Ookm175h 229 kmh rabbit is running for a total time

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Unformatted text preview: e 2 seconds later is - 5cm, what is its accelera3on? A turtle and a rabbit engage in a footrace over a distance of 4.00 km. The rabbit runs 0.500 km and then stops for a 90.0 nap. Upon awakening, he remembers the race and runs twice as fast. finishing the course in a total time 1.75h, the rabbit wins the race. Calculate the average speed of the rabbit. What was his average speed before he stopped for a nap? assume no detours or doubling back. V= total distance/total time = 4.ookm/1.75h = 229 km/h Rabbit is running for a total time of 1.75h 90m = 1.75-1.50 h = 0.250h l1/v1=l2/v2 = 0.250 => 0.500/v1 + 3.50km/2v2 = 0.250 => v1 = 9.00 km/h ti = xi - 3.00cm vo - 12.0cm/s tf=ti+2.00 deltax = 1/2at^2+vot xf-xi=1/2at^2+vot -5.00-3.00=1/2a(2.00)^2+12.0x2.00 -8.00=2.00a+24.0 => a = -16.0cm/s =- = xf- -5.00 Notes on...
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