Question

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)

Answer 1

Answer:

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:

$s = ut + \frac{1}{2}at^2$

where u = initial velocity = 0 m/s

a = acceleration = $250 m/s^2$

t = time = 0.02 s

Therefore:

$s = 0 + \frac{1}{2} * 250 * (0.02)^2\\\\\\s = 0.05 m$

PHASE 2

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

$distance = speed * time$

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:

$v = u + at\\\\=> v = 0 + (250 * 0.02)\\\\\\v = 5 m/s$

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