Motion Word Problems Students will solve motion problems by using a Guess & Check Chart and Algebra

D = R x T Review Motion Problems use the formula Distance = Rate x Time If 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 one And 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 problem It 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 = 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 TrainSlow Train 120 miles Total Distance: 360 miles Fast Distance + Slow Distance = 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 Slow Rate Slow Time Slow Dist Fast Rate Fast Time Fast Dist Distance Equation

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? Slow Rate Slow Time Slow Dist Fast Rate Fast Time Fast Dist Distance Equation SD + FD = 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? = 232 No Distance Equation SD + FD = 464 Fast Dist Fast Time Fast Rate Slow Dist Slow Time Slow Rate 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? =464 YES! =580 No = 232 No Distance Equation SD + FD = 464 Fast Dist Fast Time Fast Rate Slow Dist Slow Time Slow Rate

Using Algebra Use Guess & Check Chart to convert to an Algebra problem Place x in Guess Column. Fill in rest of row Define variables Write & Solve Equation Answer 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=464 60xX6056xX =464 YES! Distance Equation SD + FD = 464 Fast Dist Fast Time Fast Rate Slow Dist Slow Time Slow Rate Let x = # of hours56x+60x= x=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. Slow Rate Slow Time Slow Dist Fast Rate SR+10 Fast Time Fast Dist Distance Equation SD+FD= 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 = = = Distance Equation SD+FD=240 Fast Dist Fast Time Fast Rate SR+10 Slow Dist Slow Time Slow Rate KEEP GUESSING Slow Train Rate: 55Fast Train: 65

Ex. 2 2X+2(X+10)=240 2(X+10) 2X+102X2X = Dist Equ SD+FD=240 FDFTFR SR+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

Slow Rate Slow Time Slow Dist Fast Rate Fast Time Fast Dist Distance Equation 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.

SRSTSDFRFTFDDistance Equation SD=FD = = =240 45X45x60(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 D = r x t To find Distance Equation In the Guess column