3.4 Motion Problems Objective: Solve motion problems by setting up and solving an equations

What is a motion problem? A motion problem is one in which an object is moving at a specific rate for a specific period of time. A car traveling at a constant speed or a person walking at a constant speed are considered motion problems. Motion problems are often solved using the distance formula d = rt In this section, we will solve problems involving 2 rates.

How do we set them up? We let the variable be one of the unknown quantities. We represent the second unknown quantity in terms of the first unknown quantity Example: One train is traveling 20 miles per hour faster than another train. We might let r represent the rate of the slower train and r +20 represent the rate of the faster train.

Set up a table Setting up a table can help us organize the information we have and know what information we will need Item Rate Time Distance Item 1 Distance 1 Item 2 Distance 2

Basic info Distance problems will generally ask you to add two distances, subtract smaller distances from larger ones or set two distances equal to each other The set ups will look like Distance 1 + Distance 2 = total Distance Distance 1 – Distance 2 = difference in distance Distance 2 – Distance 1 = difference in distance

Example 1 Larry, Moe, and Curly go Kayaking. Larry and Moe are in one kayak and Curly is in another. Curly smacks Larry with his paddle because he knows he can go faster and get away. Larry and Moe figure they can catch him so off they go. It turns out Larry and Moe can only go 2 mph and Curly can go 4mph. Curly wants to put at least 5 miles between himself and Larry. How long will it take him?

Set up your table Let t = time when kayaks are 5 miles apart Our set up will be as follows: Distance traveled by Curly – distance traveled by Larry and Moe = 5 miles Item Rate Time Distance Curly 4 mph IDK so put “t” 4t Larry & Moe 2 mph 2t

Solve the problem 4t – 2t = 5 2t = 5 t = 2.5 It will take Curly 2.5 hours to get 5 miles away from Larry (and Moe).

You try it (4 min) Two people start walking at the same time in the same direction. One person walks 4 mph and the other person walks 6mph. In how many hours will they be 0.5 miles apart? Faster person – slower person = 0.5mi. 6t – 4t = 0.5 2t = 0.5 t = 0.25 hour or 15 minutes Person Rate Time Distance Person 1 4 mph T 4t Person 2 6 mph 6t

Example 2 Bugs Bunny and Elmer Fudd have started their own road paving business. Of course Bugs is faster than Fudd, so his crew paves 0.4 mile of road more than Fudd’s per day. Bugs’ crew works at one end of a 20 mile stretch of road, while Fudd’s works towards him on the other. They meet after 10 days. At what rate do the two crews work?

Set Up a table Let r = Fudd’s Crew Let r + 0.4 = Bug’s Crew We know they are going to cover a 20 mile stretch total. The set up is Distance of Fudd’s crew + Distance of Bug’s = 20 miles Crew Rate Time Distance Fudd’s r 10 10r Bug’s 0.4 + r 10(0.4 + r)

Solve the problem 10r + 10(0.4 + r) = 20 10r + 10r + 4 = 20 (distributive property) 20r + 4 = 20 20r = 16 r = 16/20 or 0.8 miles per day (this is Fudd’s crew) 0.8 + 0.4 = 1.2 miles per day (this is Bugg’s)

Word to the wise In example 1, Larry Moe and Curly were all going the same direction-so the problem was a subtraction problem In Example 2 Bugs and Fudd were going in opposite directions, so it was set up as an addition problem.

You try it (4 min) Two people are 6 miles apart jogging toward each other. The faster jogger jogs 0.5 mph faster than the slower jogger. Find the speed each person is jogging if they meet after 0.4 hour. 0.4r + 0.4(r + 0.5) = 6 0.4r + 0.4r + 0.2 = 6 0.8r + 0.2 = 6 0.8r = 5.8 r = 7.25mph (slower person) R = 7.25 + 0.5 = 7.75 mph (faster) Person Rate Time Distance Slower r 0.4 0.4r Faster r + 0.5 0.4(r + 0.5)