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 280m/s2 m/s^2 for 20 ms {\rm ms}, then travels at constant speed for another 30 ms {\rm ms}.

During this total time of 50 ms {\rm ms}, 1/20 of a second, how far does the tongue reach?
Express your answer to two significant figures and include the appropriate units.

Answer 1

The reach of the tongue is 23 cm.

Step-by-step explanation:

Hi there!

The equation of traveled distance is the following:

x = x0 + v0 · t + 1/2 · a · t²

Where:

x = traveled distance at time t.

x0 = initial position.

v0 = initial velocity.

a = acceleration.

t = time.

Let´s calculate the distance traveled by the tongue of the chameleon during the first 20 ms (0.020 s):

The initial position and velocity are zero (x0 = 0 and v0 = 0)

x = 1/2 · a · t²

x = 1/2 · 280 m/s² · (0.020 s)²

x = 0.056 m

Now, let´s find the distance traveled while the tongue moves at constant speed. But first, let´s find the velocity (v) of the tongue after the accelertation interval using the following equation:

v = v0 + a · t (v0 = 0)

v = 280 m/s² · 0.020 s

v = 5.6 m/s

Then, the distance traveled at constant speed can be calculated:

x = v · t

x = 5.6 m/s · 0.030 s

x = 0.17 m

The reach of the tongue is 0.17 m + 0.056 m = 0.23 m = 23 cm.