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PLZ HELP!

Two submersibles, one red and one blue, started rising toward the surface at the same time. They each rose at a constant speed. the red submersible starts rising from an altitude of 80 meters below the surface. after 30 seconds, it was 60 meters below the surface. the following equation gives the altitude (in meters relative to the surface) of the blue submersible as a function of time (in seconds). A= -90 + 0.6t

Which submersible started rising from a higher altitude?

Which submersible rose faster?

2 Answers

3 votes

Answer:

The person above is correct

Explanation:

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User Lionel
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4 votes

Answer:

  • red started higher
  • red rose faster

Explanation:

Blue started at -90 meters at t=0. Red started at -80 meters, according to the problem statement. So Red started higher.

Red rose at 20 meters in 30 seconds, or 20/30 = 2/3 meters per second.

Blue's equation has a depth term of 0.6t that indicates it rises at 0.6 meters per second. We know that 2/3 ≈ 0.6667 > 0.6, so Red rises faster.

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The graph of the red and blue depths shows red starts above blue and has a steeper rate of rise.

PLZ HELP! Two submersibles, one red and one blue, started rising toward the surface-example-1
User Tony Wang
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