I have this question about finding a quadratic equation:
Car Velocity v (km/hr) : a series of non zero positive values
Car Brake distance d (meter) : also a series of non zero positive values.
The problem asks why a quadratic equation is suitable to find the relationship between v and d?
The problem also asks if
$d = av^2 + bv + c$ where a, b and c are constants, find the function.
Here are v values (km/hr) : 20, 30, 40, 50, 60, 70, 80, 90, 100
Here are d values (meters) : 15, 20, 25, 40, 50, 65, 80, 100, 115
My observation so far:
I have seen a range of x values including zero value and corresponding y values being converted to quadratic equation using 1st difference and 2nd difference of y values. That solution may not work all the time, especially when I don't have zero value for x.
Secondly, I think the problem lacks key information such as deceleration, response time.
Thanks in advance..
Best Answer
This is just a quick answer, but you have set of values $(v_i,d_i)$ and want to fit it to an equation of the form $d(v)=av^2+bv+c$. Each term will have an associated error $r_i=d(v_i)-d_i$. You want to minimise this for all terms, in other words: $$\min\sum (r_i)^2$$ To do this we can set up a system, say: $$S=\sum r_i^2=\sum(av_i^2+bv_i+c-d_i)^2$$ and to find the minimas we want all of the partial derivatives equal to zero: $$0=\frac{\partial S}{\partial a}=2\sum v_i^2(av_i^2+bv_i+c-d_i)$$ $$0=\frac{\partial S}{\partial b}=...$$ $$0=\frac{\partial S}{\partial c}=...$$ Hopefully you get the picture. Now rearrange all of said equations, I will show the top one: $$a\sum v_i^4+b\sum v_i^3+c\sum v_i^2=\sum d_iv_i^2$$ You now have a value mutliplied by $a,b,c$ which you can work out and using all $3$ possible equations get something of the form: $$\begin{bmatrix}\alpha&\beta&\gamma\\\delta&\epsilon&\zeta\\\eta&\theta&\iota\end{bmatrix}\begin{pmatrix}a\\b\\c\end{pmatrix}=\begin{pmatrix}\kappa\\\lambda\\\mu\end{pmatrix}$$ you will know all of the greek letters' values so now just find the inverse of the matrix (online calculators for this exist) and you now have values for $a,b,c$
Use of a quick online calculator here yielded $a\approx 0.009,b\approx0.2,c\approx 6.3$