Placed at a large distance from an observer, a source emits a sound...

The time period of the emitted sound is $T = \dfrac{1}{\nu}$.

Let the source emit the first wave crest at $t = 0$. The position of this crest at time $t$ is $x_1 = vt$.

The second wave crest is emitted at $t = T$. During this time, the source moves a distance $x_s = \dfrac{1}{2}aT^2$.

The position of the second crest at time $t$ (where $t > T$) is $x_2 = x_s + v(t - T) = \dfrac{1}{2}aT^2 + v(t - T)$.

The apparent wavelength $\lambda'$ is the distance between the first and second crests:

$\lambda' = x_1 - x_2 = vt - \left[ \dfrac{1}{2}aT^2 + v(t - T) \right] = vT - \dfrac{1}{2}aT^2$

The frequency $\nu'$ heard by the observer is given by $\nu' = \dfrac{v}{\lambda'}$.

Substituting $T = \dfrac{1}{\nu}$ into the expression for $\lambda'$:

$\nu' = \dfrac{v}{\dfrac{v}{\nu} - \dfrac{a}{2\nu^2}}$

$\nu' = \dfrac{2v\nu^2}{2v\nu - a}$