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 220 m/s^2 for 20 msms, then travels at constant speed for another 30 ms.

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

Answer 1

The total distance is 0.176 m.

Step-by-step explanation:

For t = 0 s to t = 20 ms

initial velocity, u = 0

acceleration, a = 220 m/s^2

time, t = 20 ms

Let the final speed is v.

Use first equation of motion

v = u + at

v = 0 + 220 x 0.02 = 4.4 m/s

Let the distance is s.

Use second equation of motion

`s = u t + 0.5 at^2\\\\s = 0 + 0.5 * 220 * 0.02* 0.02\\\\s = 0.044 m`

Now the distance is

s' = v x t

s' = 4.4 x 0.03 = 0.132 m

The total distance is

S = s + s' = 0.044 + 0.132 = 0.176 m

Answer 2

We know,

Distance,

`$S=ut+(1)/(2)at^2$`

`$S=ut+0.5(a)(t)^2$`

For the first 20 ms,

`$S=0+0.5(220)(0.020)^2$`

S = 0.044 m

In the remaining 30 ms, it has constant velocity.

`$v=u+at$`

`$v=0+(220)(0.020)`

v = 4.4 m/s

Therefore,

`$S=ut+0.5(a)(t)^2$`

`$S'=4.4 * 0.030`

S' = 0.132 m

So, the required distance is = S + S'

= 0.044 + 0.132

= 0.176 m

Therefore, the tongue can reach = 0.176 m or 17.6 cm