Expert: Josh - 3/6/2005

Question
Matt is going to a baseball game.  If he goes 75 miles per hour, he gets there an hour before the time he needs to be there.  If he goes 45 miles per hour, he gets there an hour late.  What's the distance that he is traveling!?!  

HELP!  This is driving me crazy! Thanks!

Answer
Nikki,

The first rule for translating a word problem into a maths problem is to assign a name to each unknown quantity.

Here, our task is to work out the distance travelled, this is the unknown quantity. So, let us denote the distance travelled with D.

There is only one thing to remember. Speed is simply a change in distance over time. So, we have
distance = speed x elapsed_time.  ...[#1]

Translating this into an equation, D=S*t, where S is the speed at which you travel, t represents the time duration.

Now, what do we know. We have been given two speeds. Why not call them S1, S2. And let S1=75 [mile/hour], S2=45 [mile/hour].

The key to algebra is that everytime the numerical value of some quantity is unknown, we give it a label, like X, Y, Z etc and try to relate them.

In this problem, it really doesn't matter what time Matt starts his journey and what time he arrives at the destination. The important thing is the relative difference.

Travelling at speed S1, he arrives one hour ahead of schedule, i.e., at X-1 hour. Let's assume that the duration of the trip is Y.
Travelling at a slower speed, S2, he arrives one hour later than expected, i.e., at X+1 hour. The duration is Y+2.

Now, using the physical principle discussed in [#1], in case one, we have the distance travelled,
D=S1*Y...[#2]

In case two, we also travel the same distance,
D=S2*(Y+2)...[#3]

From [#2], Y=D/S1, putting this into expression[#3],
D=S2*((D/S1)+2)
D=(D*S2/S1)+2*S2
D-(D*S2/S1)=2*S2
D*(1-S2/S1)=2*S2
D=2*S2/[1-(S2/S1)]

But the speeds S1=75, S2=45, so
D=2*45/[1-45/75]
=90/[1-3/5]
=90/(2/5)
=90*5/2
=225 miles