Question

Eleven seconds after a deep sea diver jumps into the ocean he is 69 feet below sea level and 28 seconds later, he is195 feet below sea level. If he is descending under water at a constant rate, how many feet below sea level will hebe 1.5 minutes after his initial descent

Answer 1

7299 feet

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

`s = u t +0.5 at^2\\\\69 =11 u + 0.5 a* 11* 11\\\\69 = 11 u + 60.5 a..... (1)\\\\195 = u (28 -11) +0.5 a* (28-11)^2\\\\195 = 17 u + 144.5 a .....(2)`

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

Answer 2

Answer:
`405.7\ ft`

Explanation:

Given

After 11 sec, diver is 69 feet below sea level

after 28 s , it is 195 feet

rate of traveling

`\Rightarrow u=(195-69)/(28)\\\\\Rightarrow u=4.53\ ft/s`

after 1.5 minutes, that is 90 s, diver must have traveled

`\Rightarrow d=4.53* 90\\\Rightarrow d=407.7\ ft`