Overtaking car acceleration numerical

The acceleration falls linearly with speed, so $\dfrac{dV}{dt} = 2.01 - 0.05V$ is solved by separating variables. Assumption: the car starts at the truck's speed, $V_0 = 35$ km/h $= 9.722$ m/s at $t = 0$.

Speed as a function of time

$$\int \frac{dV}{2.01-0.05V} = \int dt \;\Rightarrow\; -20\ln(2.01-0.05V) = t + C$$

At $t=0$, $V=V_0$. This gives

$$V(t) = 40.2 - (40.2 - 9.722)\,e^{-0.05t} = 40.2 - 30.478\,e^{-0.05t}\ \text{m/s}$$

and the acceleration is $a = 2.01 - 0.05V = 0.05(40.2 - V_0)e^{-0.05t} = 1.5239\,e^{-0.05t}$.

(i) Acceleration after 20 s and 40 s

$$\begin{aligned} V(20) &= 40.2 - 30.478e^{-1} = 28.988\ \text{m/s}, & a(20) &= 2.01 - 0.05(28.988) = 0.561\ \text{m/s}^2\\ V(40) &= 40.2 - 30.478e^{-2} = 36.075\ \text{m/s}, & a(40) &= 2.01 - 0.05(36.075) = 0.206\ \text{m/s}^2 \end{aligned}$$

(ii) Maximum speed attainable

Speed is maximum when $dV/dt = 0$:

$$V_{max} = \frac{2.01}{0.05} = 40.2\ \text{m/s} = 144.7\ \text{km/h}$$

(This is approached asymptotically; a real car would be limited by the speed limit and road.)

(iii) Distance travelled in 260 s

$$\begin{aligned} x &= \int_0^{260} V\,dt = 40.2t - \frac{30.478}{0.05}\left(1 - e^{-0.05t}\right)\\ &= 40.2(260) - 609.56\left(1 - e^{-13}\right) = 9842.4\ \text{m} \end{aligned}$$

Answer: (i) $a_{20} = 0.561$ m/s$^2$, $a_{40} = 0.206$ m/s$^2$; (ii) $V_{max} = 40.2$ m/s (144.7 km/h); (iii) distance $\approx$ 9842 m (9.84 km).