A truck of mass 3000 kg has an engine that delivers a constant power of 60 kW to the wheels at…

Step 1. Level road. At the top speed, $\dfrac{P}{v} = R$, so $v_T = \dfrac{60000\ \text{W}}{1500\ \text{N}} = 40\ \text{m/s}$.

Step 2. The slope. Gravity pulls back along the road with $mg\sin\theta = 3000\ \text{kg} \times 10\ \text{m/s}^2 \times \dfrac{1}{20} = 1500\ \text{N}$. Now $\dfrac{P}{v} = 1500\ \text{N} + 1500\ \text{N}$, so $v = \dfrac{60000\ \text{W}}{3000\ \text{N}} = 20\ \text{m/s}$.

Step 3. Time. $m\dfrac{dv}{dt} = \dfrac{P}{v} - R = \dfrac{R(v_T - v)}{v}$, so $dt = \dfrac{m}{R}\,\dfrac{v\,dv}{v_T - v}$. Since $\dfrac{v}{v_T - v} = \dfrac{v_T}{v_T - v} - 1$,

Here $\dfrac{m}{R} = 2\ \text{s}^2/\text{m}$ and $\ln\dfrac{30}{20} = 0.4055$:

Step 4. Distance. With $a = v\dfrac{dv}{dx}$: $mv\dfrac{dv}{dx} = \dfrac{R(v_T - v)}{v}$, so $dx = \dfrac{m}{R}\,\dfrac{v^2\,dv}{v_T - v}$. Since $\dfrac{v^2}{v_T - v} = \dfrac{v_T^2}{v_T - v} - (v + v_T)$,