Question

Car A is 16 kilometers to the west of car B. At noon, Car A begins driving east at a constant speed 4 kilometers/hour, and car B begins driving north at a constant speed of 3 kilometers/hour. (a) (5 points) Let x be the distance that car A has traveled since noon, and let y be the distance that car B has traveled since noon. Let z be the distance between the two cars. Find an equation that relates x to x and y.

Answer 1

Answer:

The distance traveled can be found by kinematics equations

$x = vt$

$x(t) = v_At - 16= 4t - 16\\y(t) = v_Bt = 3t\\z(t) = \sqrt{x(t)^2 + y(t)^2} = \sqrt{16t^2 - 128t + 256 + 9t^2} = \sqrt{25t^2 - 128t + 256}$

Explanation:

The initial position of Car B is denoted as origin. Car A started at -16 km, and moving right. Car B started at origin and moving up. z is the magnitude of the vector with components x and y.

The relation can be found by pythagorean theorem.

For checking the solution, we can find the positions at t = 4h. Car A is at the origin, and Car B is 12 km north of origin. z = 12.