Distance, Rate and Time

 

Distance is a measurement that answers “how far?”

Time measures the length of time to travel that distance and answers “how long?”

Rate is a measurement that answers “how fast?”

The three are related this way:       D = R * T

Distance D is rate R multiplied by time T. Distance, rate and time problems require attention to units involved and  with multiple objects, deciding which one D, R, or T is the same between moving objects.

First, if we are traveling 60 mph for 3 hours, how far did we travel. How far means distance, so

directly substituting we get:

 

Notice how the unit hour in the denominator cancels the hrs next to the 3. This is very important, the units of a dimension must match, if not, unit conversions must be done first. Time units must be the same, distance units must be the same, and rate units must be the same. See further down.

Let's state this another way, suppose we were traveling 60 mph and traveled 180 miles. How long were we on the road?

Here we have the distance D, again. And we have the rate, R. We want to know how long?  We need to solve for time T.

 

One last variation, if we traveled 180 miles in 3 hours, how fast were we traveling?

D = R * T 

180 mi = R * 3 hr

dividing by 3 hr we get

180 mi/3hr = 60 mph

 

Notice how the units are carried along with their values in all calculations.

 

 

 

Unit issues:

Suppose we traveled for 8 hours at a rate of 60 feet per second. How far did we travel?

Well time is in hours, rate is in seconds. Time units must be the same. So first we need to convert one to the other. Let's convert the 8 hrs to seconds.

 

Another example:

A ball rolled 50 yards at 4 feet per second. How long did the ball roll?

Here we have yards and feet; they must be the same, so lets change the yards to feet.

This next example is a bit more advanced and a diagram can be helpful as shown.

John started to walk around a track that is 3000 feet around. He is walking at 3 mph. Sarah starts 5 minutes later at a light jog of 6 mph. How long will it take Sarah to catch up to John?

 

John is 5 minutes ahead, which means he's covered

            264 ft/min * 5 min  = 1320 feet. 

When Sarah overtakes John, both will have traveled a distance D

The distance that Sarah will have run is D = 528 ft/min * T

This distance is the same as what John has covered which is

the initial 1320 feet plus 264ft/min * T.

the distance Sarah runs = distance John walks

                528 ft/min * T   = 1320ft + 264 ft/min * T

Units are consistent, so we solve for time T

528 ft/min * T -  264 ft/min * T = 1200 ft

264 ft/min * T = 1320 ft

T = 1320 ft / (264 ft/min)

T = 5 min

The following diagram shows the distance each travels each minute.  The first shows then the head start that John has.  The vertical arrows show the progress each makes for every minute.

Sarah catches up to John at 5 minutes.  The position is D = 528ft/min * 5 min = 2640 ft

 

 

After Sarah catches up to John, how long will it take for  her to meet up with him again?

At 5 minutes they are shoulder to shoulder. Sarah will run around the track and reach this same location before catching up to John again.

This is the same as before but it's as if John had a head start of one complete track length.

John has run:   3000 ft + 264 ft/min * T

and Sarah will run 528 ft/min * T

3000 ft + 264 ft/min *T = 528 ft/min * T

3000 feet = 528 ft/min * T - 264 ft/min * T

3000 ft = 264 ft/min * T

3000 ft / (264 ft/min) = T

T = 11.36 minutes.