Question

1. a scuba diver is swimming at the depth shown, and then swims 0.5 foot toward the surface every 3 seconds. what is the location of the scuba diver, relative to the surface, after 15 seconds? - 46 feet

Explanation:

Step1: Calculate number of 3 - second intervals

The time elapsed is 15 seconds. To find the number of 3 - second intervals, we divide the total time by 3. So, the number of intervals $n=\frac{15}{3}=5$.

Step2: Calculate total upward distance

The diver swims 0.5 foot upward every 3 seconds. So the total upward distance $d$ in 15 seconds is the upward distance per interval times the number of intervals. That is $d = 0.5\times5=2.5$ feet.

Step3: Calculate final depth

The initial depth of the diver is - 46 feet (negative because it is below the surface). After moving 2.5 feet upward, the new depth $D$ is the initial depth plus the upward distance. So $D=-46 + 2.5=-43.5$ feet.

Answer:

• 43.5 feet