Transcript 8_5 _day 2_ current_ headwind TROUT 09

Warm up # 28
1. How many more days until winter break?
1.)
30t
46t
30t  72  46t t  4.5 hours
Page 383, 1 - 6
d  72 t +3 
d  120  t 
Let t = hours after fast train leaves
72 t  3  120t
72t  216  120t
216  48t
t  4.5 hours
Page 383, 1 - 6
Correct Homework
• Page 384 (4,5,6,9,10,13,14)
44t
55t
44t  55t  297
3 hours
Page 383, #4
32t
69 miles
47t
32t  69  47t
69  15t
69
t
 4.6 hours
15
Page 383, #5
192 t + 2
960 t


Let t = hours after the JET leaves
192 t  2  960t
192t  384  960t
384  768t
1
t  hour
2
Page 383, #6
d  96 t4
  384km
d  64 t +2 
96t  64 t  2
96t  64t  128
32t  128
t4
P.383, #9
Page 385 #10
It takes a small jet plane 4 hours less time
than it takes a propeller-driven plane to
travel from Glen Rock to Oakville. The jet
plane averages 637 km/h while the
propeller plane averages 273 km/h. How
far is it from Glen Rock to Oakville?
Let t = time it takes the prop plane to make the trip
d   r637t - 4 
d   r273t
637t  2548  273t
364t  2548
t=7

d = 1911 km
A motorcycle breaks down and the rider
has to walk the rest of the way to work.
Motorcycle was traveling at 45 mi/h, and he walks at 6 mi/h.
The distance from home to work is 25 miles, and the
total trip took 2 hours.
d
 45
r
t 
25-d
d  6r
t2-t 
25 miles
Let d = distance on motorcycle
Let t = hours on motorcycle
25  d  12  6t
1
d  45   15 miles
3
1 25  45t  12  6t
t
13

39t
3
Page 385 #13
Page 385 #14
A student walks and jogs to college each day.
The student averages 5 km/h walking and 9
km/h jogging. The distance from home to college
is 8 km and the trip takes one hour. How far
does he jog? Let x = distance jog
8-x
d
 r5
Let y = hours jog
t1-y 
walk
8  x  5 1  y 
8  9y  5  5y
3  4 y 0.75  y
xd
 r 9
jog
t y 
x  9y
x  9 .75
x = 6.75 km
Page 385 #14
A student walks and jogs to college each day.
The student averages 5 km/h walking and 9
km/h jogging. The distance from home to college
is 8 km and the trip takes one hour. How far
does he jog?
8 – x = 5(1 – y)
x = 9y
Let x = distance jog
x = 6.75 km
Let y = hours walk
Distance =
(Rate)
(Time)
Walking
8 - x
5
1 - y
Jogging
x
9
y
total
8
1
Tailwind & Up Stream Problems in
8.5
The tailwind and current problems affect the
rate.
How fast is the
canoe going?
Distance to tree = 12 Km
3 hours
d  r t  12  r 3
Pattern
• Tailwind and current problems talk about
going to a location and coming home. The
distance will be the same and we will set the
distances equal to each other
A motorboat took 3 hours to make a downstream trip
with a current of 6km/h.
The return trip against the same current took 5 hours.
r
=
Find the speed of the boat in still water.
d   r +6
t3 
d  r- 6
t5 
3  r  6  5  r  6
3r  18  5r  30
48  2r
r  24 km/h
An airplane flew for 5 hours with a tailwind of 25
km/h.
The return trip against the same wind took 6 hours.
r
=
Find the speed of the airplane in still air.
d   r +25 t5 
d   r -25 t6

5  r  25  6  r  25
5r  125  6r  150
275  r
r  275 km/h
3 r - 6
2  r + 6
3  r  6  2  r  6
3r  18  2r  12
r  30
=r
=r
3 r  40
4  r  40
3r  120  4r  160
280  r
r  280 km/hr
6  r  60  8  r  60
6r  360  8r  480
840  2r
r  420 km/hr
=r
6 km
4
4   r  3 t
t
distance   ratetime
r 3
distance   ratetime
10   r  3 t
10
t
 r  3
10
4

 r  3  r  3
10r  30  4r  12
6r  42 7 km / h
Assignment
Page 383
7, 8, 11, 12,15,16