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 250 m/s2 for 20 ms, then travels at a constant speed for another 30 ms. During this total time of 50 ms, how far does the tongue reach? (0.2 m)

0.2 m

Explanation:

PHASE 1

First, we calculate the distance the tongue moved in the first 20 ms (0.02 secs). We use one of Newton's equations of linear motion:

where u = initial velocity = 0 m/s

a = acceleration =

t = time = 0.02 s

Therefore:

PHASE 2

Then, for the next 30 ms (0.03 secs), we use the formula:

This speed is the same as the final velocity of the tongue after the first 20 ms.

This can be obtained by using the formula:

Therefore:

distance = 5 * 0.03 = 0.15 m

Therefore, the total distance moved by the tongue in the 50 ms interval is:

0.05 + 0.15 = 0.2 m

the answer is 12 as i may remember

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