the terminal velocity)
c. show that at any time t the velocity is goiven by v(t)=A/B91-e^-Bt

Explanation / Answer

initial velocity u =0m/s acceleration is given by a = dv/dt = A-Bv =>dv/(A-Bv) = dt integrating both sides we get, -(1/B)ln(A-Bv) = t+c at t =0 v =0 =>-(1/B)lnA = c so,-(1/B)ln(A -Bv) = t -(1/B)lnA =>1/B(lnA/(A-Bv)) = t => ln(A/A-Bv) = Bt =>A/(A-Bv) = e^Bt =>(A-Bv)/A = e^-Bt =>1-Bv/A = e^-Bt => Bv/A = 1-e^-Bt => v = (A/B)*(1-e^-Bt) a) initial acceleration = A-B*0 = A b)a = 0 =>A -Bv =0 =>v = A/B m/s c)v = (A/B)*(1-e^-Bt) already derived

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