A particle of mass m is initially at rest at the origin. It is...

Given the change in kinetic energy$\dfrac{dk}{dt} = \gammat$.

Kinetic energy is expressed as $k = \dfrac{1}{2}m{v}^{2}$.
$∴ \dfrac{dk}{dt} = mv\dfrac{dv}{dt} = \gammat$
Integrating the above relation,

$m{∫}_{0}^{v}vdv = \gamma{∫}_{0}^{t}tdt$
$\dfrac{m{v}^{2}}{2} = \dfrac{\gamma{t}^{2}}{2}$
$v = \sqrt{\dfrac{\gamma}{m}}t; υ ∝ t$ ... (i)
$a = \dfrac{dv}{dt} = \sqrt{\dfrac{\gamma}{m}} = constant$
Since $F = ma$
$∴ F = m\sqrt{\dfrac{\gamma}{m}} = \sqrt{\gammam} = constant$