A car travelling at a velocity of 50kmh\(^{-1}\), covers a distance of 20km. If it was accelerating at 6kmh\(^{-1}\), calculate, correct to one decimal place, the time the car took to cover the distance.
Take consistent units of kilometres and hours. Initial velocity \(u=50\text{ km h}^{-1},\) distance \(s=20\text{ km},\) acceleration \(a=6\text{ km h}^{-2}.\)
Using \(s=ut+\tfrac12 at^2:\)
\[20=50t+\tfrac12(6)t^2\Rightarrow 20=50t+3t^2.\]
Rearrange into a standard quadratic in \(t:\)
\[3t^2+50t-20=0.\]
Apply the quadratic formula \(t=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}\) with \(a=3,\ b=50,\ c=-20:\)
\[t=\frac{-50\pm\sqrt{2500+240}}{6}=\frac{-50\pm\sqrt{2740}}{6}.\]
Since \(\sqrt{2740}\approx52.35,\) the positive (physical) root is
\[t=\frac{-50+52.35}{6}=\frac{2.35}{6}\approx0.39\text{ h}.\]
\[t\approx0.4\text{ hours (to one decimal place)}.\]
(The negative root is rejected as time cannot be negative.)