WORK Problems 1 An old conveyor belt takes

WORK Problems 1) An old conveyor belt takes 21 hours to move one day’s coal output from the rail line. A new belt can do it in 15 hours. How long does it take when both are used at the same time?

WORK Problems Adding hours doesn’t make sense because it shouldn’t take more than 15 hours with the new belt AND the old belt. Averaging doesn’t make sense for the SAME reason.

WORK Problems Think of this in terms of speed. Find the rate of the job per time. In this case, how much of the job per hour the machines can work. Add the speeds to get the total speed for one job and convert to time.

WORK Problems X = time for first object Y= time for second object T = total time for both objects

WORK Problems 1) x = “old” belt hours, y = “new” belt hours x = 21 hours, y = 15 hours So… Therefore,

WORK Problems 2) One pipe can fill a pool 1. 25 times faster than a second pipe. When both pipes are opened, they fill the pool in five hours. How long would it take to fill the pool if only the slower pipe is used?

WORK Problems What does “ 1. 25 times faster” mean? Let x = first, faster pipe and y = second, slower pipe Then 1. 25 x = y since x is the smaller valued number in terms of time. (e. g. A is twice as fast as B makes sense as 2 A = B because if B was 6 hours then A is 3 hours, twice as fast)

WORK Problems Thus, Multiply all fractions by 5 x to get…

WORK Problems You could also solve it like this… thus…

MOTION Problems Distance = rate x time d = rt 3) Pam jogged up a hill at 6 km/hr and then jogged back down the hill at 10 km/hr. How many kilometers did she travel in all if her total jogging time was 1 hour and 20 minutes?

MOTION Problems x = time up the hill (hours) y = time down the hill (hours) total time: x + y = 1 hr, 20 min = 4/3 hr The distance up = (6 km/hr)(x hr) = 6 x The distance down = (10 km/hr)(y hr) = 10 y

MOTION Problems Distance up = Distance down so… Therefore, add the distances.

MOTION Problems Current/Wind Problems: upstream is against the flow (slows down) downstream is with the flow (speeds up) rate up/against: x – c rate down/with: x + c

MOTION Problems Manipulate d = rt to become t = d/r x is the speed with no current or no wind.