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 = 1.751.50 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
xfxi=1/2at^2+vot
5.003.00=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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 Winter '14
 Derivative, Velocity, Displacement, accelera3on

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