The acceleration-time relationship for a vehicle subjected to non-uniform acceleration is,

$$\dfrac{dv}{dt} = (\alpha\: – \beta v_0) e^{ - \beta t}$$

where, $v$ is the speed in m/s, $t$ is the time in s, $\alpha$ and $\beta$ are parameters, and $v_0$ is the initial speed in m/s. If the accelerating behavior of a vehicle, whose driver intends to overtake a slow moving vehicle ahead, is described as,
$$\dfrac{dv}{dt} = (\alpha\: – \beta v)$$

Considering $\alpha = 2 \: m/s^2, \: \beta=0.05 s^{-1}$ and $\dfrac{dv}{dt}=1.3 m/s^2$ at $t=3$s, the distance (in m) travelled by the vehicle in $35$ s is ________