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.
Explanation
Given : \(v = 50kmh^{-1} ; s = 20 km ; a = 6kmh^{-2} ; u = ?\) \(v^{2} = u^{2} + 2as\) \(50^{2} = u^{2} + 2(6)(20)\) \(2500 = u^{2} + 240 \implies u^{2} = 2500 - 240 = 2260\) \(u = \sqrt{2260} = 47.54 kmh^{-1}\) To find time, we use \(v = u + at \) \(50 = 47.54 + 6t\) \(t = \frac{50 - 47.54}{6} = 0.41h\) \(\approxeq 0.4hour\)
