# Behavior of force, velocity and acceleration when the power output becomes constant

This might look like a homework question, but really isn't. I am teaching myself some basic physics and am shamefully stuck in one place. I have designed the problem below to help myself understand the situation better.

Suppose a motorbike weighing 100 kg is at rest and begins producing a constant force of $$F=$$ 800 N. The resistance to motion is $$R=$$ 500 N, so the the net force of 300 N causes acceleration in the positive direction, $$a = (F-R)/m = 300/100 = 3$$ m/s. The acceleration is constant, so the speed is given by v = at. The power produced is given by $$P = Fv = Fat$$, so it is increasing linearly over time and in direct proportion to speed. Suppose the bike accelerates for the first 5 seconds and then reaches a certain maximum power output that remains constant. Using the numbers above, the speed reached after 5 s is 15 m/s, and so the maximum power is 12 kW.

If my understanding is correct, at an instant when $$P$$ becomes constant, $$v$$ will still be increasing, so $$F$$ will have to be decreasing in inverse proportion, $$F = P/v$$. Since $$F$$ is decreasing, the acceleration will also be decreasing, $$a = (F - R)/m$$. So we have a three-way circular relationship where to find the function of $$F$$ we need the function of $$v$$, to find the function of $$v$$ we need the function of $$a$$, and to find the function of $$a$$, we need the function of $$F$$. How we find the curves for $$F$$, $$a$$ and $$v$$?

Also, is my understanding correct in that $$v$$ will drop down until $$a = 0$$, at which point $$F = R$$, so the maximum velocity allowed by the maximum power output $$= P/F = P/R =$$ 12000/500 = 24 m/s ?

The equation you're missing is the direct relationship between power $$P$$ and kinetic energy $$K$$. Specifically, for a constant power input, we have that $$K=K_0+Pt$$ where $$K_0$$ is the initial kinetic energy. Since we also know the definition of kinetic energy, $$K=\frac{1}{2}mv^2$$, we can say that

$$\frac{1}{2}mv^2=\frac{1}{2}mv_0^2+Pt$$

so we can solve for $$v$$ directly:

$$v=\sqrt{v_0^2+\frac{2Pt}{m}}$$

Once you have $$v$$ as a function of time, you can of course solve for the net force and $$a$$ as well.

One thing you can immediately see, though, is that the vehicle is always accelerating; there is no maximum allowed velocity. This makes sense, because you have required, in your definition of the problem, that there is a constant input of power regardless of speed; in other words, you have required that the kinetic energy of the vehicle always increases at a constant rate. Increasing kinetic energy implies increasing velocity for a vehicle of constant mass, so therefore $$a>0$$ for all time (though it does approach zero as kinetic energy gets higher and higher).

• thanks for answer, but I am still confused. The function of $v(t)$ you derived implies $v$ is increasing over time. As $v$ is increasing, gross $F$ will be decreasing, since $F=P/v$. When $v$ increases up to a point when $F$ drops down to $R$, $a$ must be 0 as per Newton's 2nd law. So $v$ cannot increase after this point? Aug 23 '19 at 13:41
• Is it too much to ask to produce graphs like this? I have made the graphs above using your function of $v(t)$, but they don't look right, particularly for $a$. This is for my own understanding only. Aug 23 '19 at 14:21
• @probably_someone Not all of the power $P$ goes into increasing the kinetic energy of the bike. Some of it is lost in overcoming friction. Aug 23 '19 at 15:55