Stan is driving his $35 (metric) tonne$ truck on a horizontal road. Stand accelerates from $50 km h^-1$ to $65 km h^-1$, which is his maximum speed at $500kW$ power output. Find the maximum acceleration of the truck, assuming the total resistance is constant.
Suppose a 35000kg truck is driving along a horizontal road.
The truck accelerates from $13.889ms^{-1}$ to $18.0556ms^{-1}$.
$18.0556ms^{-1}$ is the truck's maximum speed at 500,000W power output.
Assuming the total resistance is constant, find the maximum acceleration of the truck.
The answer is $0.24ms^{-2}$ but I don't seem to be able to get to it.
OK, here are my thoughts, imagine the two cases at the start and end.
At the end the truck is at its maximum velocity so its velocity is constant. Therefore friction = driving force.
The driving force can be calculated by
$P = Fv$
As friction is constant, we therefore know the friction at the start.
Mechanical energy is conserved
so $work done = final KE + lost energy - initial energy$
Calculate the resultant using
$F_{Resultant}= F_{driving}-F_{friction}$
Finally I imagine Newton's second law is need
$F=ma$
to find acceleration assuming the acceleration is greatest initially
I expect the solution will involve the conservation of energy as well. I have given the exact wording of the question in addition to my interpretation of the wording.
It's a long time since I've done any physics, so I hope the community will correct me if something is wrong here.
You're right that at the end of the acceleration we have: $$D=R$$ where $D=\frac{P}{v_f}$ is the force of the motor, and $R$ is friction. We find that $R=\frac{500'000}{18.0556}\approx 27'692$ N. At the start of the acceleration the driving force is maximum when $D=\frac{500'000}{13.889}\approx 35'999$ N.
Thus by Newton's 2. law, the greatest acceleration of the truck is $$a=\frac{D-R}{m}=\frac{35'999-27'692}{35'000}\approx 0.237\frac{m}{s^2}$$