Math Problem Statement

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

Solution

To express the distance dd between the two cars as a function of time tt, let's start by defining the positions of the two cars.

Car 1 (South of the intersection):

  • At t=0t = 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 tt is: y1(t)=1140ty_1(t) = 11 - 40t Here, y1(t)y_1(t) represents the distance of Car 1 from the intersection (positive y1(t)y_1(t) means it is still south of the intersection).

Car 2 (East of the intersection):

  • At t=0t = 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 tt is: x2(t)=210tx_2(t) = 2 - 10t Here, x2(t)x_2(t) represents the distance of Car 2 from the intersection (positive x2(t)x_2(t) means it is still east of the intersection).

Distance d(t)d(t) between the two cars:

The distance d(t)d(t) between the two cars at time tt is given by the Pythagorean theorem: d(t)=(x2(t))2+(y1(t))2d(t) = \sqrt{(x_2(t))^2 + (y_1(t))^2} Substituting the expressions for x2(t)x_2(t) and y1(t)y_1(t): d(t)=(210t)2+(1140t)2d(t) = \sqrt{(2 - 10t)^2 + (11 - 40t)^2}

Simplifying the expression:

  1. Expand the squares: = \sqrt{(4 - 40t + 100t^2) + (121 - 880t + 1600t^2)}$$
  2. Combine like terms: d(t)=(100t240t+4)+(1600t2880t+121)d(t) = \sqrt{(100t^2 - 40t + 4) + (1600t^2 - 880t + 121)} d(t)=1700t2920t+125d(t) = \sqrt{1700t^2 - 920t + 125}

Therefore, the distance between the cars as a function of time tt is: d(t)=1700t2920t+125d(t) = \sqrt{1700t^2 - 920t + 125}

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


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.

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Math Problem Analysis

Mathematical Concepts

Algebra
Pythagorean Theorem
Functions of Time

Formulas

Distance formula: d = √(x^2 + y^2)
Position of car 1: y_1(t) = 11 - 40t
Position of car 2: x_2(t) = 2 - 10t

Theorems

Pythagorean Theorem

Suitable Grade Level

Grades 10-12