Answer problem

Question

Two cars are traveling in opposite directions on a freeway. Two and half hours after they meet they are 332 miles apart. If one car is traveling 15 miles faster than the other car find the speed of each car

Answer 1

Hi Quinton in Yuma!  Let's see if I can help out with your dilemma.

 

For starters, assume the cars are travelling on a straight road (facing perfectly East-West for convenience), and we'll begin the story from the time they're at the same spot.  We'll also assume that the cars are always perfectly travelling at a fixed speed (drivers are using cruise control).

 

To make sense of the whole problem, we have to see how each car's distance from the starting point, which is where they pass each other on the road, is related to the distance between them.  Say car A is heading west, and car B heading east.  If car A is 10 miles west and car B is 15 miles east of the starting point, what's the distance between them?  Well, it's 10 + 15 = 25 miles. 

 

So we see that the sum of their distances from the starting point equals their distance apart.  The problem tells us that they are 332 miles apart.  So a rough idea of our equation is

 

(car A distance from starting point) + (car B distance from starting point) = 332. 

 

Great!  But how do we figure out the distances?  The definition of speed is here to the rescue.  Speed = distance/time.  If we multiply both sides by time, we get

 

distance = (speed) x (time). 

 

We'll re-write this using the variables d for distance, v for speed, and t for time:

d = (v)(t).  But the problem tells us the amount of time:  2.5 hours.  So we have

d = 2.5v

 

Now for a bit of math to English translation:  "One car is traveling 15 miles an hour faster than the other car."  Let's say it's car A traveling 15 mph faster than car B.  Doesn't really matter which one we choose.  Using v_a for car A's speed and v_b for car B's speed, we have the following:  

 

v_a = v_b + 15. 

 

going back to our equation for distance, and letting d_a and d_b be the distances car A and B have gone since they passed each other, we have

 

d_a + d_b = 332.  Now, we substitute in d=2.5v for both cars:

2.5v_a + 2.5v_b = 332.  since v_a = v_b + 15, this becomes

2.5(v_b + 15) + 2.5v_b = 332.

 

It's all algebra from there.  You'll first solve for v_b, and then use that to solve for v_a.

 

Answer 2

This is a D-I-R-T (Distance-Is-Rate-times-Time) problem.  Usually, one of the items is constant or related; that way, we can solve the problem.  Here the total distance is 332 miles and one rate is 15 more than the other rate.

 

Let x = rate the first car (the speed)

 

So, the total distance is 332 = distance first car travelled    +    distance second car travelled

                               332 miles    =  (x mph) * (2.5 hrs)     +     ((x+15)mph) * (2.5 hrs)

                      332 = 2.5x + 2.5x + 37.5

                       294.5  =  5x                  (collect terms; subtract 37.5 from both sides)

                          58.9 mph = x                  (divide both sides by 5)

                        73.9 mph = x  +`15   (speed of second car)

 

Checking (very important):

  Is  (58.9)(2.5) + (73.9)(2.5) = 332    ?

            332 = 332    ?  yes