Motion, Money and Mixture Problems § 3.4 Motion, Money and Mixture Problems

Motion Problems A motion problem is one in which an object is moving at a specified rate for a specified period of time. amount = rate · time Example: A typical shower uses 30 gallons of water and lasts for 6 minutes. How much water is typically used per minute? Continued.

Motion Problems amount = rate · time Example continued: amount = rate · time Let r = the rate of water used per minute 30 = r (6 min) 30 = 6r 5 = r Water is used at the rate of 5 gallons per minute. Check: rate · time = 5 · 6 = 30 

distance = rate · time or d = r · t Distance Problems When the amount in the formula is distance, we refer to the formula as the distance formula. distance = rate · time or d = r · t Example: While swimming in the ocean, Elyse’s glasses fell off her head. If the glasses fall at a rate of 4 feet per second, how long will it take for them to fall 70 feet to the sand at the bottom? Continued.

distance = rate · time or d = r · t Distance Problems Example continued: distance = rate · time or d = r · t Let t = the time it takes the glasses to fall 70 = 4t 17.5 = t It will take Elyse’s glasses 17.5 seconds to fall to the bottom Check: rate · time = 4 · 17.5 = 70 

Problems with Two Rates Example: Two crews are laying blacktop on a road. They start at the same time at opposite ends of a 12-mile road and work toward one another. One crew lays blacktop at an average rate of 0.75 mile a day faster than the other crew. If the two crews meet after 3.2 days, find the rate of each crew. 12 miles covered in 3.2 days Continued.

Problems with Two Rates Example continued: Let r = the rate of the slower crew Rate x Time = Distance Unit Rate Time Distance Slower Crew r 3.2 3.2r Faster r + 0.75 3.2(r + 0.75) distance covered by slower crew + distance covered by slower crew = 12 3.2r + 3.2(r + 0.75) = 12 Solve the equation. Continued.

Problems with Two Rates Example continued: 3.2r + 3.2(r + 0.75) = 12 3.2r + 3.2r + 2.4 = 12 3.2r + 3.2r = 9.6 6.4r = 9.6 r = 1.5 The rate of the slower crew is 1.5 miles paved/day; the rate of the faster crew is r + 0.75 = 1.5 + 0.75 = 2.25 miles/day Check: 3.2(1.5) + 3.2(1.5 + 0.75) = 4.8 + 3.2(2.25) = 4.8 + 7.2 = 12 

Problems with Two Rates Example: Kelsey started driving to the mall at an average speed of 30 mile per hour. A short while later, she realized she forgot her wallet. She turned around and headed back to the house, driving 20 miles per hour. If it took her a total of 0.6 hours for the round trip, how far had she driven before she turned around? 30 mph 20 mph Home Mall Continued.

Problems with Two Rates Example continued: Let t = time it takes to drive one way Rate x Time = Distance Trip Rate Time Distance To 30 t 30t From 20 0.6 – t 20(0.6 – t) distance to the turn around point = distance from the turn around point 30t = 20(0.6 – t) Solve the equation. Continued.

Problems with Two Rates Example continued: 30t = 20(0.6 – t) 30t = 12 – 20t 50t = 12 t = 0.24 Distance = 30t = 30(0.24) = 7.2 Kelsey had driven 7.2 miles before turning around. Check: 30t = 20(0.6 – t) 30(.24) = 20(0.6 – .24) 7.2 = 20(0.36) 7.2 = 7.2 

interest = principal · rate · time Money Problems interest = principal · rate · time Example: Jordan invested $12,500, part at 7% simple interest and part at 6% simple interest for 1 year. How much was invested at each rate if each account earned the same interest? Continued.

principal · rate · time = interest Money Problems Example continued: Let x = the amount of money invested at 7% principal · rate · time = interest Account Principal Rate Time Interest 7% Account x 0.07 1 0.07x 6% Account 12500 – x 0.06 0.06(12500 – x) Interest from each account is the same: I.07 = I.06 0.07x = 0.06(12500 – x) Solve the equation. Continued.

Money Problems Example continued: 0.07x = 0.06(12500 – x) Jordan deposited $5769.23 at 7% and 12500 – 2769.23 = $6730.77 at 6%. Check: Interest 7%: 5769.23(0.07) = 403.85 6%: 6730.77(0.06) = 403.85 

strength · quantity = amount Mixture Problems Any problem in which two or more quantities are combined to produce a different quantity may be considered a mixture problem. strength · quantity = amount

Mixture Problems Example: At a local pet store, bird food is sold in bulk. In one barrel are sunflower seeds that sell for $1.80 per pound. In another barrel is cracked corn that sells for $1.40 per pound. If the store makes bags of a mixture of the two by mixing 2.5 pounds of sunflower seeds with 1 pound of the cracked corn, what should be the cost per pound? Continued.

strength · quantity = amount Mixture Problems Example continued: Let p = the price of the new mixture strength · quantity = amount Item Price # of pounds Value of Item Sunflower Seeds 1.80 2.5 1.8(2.5) Cracked Corn 1.60 1 1.6(1) New Mixture p 3.5 p(3.5) 1.8(2.5) + 1.6(1) = p(3.5) Solve the equation. Continued.

Mixture Problems Example continued: Check:  The new mixture should be sold for $1.69 per pound. Check: 1.8(2.5) + 1.6(1) = 1.69(3.5) 4.5 + 1.4 = 5.9 (approx) 