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

** r** for

It's not a good idea to use ** s **for

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 ** d**istance,

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

**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/h**r, 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

Their combined rate is

so 5 ×

This means

**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

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) =

**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 × t4 × 55 = 220 mi |
55 miles/hour | 4 hours |

b) | 60 t miles |
60 miles/hour | t hours |

c) | 150 miles | =r = D ÷ t150 ÷ 3 = 50 mi/hr |
3 hours |

d) | D miles |
D/7 mi/hr | 7 hours |

e) | 720 kilometers | 120 kilometers/hour | t = D ÷ =r720 ÷ 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 | 3 mi/hrr |
7 hrs | 21 r |

Barry | mi/hrr |
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 | mi/hrr |
3 hrs | 3 r |

Car 2 | ( + 10) mi/hrr |
3 hrs | 3( + 10)r |

240 miles |

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

3 *r* + 3( ** r **+ 10) = 240

3

6

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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