Question

a scuba diver descends in the water at a rate of $20\frac{1}{2}$ feet per minute for 2.8 minutes. he immediately changes course when he - sees a large fish and ascends at a rate of 25 feet per minute for 2 minutes. what is the scuba divers position relative to sea level after the 4.8 minutes? enter your answer as a decimal in the box.

Explanation:

Step1: Calculate the distance of descent

The diver descends at a rate of $20\frac{1}{2}=\frac{41}{2} = 20.5$ feet per minute for 2.8 minutes. The distance of descent $d_1$ is given by the product of the rate and time, so $d_1=20.5\times2.8 = 57.4$ feet.

Step2: Calculate the distance of ascent

The diver ascends at a rate of 25 feet per minute for 2 minutes. The distance of ascent $d_2$ is the product of the rate and time, so $d_2 = 25\times2=50$ feet.

Step3: Calculate the net - position relative to sea - level

The net position $d$ relative to sea - level is the distance of descent minus the distance of ascent. So $d=d_1 - d_2=57.4−50 = 7.4$ feet.

Answer:

7.4