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
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
