Chameleons catch insects with their tongues, which they can rapidly extend to great lengths. In a typical strike, the chameleon's tongue accelerates at a remarkable 280 for 20, then travels at constant speed for another 30.
Its velocity at the end of 0.02 s is at = 210*0.02 = 4.2 m/s.
Distance traveled during this time is average velocity * time = 2.1*0.02 = 0.042m.
Distance traveled with this speed of 4.2m/s for another 0.03 s is 4.2*0.03 = 0.126m
The total distance = 0.042+0.126=0.168m
The first part is at constant acceleration starting from rest. d = (1/2)at^2. t is 0.02 seconds and a is given. Calculate that distance in meters.
The second part is at constant velocity, d = vt, with t = 0.03 seconds, and v is given. Calculate that distance in meters.
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Chameleons catch insects with their tongues, which they can rapidly extend to great lengths. In a typical strike, the chameleon's tongue accelerates at a remarkable 230 m / s2 for 20 ms, then travels at constant speed for another 30 ms
Chameleons catch insects with their tongues, which they can rapidly extend to great lengths. In a typical strike, the chameleon's tongue accelerates at a remarkable 210 m/s2 for 20 ms , then travels at constant speed for another 30 ms . Part A During this total time of 50 ms , 1/20 of a second, how far does the tongue reach? Express your answer to two significant figures and include the appropriate units.
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