Algebra Help

Question

So, the question is:

 

Beth left the movie theater and traveled east. Heather left three hours later traveling 42 km/h faster in an effort to catch up to her. After two hours Heather finally caught up. Find Beth's average speed.

 

I don't know how to set up the equation with just a "Heather left three hours later", not any specific time.

Thanks for the help.

 

Answer 1

R = Beth's speed

 

R + 42 = Heather's speed

 

Beth travel for 5 hours

 

Heather travels for 2 hours

 

they travel the same distance

 

Rate times Time = Distance

 

5R is Beth's distance

 

(R + 42)2 is Heather's distance

 

5R = 2(R+42)

 

5R = 2R + 84

 

3R = 84

 

  R = 28

 

28 km/hr is Beth's speed

 

70 km/hr is Heather's speed

 

Answer 2

v₁ = Beth's speed

v₂ = Heather's speed = v₁ + 42

 

x₁(t) = Beth's distance from the theater measured from the time Heather left.

 

x₁(t) = v₁*3 + v₁t = v₁(3 + t)    [Note the 3v₁ accounts for Beth's 3h head start.]

 

x₂(t) = Heather's distance from the theater measured from the time Heather left.

 

x₂(t) = v₂t = (v₁ + 42)t

 

When Heather catches up (at t = 2 hours), they are at the same distance from the theater, so x₁ = x₂.

 

v₁(3 + t) = (v₁ + 42)t

 

t = 2 hours

 

v₁(3 + 2) = (v₁ + 42)*2

 

5v₁ = 2v₁ + 84

 

3v₁ = 84

 

v₁ = 84/3 = 28 km/h

 

Beth's average speed was 28 km/h.