Two cars are approaching an intersection.  One is 

11 

milemile 

south of the intersection and is moving at a constant speed of 

4040 

miles per hour.  At the same​ time, the other car is 

22 

milesmiles 

east of the intersection and is moving at a constant speed of 

1010 

miles per hour.

Question content area bottom

Part 1

​(a)  Express the distance d between the cars as a function of time t. ​ (Hint:  At 

t equals 0 commat=0, 

the cars are 

11 

milemile 

south and 

22 

milesmiles 

east of the​ intersection, respectively.)

d

equals=

StartRoot nothing t squared minus nothing t plus nothing EndRoot
 
enter your response heret2−
 
enter your response heret+
 
enter your response here

To express the distance \( d \) between the two cars as a function of time \( t \), let's start by defining the positions of the two cars.

### Car 1 (South of the intersection):
- At \( t = 0 \), Car 1 is 11 miles south of the intersection.
- The car is moving north at 40 miles per hour.
- So, the position of Car 1 as a function of time \( t \) is:
  \[
  y_1(t) = 11 - 40t
  \]
  Here, \( y_1(t) \) represents the distance of Car 1 from the intersection (positive \( y_1(t) \) means it is still south of the intersection).

### Car 2 (East of the intersection):
- At \( t = 0 \), Car 2 is 2 miles east of the intersection.
- The car is moving west at 10 miles per hour.
- So, the position of Car 2 as a function of time \( t \) is:
  \[
  x_2(t) = 2 - 10t
  \]
  Here, \( x_2(t) \) represents the distance of Car 2 from the intersection (positive \( x_2(t) \) means it is still east of the intersection).

### Distance \( d(t) \) between the two cars:
The distance \( d(t) \) between the two cars at time \( t \) is given by the Pythagorean theorem:
\[
d(t) = \sqrt{(x_2(t))^2 + (y_1(t))^2}
\]
Substituting the expressions for \( x_2(t) \) and \( y_1(t) \):
\[
d(t) = \sqrt{(2 - 10t)^2 + (11 - 40t)^2}
\]

### Simplifying the expression:
1. Expand the squares:
   \[
   d(t) = \sqrt{(2 - 10t)^2 + (11 - 40t)^2}
   = \sqrt{(4 - 40t + 100t^2) + (121 - 880t + 1600t^2)}
   \]
   
2. Combine like terms:
   \[
   d(t) = \sqrt{(100t^2 - 40t + 4) + (1600t^2 - 880t + 121)}
   \]
   \[
   d(t) = \sqrt{1700t^2 - 920t + 125}
   \]

Therefore, the distance between the cars as a function of time \( t \) is:
\[
d(t) = \sqrt{1700t^2 - 920t + 125}
\]

Would you like more details or have any questions about this?

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**Related Questions:**

1. How does the speed of the cars affect the rate at which the distance between them changes?
2. At what time will the two cars be closest to each other?
3. What is the minimum distance between the two cars?
4. How would the function change if the cars were moving in the opposite directions?
5. How does the expression change if the second car starts at a different distance from the intersection?

**Tip:** When solving problems involving motion, always break down the motion into components and consider each dimension separately before combining the results.

Two cars are approaching an intersection. One is 11 miles south of the intersection and is moving at a constant speed of 40 miles per hour. At the same time, the other car is 2 miles east of the intersection and is moving at a constant speed of 10 miles per hour. Express the distance d between the cars as a function of time t.

Learn how to calculate the distance between two cars approaching an intersection using algebra and the Pythagorean theorem. This problem involves creating a function of time based on the cars' speed and direction. Suitable for high school students in Grades 10-12.

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