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A bicycle path starts 5 miles due east of an intersection and continues in a straight line to finish 8 miles due north of the same intersection. Let the intersection be represented by the point [tex]$(0,0)$[/tex]. Suppose a cyclist rides the path from start to finish in 2 hours. Which parametric equations model the path of the rider?

A. [tex]$x(t)=-\frac{5}{2} t+5$[/tex] and [tex]$y(t)=4$[/tex]

B. [tex]$x(t)=-\frac{5}{2} t+5$[/tex] and [tex]$y(t)=4 t+8$[/tex]

C. [tex]$x(t)=5 t+2$[/tex] and [tex]$y(t)=8 t$[/tex]

D. [tex]$x(t)=5 t$[/tex] and [tex]$y(t)=8 t+2$[/tex]

Sagot :

To determine the parametric equations for the path of the cyclist, let's break down the problem step-by-step:

1. Identify Initial and Final Positions:
- The intersection is at the origin [tex]\((0,0)\)[/tex].
- The start of the bicycle path is 5 miles east of the intersection, so the starting point is [tex]\((5,0)\)[/tex].
- The end of the bicycle path is 8 miles north of the intersection, so the ending point is [tex]\((0,8)\)[/tex].

2. Determine the time required:
- The cyclist rides from the start to the end in 2 hours. Thus, the total time is 2 hours.

3. Derive Parametric Equations:
The parametric equations for a straight-line motion between two points typically take the form:
[tex]\[ x(t) = x_{\text{initial}} + (x_{\text{final}} - x_{\text{initial}}) \frac{t}{T} \][/tex]
[tex]\[ y(t) = y_{\text{initial}} + (y_{\text{final}} - y_{\text{initial}}) \frac{t}{T} \][/tex]
where:
- [tex]\(x_{\text{initial}} = 5\)[/tex]
- [tex]\(y_{\text{initial}} = 0\)[/tex]
- [tex]\(x_{\text{final}} = 0\)[/tex]
- [tex]\(y_{\text{final}} = 8\)[/tex]
- [tex]\(T = 2\)[/tex]

Using these values,
[tex]\[ x(t) = 5 + (0 - 5) \frac{t}{2} \][/tex]
[tex]\[ y(t) = 0 + (8 - 0) \frac{t}{2} \][/tex]
Which simplifies to,
[tex]\[ x(t) = 5 - \frac{5}{2} t \][/tex]
[tex]\[ y(t) = 4 t \][/tex]

Thus, the parametric equations that model the path of the cyclist are:
[tex]\[ x(t) = 5 - \frac{5}{2} t \][/tex]
[tex]\[ y(t) = 4 t \][/tex]

Therefore, the correct answer is:
[tex]\[ x(t)=5 - \frac{5}{2}t \quad \text{and} \quad y(t)=4t \][/tex]