PRECAL HW MATH HELP

Question

Two particles are moving in the x-y plane. They move along straight lines at constant speed. At time t, particle A’s position is given by x = t + 2, y = 1/ 2 t − 3 and particle B’s position is given by x = 12 − 2t, y = 6 − 1/ 3 t

(a) Find the equation of the line along which particle A moves. Sketch this line, and label A’s starting point and direction of motion.

(b) Find the equation of the line along which particle B moves. Sketch this line on the same axes, and label B’s starting point and direction of motion.

(c) Find the time (i.e., the value of t) at which the distance between A and B is minimal. Find the locations of particles A and B at this time, and label them on your graph.

Answer 1

The two line equations are easy,

line 1 y=(x/2)-4

line 2 y=(x/6)+4

What is interesting is that because time≥0

the lines start at t=0 and have a direction for A and B

t=0 (x₁,y₁)=(2,-3) t=1 (x₁,y₁)=((3,-5/2)

t=0 (x₂,y₂)=((12,6) t=1 (x₂,y₂)=10,17/3)

The lines intersect if the domain for t is (-infinity,infinity)

but it would be unlikely that A and B would be

at intersection at same time.

As it is,

time starts at zero and direction is away from intersection point for B.

Using calculus, at t=3.9 the distance from A to B is minimum.

A at (5.9,-1.05)

B at (4.2,4.7)

Answer 2