a) Distance moved by the top lid during a blink: 1 cm (estimate)
b) The acceleration is [tex]34.7 m/s^2[/tex]
c) The final speed is 0.83 m/s
Explanation:
a)
For the purpose of this problem, we can estimate the size of the eye from the top lid to the bottom lid to be 1 cm, so this is the distance moved by the top lid during a blink.
b)
Assuming the motion of the eyelid to be at constant acceleration, we can find the acceleration by using the following suvat equation:
[tex]s=ut+\frac{1}{2}at^2[/tex]
where
s is the distance covered
u is the initial velocity
t is the time
a is the acceleration
For the eyelid, we have:
u = 0 (it starts from rest)
s = 1 cm = 0.01 m is the distance covered
t = 0.024 s is the time taken
Solving for a, we find the acceleration:
[tex]a=\frac{2s}{t^2}=\frac{2(0.01)}{0.024^2}=34.7 m/s^2[/tex]
c)
The final speed of the upper eyelid can be found by using another suvat equation:
[tex]v=u+at[/tex]
where
v is the final speed
u is the initial speed
a is the acceleration
t is the time
For the eyelid in this problem, we have
u = 0
[tex]a=34.7 m/s^2[/tex]
t = 0.024 s
Substituting, we find the final speed:
[tex]v=0+(34.7)(0.024)=0.83 m/s[/tex]
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