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

The relationship between v and t is given by v(t) = (Ae)^-bt + C

Given equation - a = k - bv

a is acceleration, v is speed and k and b are positive constants.

a = dv/dt ( acceleration is equal to the change in velocity with respect to time )

=> dv/dt = k - bv

=> ( dv / dt ) + bv = k

Solving the first-order, linear, constant coefficient, non-homogeneous differential equation differential equation we get -

=> v(t) = A(e)^-bt + C

A and C are arbitrary constants in the equation .