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.
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Correct Answer: Option n
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
\(v = u + at \)
\(50 = 47.54 + 6t\)
\(t = \frac{50 - 47.54}{6} = 0.41h\)
\(\approxeq 0.4hour\)
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
\(v = u + at \)
\(50 = 47.54 + 6t\)
\(t = \frac{50 - 47.54}{6} = 0.41h\)
\(\approxeq 0.4hour\)