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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
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