Motion Word Problems

Students will solve motion problems by using a Guess & Check Chart and

Algebra

D = R x T ReviewMotion Problems use the formula

Distance = Rate x TimeIf the rate is 60 mph and the time is 3 hours, what is the distance?

If the distance is 200 miles and the rate is 40 mph, what is the time?

If the distance is 300 miles and it took 4 hours, what was the rate of speed?

If the time was 2.5 hours and the rate was 40 mph, what was the distance?

What’s involved in a motion problem?Two vehicles

A slower oneAnd a faster one!

What’s involved in a motion problem?Going in the same

direction.

Or opposite directions!!

What’s involved in a motion problem?Leaving at the

same time.

Leaving at different times

Step One: Draw the problemIt helps to find the distance needed to solve the problem.

Look for: Starting Point,Direction, Departure Time

Slow Car (56 mph) Fast car (60 mph)

Slow Car Distance + Fast Car Distance = 464Slow Car Distance + Fast Car Distance = 464

Draw: Ex 1. Two cars start at the same point, at the same time, and travel in opposite directions. The slow car travels at 56 mph and the fast car travels at 60 mph. In how many hours, will they be 464 miles apart?

Ex. 2 Two trains start at the same time from stations that were 360 miles apart and travel towards each other. The rate of the fast train exceeded the rate of the slow train by 10 miles per hour. At the end of 2 hours, the trains were still 120 miles apart. Find rate of each train.

Drawing Practice

Starting Pt: 360 miles apart.

Direction: Towards each other

Starting Time: Same

Fast Train Slow Train120 miles

Total Distance: 360 miles

Fast Distance + Slow Distance + 120 = 360 or

Fast Distance + Slow Distance = 240

Drawing Practice

Starting Pt: Home & Unknown

Direction: Away and back home

Starting Time: A total of 7 hours

Fast Rate

Slow Rate

Fast Distance = Slow Distance

Ex 3. How far can a man drive out into the country at the average rate of 60 miles per hour and return over the same road at 45 miles per hour if he travels a total of 7 hours.

Step 2: Make A Chart

SlowRate

Slow Time

Slow Dist

FastRate

FastTime

FastDist

DistanceEquation

Ex 1. Two cars start at the same point, at the same time, and travel in opposite directions. The slow car travels at 56 mph and the fast car travels at 60 mph. In how many hours, will they be 464 miles apart?

SlowRate

Slow Time

Slow Dist

FastRate

FastTime

FastDist

DistanceEquation

SD + FD = 464

56 60

Fill in: What you know

Which is the “GUESS” column? Hours

Ex 1. Two cars start at the same point, at the same time, and travel in opposite directions. The slow car travels at 56 mph and the fast car travels at 60 mph. In how many hours, will they be 464 miles apart?

112+120 = 232No

120260112256

DistanceEquation

SD + FD = 464

FastDist

FastTime

FastRate

Slow Dist

Slow Time

SlowRate

Guess: 2 hours and fill in rest of chart

Try more Guesses

IN 4 HOURS, THEY WILL BE 464 MILES APART

Ex 1. Two cars start at the same point, at the same time, and travel in opposite directions. The slow car travels at 56 mph and the fast car travels at 60 mph. In how many hours, will they be 464 miles apart?

224+240=464YES!

240460224456

280+300=580No

300560280556

112+120 = 232No

120260112256

DistanceEquation

SD + FD = 464

FastDist

FastTime

FastRate

Slow Dist

Slow Time

SlowRate

Using Algebra

Use Guess & Check Chart to convert to an Algebra problem

Place x in Guess Column. Fill in rest of row

Define variablesWrite & Solve EquationAnswer Question!

Ex 1. Two cars start at the same point, at the same time, and travel in opposite directions. The slow car travels at 56 mph and the fast car travels at 60 mph. In how many hours, will they be 464 miles apart?

56x+60x=46460xX6056xX56

224+240=464YES!

240460224456

DistanceEquation

SD + FD = 464

FastDist

FastTime

FastRate

Slow Dist

Slow Time

SlowRate

Let x = # of hours 56x+60x=464

116x=464

X=4

Answer: It took four hours.

Ex. 2 Two trains start at the same time from stations that were 360 miles apart and travel towards each other. The rate of the fast train exceeded the rate of the slow train by 10 miles per hour. At the end of 2 hours, the trains were still 120 miles apart. Find rate of each train.

SlowRate

Slow Time

Slow Dist

FastRateSR+10

FastTime

FastDist

DistanceEquationSD+FD=240

2 2

Fill in what you know and find Guess Column

Guess: Slow Rate

Fast Rate = Slow Rate + 10

Ex. 2 Two trains start at the same time from stations that were 360 miles apart and travel towards each other. The rate of the fast train exceeded the rate of the slow train by 10 miles per hour. At the end of 2 hours, the trains were still 120 miles apart. Find rate of each train.

110+130=240130265110255

120+140=260140270120260

100+120=220120260100250

DistanceEquationSD+FD=240

FastDist

FastTime

FastRateSR+10

Slow Dist

Slow Time

SlowRate

KEEP GUESSING

Slow Train Rate: 55 Fast Train: 65

Ex. 2

2X+2(X+10)=2402(X+10)2X+102X2X

110+130=240130265110255

Dist EquSD+FD=240

FDFTFRSR+10

SDSTSR

On to ALGEBRA

Let x = slow rate

Let x + 10= Fast rate

2x+2(x+10) = 240

2x + 2x + 20 = 240

4x+20 = 240

4x = 220

X = 55

Answer:

Slow Train: 55 mph

Fast Train:

65 mph

X is in Guess column

SlowRate

Slow Time

Slow Dist

FastRate

FastTime

FastDist

DistanceEquation

SD=FD

Fill in what you know and find Guess Column

Guess: Slow Time

Slow Time + Fast Time = 7

Ex 3. How far can a man drive out into the country at the average rate of 60 miles per hour and return over the same road at 45 miles per hour if he travels a total of 7 hours.

SR ST SD FR FT FD DistanceEquation

SD=FD

45 5 225 60 2 120 225=12045 4 180 60 3 180 180=18045 3 135 60 4 240 135=24045 X 45x 60 (7-x) 60(7-x) 45x=60(7-x)

Ex 3.

Let x = Slow Time

Let 7-x = Fast Time

45x=60(7-x)

45x=420-60x

105x=420

X=4

Answer the Question!

The man drove out 180 miles

Wrapping It UP- Motion

Distance Formula :

Why is it necessary to draw picture?

Where does the x go in the Guess and Check chart?

D = r x t

To find Distance Equation

In the Guess column