# Thread: need help setting up a uniform motion problem

1. ## need help setting up a uniform motion problem

I can't figure out how to set this up. Would someone help please.

Two cars enter the florida turnpike at Commercial Blvd. at 8am each heading toward Wildwood. One cars average speed is 10mph more than the other's. The faster car arrives at Wildwood at 11am, 1/2 hour before the other car. What is the average speed of each car? How far did each travel?

2. Originally Posted by dsgarr
I can't figure out how to set this up. Would someone help please.

Two cars enter the florida turnpike at Commercial Blvd. at 8am each heading toward Wildwood. One cars average speed is 10mph more than the other's. The faster car arrives at Wildwood at 11am, 1/2 hour before the other car. What is the average speed of each car? How far did each travel?
ok, so you should know that

$\text{Average Speed} = \frac {\text{Total distance traveled}}{\text{Total time taken}}$

Here, the total time for the faster car 3 hrs. the total time for the slower car is 3.5 hrs.

Now, let the average speed of the faster car be $v$, then the average speed of the slower car is $(v - 10)$.

also, they travel the same distance, $d$, which is yet an unknown.

so, we have:

$v = \frac d3$ ...........................(1) this is for the faster car

and

$v - 10 = \frac d{3.5}$ ..............(2) this is for the slower car

you have two equations and two unknowns, all you want to find is $v$ and you will know the average speed of the faster car, and hence, by subtracting 10, the average speed of the slower car

3. Hello, dsgarr!

Two cars enter the florida turnpike at Commercial Blvd. at 8am, each heading
toward Wildwood. One car's average speed is 10mph more than the other's.
The faster car arrives at Wildwood at 11am, 1/2 hour before the other car.
What is the average speed of each car? How far did each travel?

Let $x$ = speed of the slower car.
Let $x+10$ = speed of the faster car.

Let $D$ = distance to Wildwood.

The faster car reached Wildwood in 3 hours.
. . Its time is: . $\frac{D}{x+10}$ hours.
We have: . $\frac{D}{x+10} \:=\:3 \quad\Rightarrow\quad D \:=\:3x + 30$ .[1]

The slower car reached Wildwood in 3.5 hours.
.Its time is: . $\frac{D}{x}$ hours.
We have: . $\frac{D}{x} \:=\:\frac{7}{2} \quad\Rightarrow\quad D \:=\:\frac{7}{2}x$ .[2]

Equate [2] and [1]: . $\frac{7}{2}x \:=\:3x + 30 \quad\Rightarrow\quad \frac{1}{2}x \:=\:30 \quad\Rightarrow\quad x \:=\:60$

Substitute into [2]: . $D \:=\:\frac{7}{2}(60) \:=\:210$

The slower car went 60 mph; the faster car went 70 mph.

The distance to Wildwood is 210 miles.