rate distance and time

Question

Two jets leave an air base at the same time and travel in opposite directions. One jet travels
85

/mih
slower than the other. If the two jets are
6068mi
apart after
4
hours, what is the rate of each jet?

Answer 1

d = r · t

 

d₁ = r(4)

d₂ = (r - 85)(4)

 

d₁ + d₂ = 6068

r(4) + (r - 85)(4) = 6068

 

Can you solve for r?

Answer 2

If you set this problem up as two equations in two unknowns, and rearrange one equation to define one of the unknowns in terms of the other unknown, you can then substitute the "defined" unknown into the other equation, leaving you with one equation in one unknown, which is a solvable situation. So what are the two unknowns, and the two equations?

Let's let X represent the speed of the faster jet, and

let Y represent the speed of the slower jet. These are our two unknowns.

 

Our first equation gives us the relationship between these two speeds:  X  =  Y + 85  mi/h

 

The second equation steps from the function of the distance between the two jets with respect to time:

D(t)   =    X * t   +   Y * t     =     (X + Y) * t 

D(4)  =   (X + Y) * 4  =  6068 mi.

 

If we substitute the first equation into the second, we get    6068 mi  =  ((Y + 85) mi/h + Y mi/h) * 4 h, or

 

6068 mi  =  ( 2 * Y   +   85 ) * 4  mi/h * h  =  8*Y + 340 mi.

 

Solving for Y, we get  Y  =  ( 6068 - 340 ) / 8  =  716 mi/h.

 

Substituting this value for Y into the first equation and sloving for X, we get  X  =  716 mi/h  +  85 mi/h  =  801 mi/h

 

If you have any questions about how to set the equations for problems like this, please do not hesitate to post them.

 

Chris J.