SOLVING WORD PROBLEMS: MOTION

The Holy Trinity of Motion: Distance, Speed and Time

Thoroughbred Horses Speeding around a 1¼ - mile Racetrack

Since the first Saturday in May, 1973, horse racing fans have talked about Secretariat's Kentucky Derby win. First, because his final time of 1:59 2/5 broke the track record for 1¼ miles in the world's most important horse race. And second, because The Big Red Horse ran faster every quarter mile of the race, just building power like a locomotive and finishing like a jet.

Speed: Distance ÷ Time

Let's calculate Secretariat's average RATE OF SPEED for this race from the given information. Speed for vehicles is measured in either miles or kilometers per hour, but we'll find this one in feet per second.

We know he ran 1¼ miles in 1 minute, 59 and 2/5 seconds.
We convert miles to feet and write time in seconds in decimal form:

1¼ miles × 5280 feet/mile = 6 600 feet,
and
1 minute, 59 and 2/5 seconds = (60 + 59 + 0.4) = 119.4 seconds.
His speed in feet/second = 6 600 feet ÷ 119.4 seconds = 55.28 feet/second.

We found Secretariat's speed by dividing the distance he ran by the time he took to run it.

Speed, also called rate or velocity, is a rate -- a comparison of numbers in different units. The units for speed tell us that it is distance divided by time because speed is measured in miles per hour, kilometers per hour, feet per second, yards per minute etc. Every one of these rates is composed of a distance (miles, feet, yards) divided by time (hours, minutes, seconds).

Example: If we travel 200 miles in 4 hours, what is our average speed or velocity?

Solution: Speed = distance ÷ time = 200 mi ÷ 4 hrs = 50 mi/hr.

The units tell us we divided distance by time.

Notation Note: In the formulas we've developed for solving motion problems, we always use D or d for distance, and t for time but we don't always use the same letter to represent speed because we call it different things. The letters we use to denote speed are:

r for rate, s for speed, and v for velocity.

It's not a good idea to use s for speed because " s " can look like 5 when written by hand. Generally, we don't use v for velocity unless we're doing physics problems, so we'll use r for rate in our formulas and questions in this lesson.

We see from the examples we've done that the formula for speed or rate is:

rate equals distance ÷ time

If we multiply both sides of this equality by t, we get:

D = r × t
which says distance equals rate × time,

and now, we'll divide both sides of this formula by r to get:

time equals distance ÷ rate.

Useful Units

The units we use for the 3 variables in our motion problems tell us all we need to know. They help us recognize what operations to perform when we need to find distance, rate or time.
It's obvious that speed or rate is distance / time -- we measure it in units of distance over units of time.

Now let's investigate the units in the 2nd formula, the one for distance that says D = r × t.
If the rate is in miles / hour and the time is in hours, we multiply them to get Distance or miles. We can see that the hours cancel each other out and we're left with miles.

Example: Driving at 80 kilometers per hour for 6 hours, we travel 80 × 6 = 480 kilometers.

The 3rd formula says time = Distance ÷ rate. Using the same units as before, we should get hours when we divide miles by miles per hour. (Remember, to divide, we invert and multiply.)
Let's see how this one works:

Example:

If we traveled 480 kilometers at a constant rate of 80 km / hr, it took us 480 ÷ 80 = 6 hours.

Solving Motion Problems

Since motion involves 3 variables -- Distance, rate and time -- there are 3 kinds of motion problems: a) the distance is unknown, b) the time is unknown and c) the rate is unknown.

1 - Finding the Distance:

Mary leaves Toronto heading for Montreal at an average speed of 50 km/h at the same time that Charles leaves Montreal heading for Toronto. He averages 40 km/hr, and they meet in 5 hours.
How far apart are the two cities?

To help us "see the picture" we make a diagram.

There are 2 ways to work this one out:

First, we know that together, they travel 90 kilometers in an hour. If they drive for 5 hours, they must have traveled 450 kilometers.

The other way is to work in parts -- find the distance each one travels, then add them together.
Mary travels 5 × 50 = 250 kilometers and Charles travels 5 × 40 = 200 kilometers, so together, they traveled 250 + 200 = 450 kilometers.

Speedometer with miles/hour and kilometers/hour.

2 - Rate:

Now Mary and Charles are driving towards each other from cities that are 300 miles apart. If Mary travels twice as fast as Charles and they meet in 5 hours, how fast is each one going?

Here's the picture:

We let r = Charles' rate, so 2r = Mary's rate.
Their combined rate is r + 2r or 3r.
distance = rate × time = 300 miles
so 5 × 3r = 300 miles.
This means 15r = 300, so r = 300 ÷ 15 = 20.

Charles crawls at 20 miles per hour and Mary, the speeder, moves at 40 miles per hour.

When we check, we see that 5 × 20 + 5 × 40 = 100 + 200 = 300 miles

Stopwatch to measures seconds, minutes and hours.

3 - Time:

Montreal and Detroit are 600 km apart. Ryan leaves Detroit for Montreal travelling at a constant speed of 120 km/hr. At the same time, Norman leaves Montreal heading for Detroit. He averages 130 km/hr. How long does it take them to meet? How far from Detroit do they meet?

We need to find the time.
time = distance ÷ rate
their combined rate is 250 km/hr.
We divide 600 by 250 to get 2.4 hours (2 hours 24 minutes).

It takes 2.4 hours for them to meet.

In 2.4 hours, at 120 km/hr, Ryan will be 120 × 2.4 or 288 kilometers from Detroit.

There's another way to solve this one. We could make a data table, then write and solve an equation that says the sum of their distances (rate × time) is 600 km.
We let t = the time it takes to meet. Here's the data table.

 rate time distance ( r × t ) Ryan 120 km/hr t 120 t Norman 130 km/hr t 130 t

120 t + 130 t = 600
250 t = 600

t = 2.4 hours. D = 2.4 (120) = 288 km

They meet after 2.4 hours at a distance of 288 km from Detroit.

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Now get a pencil, an eraser and a note book, copy the questions,
do the practice exercise(s), then check your work with the solutions.
If you get stuck, review the examples in the lesson, then try again.

Practice Exercises

1) Write the formula we use to find the missing variable
then fill in the blanks with a number value or an algebraic expression.

 Distance Rate Time a) ? 55 miles/hour 4 hours b) ? 60 miles/hour t hours c) 150 miles ? 3 hours d) D miles ? 7 hours e) 720 kilometers 120 kilometers/hour ? f) 260 feet r feet/second ?

2) Solve these word problems. Make a diagram or data table if needed.
Show all your work and remember the units. Answers should be complete sentences.

a) Two cars start at exactly the same place at the same time and drive in opposite directions. One travels at 70 km/hour, the other at 90 km/hour. How long will it take them to be 800 kilometers apart? (make a diagram!)

b) Chuck and his friend Barry live in towns that are 560 miles apart. Last Saturday, they both left home at the same time and drove towards each other to meet for the long weekend. Chuck has a new car so he drove 3 times as fast as Barry did in his old clunker. If they met after 7 hours of driving, what was Chuck's rate of speed?

c) Two cars start at exactly the same place at the same time and drive in opposite directions. One travels 10 mi/hour faster than the other. After 3 hours, they are 240 miles apart. How fast is each car going?

d) Speedy walks very fast. He takes 70 strides/minute and each stride is about 1 yard long.
About how long will he take to walk a mile at this pace?

Learning Activity:

A standard American football field is 120 yards long if measured from the back of one end zone to the back of the other.

a) How many times do we have to go from end to end to walk ½ mile?

b) How many yards are there in ½ mile?

c) Use a stopwatch or a wristwatch with a second hand, to find the time it takes to run, walk, limp or hop (your choice) that ½ mile. Find your speed in feet per second and yards per minute. Compare your speed to Secretariat's speed in the 1973 Kentucky Derby (at the start of the lesson)

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Solutions

1) Write the formula we use to find the missing variable
then fill in the blanks with a number value or an algebraic expression.

 Distance Rate Time a) D = r × t = 4 × 55 = 220 mi 55 miles/hour 4 hours b) 60 t miles 60 miles/hour t hours c) 150 miles r = D ÷ t = 150 ÷ 3 = 50 mi/hr 3 hours d) D miles D/7 mi/hr 7 hours e) 720 kilometers 120 kilometers/hour t = D ÷ r = 720 ÷ 120 = 6 hours f) 260 feet r feet/second 260/ r seconds

2) a) One travels at 70 km/hour, the other at 90 km/hour. How long will it take them to be 800 kilometers apart?

b) 560 miles apart. Chuck drove 3 times as fast as Barry. They met after 7 hours

We let r = Barry's rate. Here's the data table.

 rate time distance ( r × t ) Chuck 3r mi/hr 7 hrs 21 r Barry r mi/hr 7 hrs 7 r 560 miles

The sum of their distances must = 560 miles, so our equation is:

21 r + 7 r = 560
28 r = 560 so r = 560 ÷ 28 = 20
When we check our solution in the original equation, it is correct.

Chuck's rate of speed was 60 miles per hour.

c) opposite directions. One 10 mi/hour faster. In 3 hours, 240 miles apart. Find rates.

We let r = first car's rate. Here's the data table.

 rate time distance ( r × t ) Car 1 r mi/hr 3 hrs 3 r Car 2 ( r + 10) mi/hr 3 hrs 3( r + 10) 240 miles

The sum of their distances must = 240 miles, so our equation is:

3 r + 3( r + 10) = 240
3 r + 3 r + 30 = 240
6 r = 210, so r = 210 ÷ 6 = 35
When we check our solution in the original equation, it is correct.

The first car drove at 35 miles per hour, the second at 45 miles per hour.

d) He walks 70 yards/minute. 1 mile = 1760 yards.
t = D ÷ r so it takes him 1760 ÷ 70 = 25.14 minutes.

Speedy will take 25.14 minutes to walk a mile.

Learning Activity:

a) How many times do we have to go from end to end to walk ½ mile?

½ mile = 1760 yards ÷ 2 = 880 yards, field measures 120 yds: 880 ÷ 120 = 7.33

b) How many yards are there in ½ mile? 880 yards = ½ mile.

c) 880 yards = 880 yards × 3 feet/yard = 2640 feet in ½ mile.
To find your speed in yards/minute, divide 880 yards by your time in minutes.

To find your speed in feet/second, divide 2640 feet by your time in seconds.

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