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Light-rail passenger trains that provide transportation within andchameleons 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 290 m/s2 for 20 ms, then travels at constant speed for another 30 ms. you may want to review (pages 46 - 49) . part a during this total time of 50 ms, 1/20 of a second, how far does the tongue reach? between cities speed up and slow down with a nearly constant (and quite modest) acceleration. a train travels through a congested part of town at 5.0 m/s . once free of this area, it speeds up to 14 m/s in 8.0 s. at the edge of town, the driver again accelerates, with the same acceleration, for another 16 s to reach a higher cruising speed. you may want to review (pages 42 - 45) . part a what is the final speed?

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W0lf93
Chameleon tongue reaches 23 cm. 
Train's final speed is 32 m/s. 
 The distance the tongue travels is divided into 2 phases.
 1. The acceleration phase.
 2. The coasting phase.
 For the acceleration phase, the formula d = 0.5AT^2 determines how far the tongue travels while accelerating. The during the coasting phase, the tongue continues onward without changing its velocity, so it's formula is d = VT. The peak velocity of the tongue will be reached after 20 ms. So let's calculate it.
 d = 0.5AT^2
 d = 0.5*290 m/s^2 * (0.020 s)^2
 d = 145 m/s^2 * 0.0004 s^2
 d = 0.058 m 
 Now to handle coasting
 d = 0.058 m + 0.030 s * 290 m/s^2 * 0.20 s
 d = 0.058 m + 0.174 m
 d = 0.232 m 
 Rounding to 2 significant digits gives 0.23 meters, or 23 cm. 
 For the train, we need to determine the acceleration. We know the velocity changed from 5.0 m/s to 14.0 m/s over a period of 8.0 seconds. So the acceleration is:
 (14.0 m/s - 5.0 m/s)/8.0 s = (9.0 m/s)/8.0 s = 1.125 m/s^2 
 At the edge of town, the train is traveling at 14 m/s and accelerates at 1.125 m/s^2 for 16 seconds, so:
 14 m/s + 1.125 m/s^2 * 16 s = 14 m/s + 18 m/s = 32 m/s 
 So the final speed of the train is 32 m/s

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