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Difficulty: MediumBasic Algebraic Expressions and One-Step Equations

A commercial diver descends at a constant rate of 2.52.5 meters per second. If the diver's total change in depth after tt seconds is −85-85 meters, which of the following equations can be solved to find tt?

  1. −2.5t=−85-2.5t = -85Answer
  2. B
    t−2.5=−85\frac{t}{-2.5} = -85
  3. C
    t−2.5=−85t - 2.5 = -85
  4. D
    −2.5+t=−85-2.5 + t = -85
  5. E
    −85t=−2.5-85t = -2.5

Answer

−2.5t=−85-2.5t = -85
The rate of depth change is −2.5-2.5 meters per second. Over tt seconds, the total distance descended is given by the product of the unit rate and time, −2.5t-2.5t. Equating this to the given total change in depth of −85-85 meters gives −2.5t=−85-2.5t = -85.

Step-by-Step Solution

1
Identify the given rate, total quantity, and variable.
Rate of change = −2.5-2.5 m/s, time = tt seconds, total depth change = −85-85 m.
Establishing knowns and unknown variables is the first step in setting up a one-step algebraic equation.
2
Formulate the relationship using the rate formula (Rate)×(Time)=Total Change(\text{Rate}) \times (\text{Time}) = \text{Total Change}.
−2.5×t=−85-2.5 \times t = -85, which simplifies to −2.5t=−85-2.5t = -85.
Since the descent occurs at a constant rate per second, total change is the product of the rate per second and total seconds.

Key Concept

Formulating one-step linear equations from rate word problems
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Topics

Basic Algebraic Expressions and One-Step Equations

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