Answer: [tex]405.7\ ft[/tex]
Step-by-step explanation:
Given
After 11 sec, diver is 69 feet below sea level
after 28 s , it is 195 feet
rate of traveling
[tex]\Rightarrow u=\dfrac{195-69}{28}\\\\\Rightarrow u=4.53\ ft/s[/tex]
after 1.5 minutes, that is 90 s, diver must have traveled
[tex]\Rightarrow d=4.53\times 90\\\Rightarrow d=407.7\ ft[/tex]
Answer:
7299 feet
Step-by-step explanation:
At 11 second the depth is 69 feet
At 28 seconds the depth is 195 feet
Let the initial velocity at the time t = 0 is u.
Use second equation of motion
[tex]s = u t +0.5 at^2\\\\69 =11 u + 0.5 a\times 11\times 11\\\\69 = 11 u + 60.5 a..... (1)\\\\195 = u (28 -11) +0.5 a\times (28-11)^2\\\\195 = 17 u + 144.5 a .....(2)[/tex]
By soling (1) and (2)
a = 1.73 m/s^2, u = 3.25 m/s
So, the distance in 1.5 minutes is
h = 3.25 x 1.5 x 60 + 0.5 x 1.73 x 1.5 x 1.5 x 60 x 60
h = 292.5 + 7006.5 = 7299 ft
