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Two objects are moving along separate linear paths where each path is described by position [tex]\( d \)[/tex] (in meters) and time [tex]\( t \)[/tex] (in seconds). The equation for the first object is [tex]\( d = 2.5t + 2.2 \)[/tex]. The second object moves along a parallel path passing through [tex]\((t=0, d=1)\)[/tex]. What is the equation of the second path?

A. [tex]\( d = 2.5t + 1 \)[/tex]
B. [tex]\( d = t + 2.5 \)[/tex]
C. [tex]\( d = -0.4t + 1 \)[/tex]
D. [tex]\( d = 2.5t \div 3.2 \)[/tex]


Sagot :

Let's walk through the steps to determine the correct equation for the path of the second object.

### Step-by-Step Solution:

1. Equation of the First Object:
The position of the first object with respect to time is given by the equation:
[tex]\[ d = 2.5t + 2.2 \][/tex]
Here, the slope of the equation (rate of change of distance with time) is [tex]\(2.5\)[/tex].

2. Condition of Parallel Lines:
Since the second object's path is described by a line parallel to the first object, its slope must be the same. Therefore, the slope of the second object's path is also [tex]\(2.5\)[/tex].

3. Identifying the y-intercept:
The second object’s path passes through the point [tex]\((t=0, d=1)\)[/tex]. This point gives us the y-intercept of the line.

4. General Form of the Equation:
For a linear equation in the slope-intercept form [tex]\(d = mt + b\)[/tex], where [tex]\(m\)[/tex] is the slope and [tex]\(b\)[/tex] is the y-intercept:
- Slope ([tex]\(m\)[/tex]) is [tex]\(2.5\)[/tex].
- Y-intercept ([tex]\(b\)[/tex]) is [tex]\(1\)[/tex] (since it passes through the point [tex]\((0, 1)\)[/tex]).

5. Equation of the Second Object:
Plugging the slope and the y-intercept into the slope-intercept form gives us:
[tex]\[ d = 2.5t + 1 \][/tex]

6. Selecting the Correct Answer:
We compare this derived equation to the given choices:
- A. [tex]\(d = 2.5t + 1\)[/tex]
- B. [tex]\(d = t + 2.5\)[/tex]
- C. [tex]\(d = -0.4t + 1\)[/tex]
- D. [tex]\(d = 2.5t \div 3.2\)[/tex]

The correct answer matches the derived equation:

A. [tex]\(d = 2.5t + 1\)[/tex]

So, the equation of the second graph is:
[tex]\[ \boxed{d = 2.5t + 1} \][/tex]