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 220 m/s2 for 20 ms, then travels at constant speed for another 30 ms.

We can assume the tongue begins to reach. We need to solve this using one of the kinematic equations of motion, solving in two parts. We assume the tongue begins at rest.

Firstly, the accelerating part. We have the following informtion:
u = initial velocity = 0 m/s
a = acceleration = 270 m/s²
t = time = 0.020 s
d = distance = ?

d = ut + 1/2 at². Since u = 0, this reduces to d = 1/2 at² = 135 * 0.020² = 0.054 m.

Now what about the constant speed part? First let's find the speed. Using u, a and t as before, we use the equation v = u + at. Since u = 0, this reduces to v = at = (270 m/s)(0.02 s) = 5.4 m/s. And when speed is constant, we know distance travelled is speed times time. We know it travels at 5.4 m/s for 0.030 s, so during that time it travels (5.4 m/s)(0.03 s) = 0.162 m.

So the total distance it travels, and therefore the reach of the tongue, is 0.054 m + 0.162 m = 0.216 m = 21.6 cm.

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