Algebra (Period 1)

Motion Word Problems 8-5 & 10-7
rt=d problems
(rate)(time)=distance
Back in Pre-Algebra, these were fairly simple word problems:
1) If you go 60 mph for 3 hours, how far have you gone? (180 miles)
2) You've gone 100 miles in 2 hours. What was your average speed? 
(100/2 = 50 mph)
 
3)You've driven 1000 miles at an average speed of 25 mph. How long did it take you? (1000/25 = 40 hours)
 
Now the problems become more difficult. Usually they involve 2 cars, trains, planes, etc. One car is the "slow" car and the other is the "fast" car.
 
Just like the mixture problems, it helps to make a matrix and also draw a picture to help you understand the words.
 
SLOW CAR/FAST CAR LEAVE FROM SAME PLACE, IN THE SAME DIRECTION, AT DIFFERENT TIMES, WHEN WILL THE 2ND CAR CATCH UP WITH THE FIRST CAR? 
 
2 cars leave 3 hours apart. One travels 72 mph. The other travels 120 mph. The slower car leaves first. When with the faster car catch up with the slower car?
 
Use the set up forms from class... you can find more blank forms online!!

SLOW     72           t              72t  
FAST     120        t - 3        120(t - 3) 
         (fast car left 3 hours later so driving 3 less hours or t - 3)
 
At the point when the fast car catches and overtakes the slow car, what is true of the distances of the 2 cars at that exact moment???
They are equal!
WHAT IS THE EQUATION?
Set the 2 cars' distances equal:
72t = 120(t - 3)
 
72t = 120t - 360
-48t = -360
t = 7.5 hours (slow car's time on the road)
t - 3 = 7.5 - 3 = 4.5 hours (fast car's time on the road)
 
CHECK:
The 2 cars should have traveled the same distance because one car catches up with the other car:
slow: (72)(7.5) = 540
fast: (120)(4.5) = 540
 
 
SLOW CAR/FAST CAR LEAVE FROM SAME PLACE, GOING IN DIFFERENT DIRECTIONS, LEAVING AT THE SAME TIME, WHEN WILL THEY BE A CERTAIN DISTANCE APART?
 
2 cars leave going in different directions. One travels 60 mph. The other travels 50 mph. In how many hours will the cars be 605 miles apart?
 
                      CAR     (RATE)    (TIME)   =   DISTANCE
                     SLOW     50            t                  50t 
                     FAST      60            t                  60t
                      (They travel the same amount of time)
 
What is true of the distances the 2 cars travel?
Together they travel 605 miles because they are 605 miles apart.
WHAT IS THE EQUATION?
Set the 2 cars' distances as a SUM to 605.
50t + 60t = 605
110t = 605
t = 5.5 hours
So in 5 1/2 hours the 2 cars will be 605 miles apart.
 
CHECK:
If you plug in 5.5 hours for each car to find each cars distances, they should add to 605 miles.
slow: (50)(5.5) = 275 miles traveled
fast: (60)(5.5) = 330 miles traveled
275 + 330 = 605 miles
 
SLOW CAR/FAST CAR LEAVE FROM SAME PLACE, GOING IN SAME DIRECTION, LEAVING AT THE SAME TIME, WHEN WILL THEY BE A CERTAIN DISTANCE APART?
 
2 cars leave going in the same direction. One travels 45 mph. The other travels 35 mph. In how many hours will the cars be 15 miles apart?
 
                      CAR     (RATE)    (TIME)   =   DISTANCE
                SLOW     35            t             35t 
                FAST      40            t             40t
                      (They travel the same amount of time)
 
What is true of the distances the 2 cars travel?
They are getting further and further apart as the minutes go by.
The DIFFERENCE of the 2 cars is 15 miles after a certain amount of time.
WHAT IS THE EQUATION?
40t - 35t = 15
5t = 15
t = 3 hours
So in 3 hours the 2 cars will be 15 miles apart.
 
CHECK:
If you plug in 3 hours for each car to find each cars distances, their distances should subtract to 15 miles.
slow: (35)(3) = 105 miles traveled
fast: (40)(3) = 120 miles traveled
120 - 105 = 15 miles
 
SLOW CAR/FAST CAR LEAVE FROM SAME PLACE, GOING IN DIFFERENT OR SAME DIRECTION, IN THE SAME AMOUNT OF TIME, EACH TRAVELS A  DIFFERENT DISTANCE, WHAT IS THEIR SPEED? 
One car travels 20 mph faster than the other car. One car travels 240 miles while the other travels 180 miles. Find their average speeds.
 
                      CAR     (RATE)       (TIME)   =   DISTANCE
               SLOW        r           180/r             180 
               FAST      r + 20    240/(r + 20)        240
This time you have the distance and know that the faster car is 20 mph faster than the slower car. To find the time for each car, use the fact that d/r = t so divided each car's distance by their rates.
 
WHAT IS THE EQUATION?
The times of these 2 cars is equal (left at same time and stopped at same time) so set their times equal:
180/r = 240/(r + 20)
Multiply both sides equally by the LCM:
(r)(r + 20)(180/r) = (r)(r + 20)240/(r + 20)
Cross cancel:
180(r + 20) = 240(r)
180r + 3600 = 240r
3600 = 60r
r = 60 mph (the slower car)
r + 20 = 60 + 20 = 80 mph (the faster car)
 
CHECK:
If you plug in the speeds, you should find that both cars traveled the same amount of time:
slow: (60)t = 180   t = 3 hours
fast: (80)t = 240   t = 3 hours
 
A BOAT/PLANE TRAVELS WITH THE CURRENT ON THE DEPARTING LEG OF THE JOURNEY AND TRAVELS AGAINST THE CURRENT ON THE RETURN LEG. WHAT IS THE SPEED OF THE BOAT/PLANE IN STILL WATER/AIR? 
 
A boat travels with a current of 6 mph for 3 hours and then returns home against the same current. The trip home takes 5 hours. What is the speed of the boat in still water?
 
           BOAT/PLANE        (RATE)       (TIME)   =   DISTANCE
      WITH CURRENT        r + 6           3            3(r + 6)  
      AGAINST CURRENT    r - 6           5            5(r - 6) 
r is the speed in still water and 6 is the speed of the current. The boat with need less time to go the same distance with the current than against it. 
 
WHAT IS THE EQUATION?
The distance to the boat's location and the distance home must be equal.
Set the distances equal:
3(r + 6) = 5(r - 6) 
3r + 18 = 5r - 30
48 = 2r
r = 24 mph (speed in still water)
r + 6 = 24 + 6 = 30 mph (speed with the current)
r - 6 = 24 - 6 = 18 mph (speed against the current)
 
CHECK:
If you plug in the speed with and against the current with the hours traveled, you get the distances to and home. Those distances should be equal:
3(30) = 5(18)
90 = 90