Question

which of the following equations can be used to calculate the values of t for which the distance between Chris and any is 300 miles

Answer 1

The first thing we must know in this case is that the distance is equal to the speed for time.
 We have then:
 For Chris:
 d1 = 40 * t
 For Amy:
 d2 = 60 * (t-2)
 The distance between both will be given by the hypotenuse of the rectangle triangle:
 300 = root ((40 * t) ^ 2 + (60 * (t-2)) ^ 2)
 Answer:
 300 = root ((40 * t) ^ 2 + (60 * (t-2)) ^ 2) 
 option 3

Answer 2

Time that Chris has been driving: t
Velocity of Chris: V1=40 mph
Distance traveled by Chris: d1 (The horizontal side of the right triangle in the figure)

We have:
V=d/t

Applying this formula for the case of Chris:
V1=d1/t
Replacing the known values:
40 mph = d1/t
Solving for d1:
t (40) = t (d1/t)
40t = d1
d1 = 40t

Amy leaves two hours later, then the time for Amy is the time for Chirs minus 2 hours:
Time for Amy: t2=t-2
Velocity of Amy: V2=60 mph
Distance traveled by Amy: d2 (The vertical side of the right triangle in the figure) 

Applying the formula of Velocity for Amy:
V2=d2 / t2
Replacing the known values:
60 mph = d2 / (t-2)
Solving for d2:
 (t-2) (60) = (t-2) [ d2 / (t-2) ]
60(t-2) = d2
d2=60(t-2) 

The Distance D between Chris and Amy is the hypothenuse of the right triangle, then we can use the Pythagorean Theorem to calculate it:
c^2=a^2+b^2, where c is the hypohenuse and a and b are the legs, the sides that form the right angle (the angle of 90°)

According with the figure:
c=D=300 miles
a=d1=40t
b=d2=60(t-2)

Replacing in the Pythagoras Theorem:
300^2 = (40t)^2 + [ 60(t-2) ]^2
Square root both sides of the equation:
sqrt(300^2) = sqrt { (40)^2 + [ 60(t-2) ]^2 }
300 = sqrt { (40t)^2 + [ 60(t-2) ] ^2 }
sqrt { (40t)^2 + [ 60(t-2) ]^2 } = 300

Answer: Third option