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)