Question

Amy and Reno are on a flat surface. Amy is 2 km due North of Reno. At midday, Amy starts walking slowly due East at 1 km/hr, while Reno starts walking slowly due North at 1 km per hour.

a) Find an expression for the square of the distance between them after t hours.
A) (2 + t)^2
B) (2 - t)^2
C) (2t)^2
D) (2t - 2)^2

Answer 1

The expression for the square of the distance between Amy and Reno after
`\( t \)` hours is option D)
`\( (2t - 2)^2 \)`.

Therefore the correct answer is option D)
`\( (2t - 2)^2 \)`

Step-by-step explanation:

Certainly!

After
`\( t \)` hours, Amy moves
`\( t \)` kilometers due East, adding to her initial position of 2 kilometers North of Reno. Meanwhile, Reno moves
`\( t \)` kilometers due North. Considering their movements, the distance between them forms the hypotenuse of a right-angled triangle.

Using the Pythagorean theorem
`\( \text{hypotenuse}^2 = \text{opposite side}^2 + \text{adjacent side}^2 \)`, we square the sum of their respective distances from their initial positions. Amy's Eastward movement combined with her initial position gives
`\( 2 + t \)` kilometers. Thus, the expression becomes
`\( (2 + t)^2 = (2t + 2 - 2)^2 = (2t - 2)^2 \)`.

Therefore, after
`\( t \)` hours, the square of the distance between Amy and Reno is accurately represented by
`\( (2t - 2)^2 \)`. This derivation considers their individual movements, the relationship between their positions, and time elapsed, ensuring a precise depiction of their distance after
`\( t \)` hours.

Therefore the correct answer is option D)
`\( (2t - 2)^2 \)`.