Average Speed

Question

It took Gina 3.6 hours to drive to her mother's house on Tuesday. When returning home on Wednesday night, traffic was heavier, so the trip took her 4 hours. Her average speed on Wednesday was 4 MPH slower than it was on Tuesday. What was her average speed on Wednesday?

Answer 1

We know that speed times time equals distance

 

st = d

 

and we know in the scenario, that the distance to he mother's house was the same, so:

 

s₁t₁ = d

 

and

 

s₂t₂ = d

 

Where s₁ and s₂ are her speed on Tuesday and Wednesday, respectively, and t₁ and t₂ are her corresponding times. So, since d is the same in both equations, we can equate the two equations:

 

s₁t₁ = s₂t₂

 

We know what t₁ is (3.6 hours), and we know what t₂ is (4 hours). That leaves us with 2 unknowns. However, we do know that her speed on Wednesday was 4 mph slower (or less) than on Tuesday, so we can say:

 

s₂ = s₁ - 4

 

So now we can plug in our numbers:

 

s₁(3.6) = (s₁ - 4)(4)

 

Now we can solve for s₁, and we get:

 

3.6s₁ = 4s₁ - 16

-0.4s₁ = -16

s₁ = 40

 

So now we know her speed on Tuesday, and since her speed on Wednesday was 4 mph slower, then it was 36 mph

Answer 2

Start with what you know...

Tuesday:
Rate: x
Time: 3.6hours

Wednesday:
Rate: x -4
Time: 4hours

Distance are equal

 

Thus:
Rate x time = rate x time

 

Plug in what you know from above:

(x)3.6 = (x-4)4

Solve
3.6x = 4x – 16
-0.4x = -16

x=40mph
Rate on Tuesday

x-4 = 36mph
Rate on Wednesday