   SEARCH HOME Math Central Quandaries & Queries  Question from lanie, a student: Two cars enter the Florida Turnpike at Commercial Boulevard at 8:00am, each heading for Wildwood. One car's average speed is 10 miles per hour more than the other's. The faster car arrives at Wildwood at 11:00 am, 1/2 hour before the other car. What is the average speed of each car? How far did each travel? Hi Lanie.

Assume that both cars took the same route, so they went the same distance.

Remember that velocity (speed) is the distance divided by the time. So we can use algebra to solve this problem.

v = d / t.

In particular, let's call the two cars F and S (fast and slow). Then we can double up our equation with subscripts:

vS = dS / tS

vF = dF / tF

If we solve each of these for distance, we have:

dS = vS × tS

dF = vF × tF

But we said that both cars went the same distance, so dS = dF and so

vS × tS = vF × tF

Now look at the times involved. The fast car took 3 hours and the slow car took 3.5 hours, so we can substitute those into the equation:

vS × (3.5) = vF × (3)

And since we know the fast car's speed is 10 mph faster than the slow car, we know that vF = vS + 10, so we can substitute vS + 10 in for vF:

vS × (3.5) = (vS + 10) × (3)

Now, you can easily solve for vS and the rest of the questions you are asked.

Cheers,
Stephen La Rocque.     Math Central is supported by the University of Regina and The Pacific Institute for the Mathematical Sciences.