Question

The pressure (P) exerted on an object in the ocean at a depth (d) of 100 meters is 11 atmospheres. Supposing that the object is descending at a rate of 2 m/s, find the rate of change of pressure at a depth of 100 meters.
(P = kd + 1, where k is a constant.)

Answer 1

\[P(t)=kd(t)+1\]
\[P'(t)=kd'(t)\]
But
\[d'(t)=2\]
So
\[P'(t)=2k\]
for any depth.

Answer 2

from P=kd+1
11atm =k(100)+1, k=10/100=0.1= constant
DP(t)=kDd(t), Dd(t)=2m/s,
DP(t)=kDd(t)=(0.1)(2)
= 0.2 atmospheric pressure per sec ans