Compare linear and squared-speed models of drag

The source puts $v$ in L1 and $F$ in L2, and labels axes with extrema 0,36 and 0,33.9. Preserve the origin observation as well as the other nine points. The increasing steepness suggests examining a transformed model, rather than choosing a line from correlation alone.

Use 8:LinReg(a+bx) with Xlist=L1, Ylist=L2. The source screen gives intercept $-1.990909091$, slope $0.9022727273$, $r^2=0.9722440983$ and $r=0.9860243903$. Retain four decimal places as requested, including the trailing zero.

Highlight L3, press ENTER, enter L3=L1², then ENTER. The source displays squared inputs $0,16,64,144,256,400,576,784,1024,1296$. Use Xlist=L3, Ylist=L2: calculator x represents $v^2$, calculator y represents $F$.

The printed output is $a=3.195652174$, $b=0.0242419908$, $r^2=0.9814487724$, $r=0.990680964$. Hence $r=0.9907$ to 4 d.p.

The source observes that drag increases at an increasing rate, and the squared-speed correlation is closer to 1 than the original-speed correlation. Both the curvature and the transformed coefficient favour the quadratic candidate. This comparison does not assert that the fitted intercept is a physically exact zero-speed drag law or establish validity beyond the measured range.

Speed is the prescribed explanatory variable in this wind-tunnel model, so fit force as the response to $v^2$ and solve that model backwards:

$\begin{aligned}&F\\ & =3.195652174\\ &{}+0.0242419908v^2.\end{aligned}$

$\begin{aligned}&v\\ & =\sqrt{\frac{26-3.195652174}{0.0242419908}}.\end{aligned}$

$v=30.6707705\ldots\approx30.7.$

The source prints $F=3.19565+0.0242199v^2$ in the final substitution, while its preceding calculator screen gives $0.0242419908$. The final printed slope omits a digit; use the full screen coefficient, preserving this discrepancy in review. Both produce the printed final $30.7$ after 3-s.f. rounding. Reject the negative root because speed is nonnegative.

The blank explanation is completed from the source’s modelling convention: regressions of $v$ on $F$ or $v^2$ on $F$ minimise a different response error and would change the prescribed explanatory role. They are not the calibration of the fitted force model requested here.