Answer:
Distance, d = 101.388 meters
Explanation:
It is given that,
The average velocity of plane, v = 3650 km/h = 1013.88 m/s
The time for which eye blinks, [tex]t=100\ ms=0.1\ s[/tex]
Let d is the distance covered by the jet. It can be calculated as :
[tex]d=v\times t[/tex]
[tex]d=1013.88\ m/s\times 0.1\ s[/tex]
d = 101.388 meters
So, the distance covered by a fighter jet during a pilot's blink is 101.388 meters. Hence, this is the required solution.