1 / 10

Section 6.5

Section 6.5. Solving Applications Using Rational Equations. Example 3, page 443. A racer is bicycling 15 km/h faster than a person on a mountain bike. In the time it takes the racer to travel 80 km, the person on the mountain bike has gone 50 km. Find the speed of each bicyclist.

hedva
Download Presentation

Section 6.5

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. Section 6.5 Solving Applications Using Rational Equations

  2. Example 3, page 443 A racer is bicycling 15 km/h faster than a person on a mountain bike. In the time it takes the racer to travel 80 km, the person on the mountain bike has gone 50 km. Find the speed of each bicyclist. Motion (Rate) problems : d = rt d = distance r = rate (distance traveled in some amount of time) t = time

  3. Problem Solving Steps Familiarize. The given facts: Rates: rmtn = x (not specified) rracer = x + 15 km/h Times: tmtn =tracer Distances: dmtn = 50km dracer = 80km

  4. Problem Solving Steps Translate. Know d = rt. Know the t’s are equal. Rewrite d = rt as t = d/r. tmtn = tracer therefore dmtn=dracer rmtn=rracer 50 = 80 x x+15 Carry-Out (solve for x) and Check.

  5. Example 4, page 445 A Hudson River tugboat goes 10 mph in still water. It travels 24 mi upstream and 24 mi back in a total time of 5 hr. What is the speed of the current? Again, Motion problems : d = rt d = distance r = rate (distance traveled in some amount of time) t = time But with water current: The rate upstream = rate of boat – rate of current The rate downstream = rate of boat + rate of current Boat downstream Current Boat upstream

  6. Problem Solving Steps Familiarize: List the given facts: Rate of boat: rb = 10 mi/h Rate of Current: rc = x (unknown) Rate Upstream: ru = rb – rc Rate Downstream: rd = rb + rc Distance Upstream = Distance Downstream: du = dd = 24 mi Time Upstream = tu Time Downstream = td Total Time = 5hr = tu + td Translate: Use t = d/r Total time = 5 = du/ru + dd/rd = 24 + 24 rb – rc rb + rc

  7. Example 1, page 439 Sue can mow a lawn in 4 hr. Lenny can mow the same lawn in 5 hr. How long would it take both of them, working together, to mow the lawn? Problem Involving Work, Similar to Motion (Rate) In motion, rate = distance per time (miles per hour) In work, rate = work per time: r = work/t, or work=rt Sue’s rate is 1 job per 4 hours (a job = mow a lawn) Lenny’s rate is 1 job per 5 hours

  8. Problem Solving Steps Familiarize and Translate: Rates: Sue’s rate: rS = 1/4 lawn per hr = 1 lawn 4 hours Lenny’s rate: rL = 1/5 lawns per hour Total rate (working together): rS + rL Time is same for both: tS = tL = x (unknown) Total work is 1 lawn Work = rt, 1 = (rS + rL)x 1 = (1/4 + 1/5)x

  9. Example 2, page 441 text It takes Ruth 9 hr more than Willie to repaint a car. Working together, they can do the job in 20 hr. How long would it take, each working alone, to repaint a car?

  10. Problem Solving Familiarize and Translate: Willies’s rate is unknown: rW = 1car in x hrs = 1 cars per hr x It takes Ruth 9 hours more than Willie: Ruth’s rate is one car in 9+x hours: rR = 1 cars per hr 9+x Total time working together = 20 hr Total work = 1 job Work = rt 1 = (rW + rR)20 = 1 + 1 20 x 9+x

More Related