when a chameleon captures an insect its tingue car extend 16 cm in 0.10 s. (a) Find the magnitude of the tongue's acceleration, assuming it to be constant.(b) In the first 0.050 s, dones the tongue extend 8.0 cm, more than 8.0 cm, or less than 8.0 cm? Support your conclusion with a calculation.

d=1/2 a t^2 solve for a.

b) at t=.05s, d=1/2 a t^2 you know a.

To find the answers to these questions, we can use the formulas of constant acceleration. Let's break it down step by step.

(a) To find the magnitude of the tongue's acceleration, we can use the equation:

d = v₀t + (1/2)at²

Where:
- d is the distance covered by the tongue (16 cm),
- v₀ is the initial velocity (0 cm/s, as it starts from rest),
- t is the time taken (0.10 s), and
- a is the acceleration we need to find.

Rearranging the equation gives:

16 cm = 0 cm/s * 0.10 s + (1/2) * a * (0.10 s)²

16 cm = 0 + (1/2) * a * 0.01 s²

16 cm = (1/2) * 0.01 s² * a

Now, we can solve for a:

a = (16 cm) / [(1/2) * 0.01 s²]
a = (16 cm) / (0.005 s²)
a = 3200 cm/s²

Therefore, the magnitude of the tongue's acceleration is 3200 cm/s².

(b) To determine if the tongue extends more than, less than, or exactly 8.0 cm in the first 0.050 s, we can use the same formula as in part (a) but with new values. Here, we want to find the distance covered, so we rearrange the formula:

d = v₀t + (1/2)at²

Given:
- v₀ = 0 cm/s (since it starts from rest),
- t = 0.050 s, and
- a = 3200 cm/s² (as found in part a).

Substituting the values, we have:

d = 0 cm/s * 0.050 s + (1/2) * (3200 cm/s²) * (0.050 s)²

Simplifying,

d = 0 + (1/2) * (3200 cm/s²) * (0.0025 s²)

d = (2) * (3200 cm/s²) * (0.0025 s²)

d = 16 cm

Therefore, the tongue extends exactly 8.0 cm in the first 0.050 s.