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 280 for 20 , then travels at constant speed for another 30 .

Answer 1

Answer:

$d_{total}=22.4cm$

Explanation:

The complete question is

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 for 20 ms, then travels at constant speed for another 30 ms .

During this total time of 50 ms , 1/20 of a second, how far does the tongue reach?

Express your answer to two significant figures and include the appropriate units.

The movement of the tongue is divided in two different motions, with acceleration and without acceleration.

The accelerated movement is defined as

$d=\frac{1}{2}at^{2}$

Where $a=280 m/s^{2}$ and $t=20ms=0.02 sec$, because if 50ms equals 1/20 seconds, then 20ms equals 0.02 seconds.

So, the distance is

$d=\frac{1}{2}280(0.02)^{2} =0.056m=5.6cm$

And the speed is $s=at$

$s=280(0.02)=5.6m/s$

This speed is the same for the constant movement, which actually won't change.

Now, the non accelerated movement is defined as

$d=st$

Where

$s=5.6 m/s$ and $t=30 ms = 0.03sec$

So, the distance in the constant movement is

$d=(5.6m/s)(0.03)=0.168m=16.8cm$

Therefore, the total distance traveled by the tongue is

$d_{total}=5.6cm+16.8cm=22.4cm$

Answer 2

The first part is an acceleration that starts from rest. Distance is determined using the formula d=(1/2)at², where a is acceleration and t is time. Next is to determine the distance using the constant velocity.  The formula that can be used is d=vt wherein v is velocity.