Motion of a body given acceleration as a function of velocity

Question

I'm having trouble figuring this out. I'm trying to determine the final time and final velocity for the motion of a vehicle over a given distance along a straight line.

$t$ is time
$v$ is velocity
$a$ is acceleration
$r$ is displacement
$m$ and $c$ are constant

Known:
$t_0$, $v_0$, $r_0$, $\Delta r$

Acceleration of the vehicle is described by:
$a=\frac{f(v)-c}{m}$

I want to derive expressions for:
$\Delta t$, and $v$

$f(v)$ is a function which returns the force which the engine can apply at a given velocity. This is essentially a lookup against a table of values from the vehicle manufacturer. I can perform an accumulation over this if required.

I'm currently estimating a result to this problem by iteratively calculating using small intervals of time.

Any help would be greatly appreciated.

Are these vectors or scalars? For example, is displacement the same as the distance travelled? Is the motion in a straight line? — Paresh, Feb 12 '13 at 14:59
@Jack What do you mean by final time? Do you mean the time at which velocity reaches some value? — oks, Feb 12 '13 at 15:53
@oks The time at which the displacement has reached some value; i.e. the travel time — Jack, Feb 12 '13 at 16:13
How are you interpolating $f(v)$ between the points in the lookup table? Is it just linear interpolation? — , Feb 12 '13 at 17:33

score 2 · Answer 1 · answered Feb 12 '13 at 17:05

Since $a = \dot{v}$, your equation of motion is the ODE $$m\dot{v} = f(v) - c$$

This is an autonomous first-order ODE; separating the variables gives $$m\int_{v(0)}^{v(t)} \frac{1}{f(w)-c}dw = t,$$ and if you had a formula for $f$ you could try to perform the integration on the LHS to get $mg(v) = t$, and then solve for $v$.

Since you only have numerical data, though, your best bet is numerical quadrature (which it sounds like you're already doing). For instance, pick a time step $h$ a good bit smaller than your estimated $\Delta t$, and iterate

\begin{align*} r_{i+1} &= r_i + h v_i\\ v_{i+1} &= v_i + h \frac{f(v_i) -c}{m}\\ t_{i+1} &= t_i + h \end{align*}

until $r_i-r_0 > \Delta r$.

score 2 · Answer 2 · answered Feb 12 '13 at 17:06

You are trying to calculate $\frac 1m \int f(v)-c \; dt$ and unless you have a nice formula for $f(v)$ you will be doing it numerically, which basically amounts to what you are doing-add up accelerations times small timesteps. If $f(v)$ is reasonably smooth, you can reduce the calculation time and make it more robust by using an adaptive integration routine. Any numerical analysis text will have a discussion. I like Numerical Recipes, which is free online and has routines in C.

score 2 · Accepted Answer · edited Apr 13 '17 at 12:20

I think you can get much farther with explicit methods than the other answers suggest. First, given $a$ as a function of $v$, we can find $t$ as a function of $v$ by integration: $$\begin{gather} \frac{\mathrm dv}{\mathrm dt}=a=\frac{f(v)-c}m.\\ \frac m{f(v)-c}\,\mathrm dv=\mathrm dt.\\ \int_{v_0}^{v_1}\frac{m\,\mathrm dv}{f(v)-c}=\int_{t_0}^{t_1}\mathrm dt=t_1-t_0. \end{gather}$$ Since $f(v)-c$ is piecewise linear, $\int_{v_0}^{v_1}m\,\mathrm dv/(f(v)-c)$ is piecewise logarithmic and you can compute it explicitly. This gives us $t$ as an increasing function of $v$, which I'll call $\tau$: $$t_1=\tau(v_1):=t_0+\int_{v_0}^{v_1}\frac{m\,\mathrm dv}{f(v)-c}.$$ Now we get $r$ by integrating again: $$\begin{gather} \tau(v)=t.\\ \frac{\mathrm dr}{\mathrm dt}=v=\tau^{-1}(t).\\ r_1-r_0=\int_{t_0}^{t_1}\tau^{-1}(t)\,\mathrm dt. \end{gather}$$ It appears we have to integrate $\tau^{-1}$, which seems complicated. But not too complicated: let $t_0=\tau(v_0)$ and $t_1=\tau(v_1)$, and then we have $$\int_{t_0}^{t_1}\tau^{-1}(t)\,\mathrm dt=(t_1v_1-t_0v_0)-\int_{v_0}^{v_1}\tau(v)\,\mathrm dv,$$ and integrating the piecewise logarithmic $\tau$ can also be done explicitly. So now we have $r$ as another explicitly known, increasing function of $v$. Your problem amounts to finding the $v$ for the given $r$, which you will have to do via a numerical procedure, but something really simple like bisection search should work just fine.

This is brilliant, I'll implement this and let you know how it goes. I also have some termination conditions (like braking distance requirements), so a bisection search would have been required regardless... I think. — Jack, Feb 13 '13 at 07:46

score 1 · Answer 4 · answered Feb 12 '13 at 18:15

As user7530 says, you could integrate numerically by using a first order Euler method to integrate numerically

\begin{align*} r_{i+1} &= r_i + h v_i\\ v_{i+1} &= v_i + h a_{i+1} \\ a_{i+1} & = \frac{f(v_i) -c}{m}\\ t_{i+1} &= t_i + h \end{align*}

until $r_i-r_0 > \Delta r$.

But you could use different methods to integrate numerically. One way to test whether it's worth using a more complicated method would be to replace your looked up acceleration function with something that has a nice formmula for f(v), and then compare the numerical result with the exact result.

Motion of a body given acceleration as a function of velocity

4 Answers4