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

0.224 m

Step-by-step 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+(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*20*10^(-3) + (1)/(2) * 280 * (20*10^(-3))^2`

`s = 140* 0.0004 = 0.056 \text{ m}`

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

`v=u+at=0+280*20*10^(-3) = 5.6 \text{ m/s}`.

This velocity is maintained for 30 ms through a distance of

`5.6*30*10^(-3)=0.168 \text{ m}`

The total distance is 0.056 + 0.168 = 0.224 m