Kinematics problem for an object

Question

While solving kinematics problem for the motion of an object, a student obtained the

following result: a = k –bv, where, a is acceleration, v is speed and k and b are positive

constant. Given that at t = 0, v = 0 find the relation between v and t then draw the v – t

graph for the equation.

Answer 1

Given: a = k – bv

=> dv/dt = k - bv since a = dv/dt by definition of acceleration1

1/(k - bv) dv = dt

∫1/(k - bv) dv = ∫ dt integrating both sides

(-1/b) ln (k - bv) = t + C for some constant, C . . . . . . . . . . . . . . eqn (1)

(-1/b) ln k = 0 + C plugging in the initial condition v₀ = 0

C = (-1/b) ln k

(-1/b) ln (k - bv) = t - (1/b) ln k plugging C back into eqn (1)

ln (k - bv) = -b ( t - (1/b) ln k)

ln (k - bv) = -bt + ln k

k - bv = e^{-bt + ln k} by the laws of logs

k - bv = e^-bt . e^{ln k}

k - bv = ke^-bt

bv = k - ke^-bt

v = (k/b) (1 - e^-bt)

This is the required v-t equation.

I leave you to draw the graph.