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 m/s^2 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.

Answer 1

Answer:

0.224 m

Explanation:

The motion of the first 20 ms is a uniformly accelerated motion. With an initial velocity of 0 m/s, the equation of motion used is

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

where $s$ is the distance, $u$ is the initial velocity, $a$ is the acceleration and $t$ is the time.

$s = 0\times20\times10^{-3} + \frac{1}{2} \times 280 \times (20\times10^{-3})^2$

$s = 140\times 0.0004 = 0.056 \text{ m}$

The velocity, $v$, at the end of this part of the motion is

$v=u+at=0+280\times20\times10^{-3} = 5.6 \text{ m/s}$.

This velocity is maintained for 30 ms through a distance of

$5.6\times30\times10^{-3}=0.168 \text{ m}$

The total distance is 0.056 + 0.168 = 0.224 m