Physics Kinematics

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

Acceleration is the time rate of change of velocity:

a = dv / dt

Therefore:

( dv / dt ) = k - bv

or

( dv / dt ) + bv = k

This equation is a first-order, linear, constant coefficient, non-homogeneous differential equation. Since the problem as stated is an initial value problem, with the initial conditions given, we will assume that this equation is valid for t >= 0 only.

The solutions take the form

v(t) = Ae^-bt + C

for t >= 0

where A and C are arbitrary constants determined by the initial conditions and the values of b and k.

First, plugging this solution in to the differential equation, we have:

bC = k

or

C = k / b

Then, from the initial condition, we have

A = -C

So,

A = -k / b

So, the complete solution is

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

for t >= 0