2. Flying to Kampala with a tailwind a plan averaged 158km/h. On the return trip, the plane only averaged 112 km/h while flying back into the same wind. Find the speed of the wind and the speed of the plane in still air.
Reiny
answered
speed of boat in still water --- x mph
speed of current ---- y mph
10(x+y) = 210 ----> x+y = 21
70(x-y) = 210 ----> x-y = 3
add them
2x = 24
x=12
then y = 9
speed of boat = 12mph , speed of current = 9 mph
#2
even easier ....
define x and y as above
x+y = 158
x-y = 112
etc
Kian
answered
x+y=158
x-y=112
and then i subtract them:
2y=46 -> y=23
why do I get the speed of plane as 23?
Reiny
answered
in x-y = 112
x must be the plane's speed and y the wind's speed
so sub y = 23 into
x+y = 158
x+23=158
x = 155
Kian
answered
thank you so much! i also have another question:
does it matter whether we add or subtract simultaneous equations? do we have to either add or subtract or are both fine?
Reiny
answered
Remember that if we add two opposites we get zero
and when we subtract two "likes" we also get zero
so for your two equations
x+y=158
x-y=112
if I add them ---> 2x = 270 , x = 135
if I subtract ----> 2y = 46 , y = 23
in general, if the signs are opposite, we add
if the signs are the same, we subtract
btw, how did you get 155 ????
Kian
answered
I got 135 not 155...
but does that apply everytime or is that just a shortcut type of thing? it's not wrong if I subtract x+y=158 and x-y=112 right?
Reiny
answeredBtw, look up there , you have x = 155
Kian
answered
in x-y = 112
x must be the plane's speed and y the wind's speed
so sub y = 23 into
x+y = 158
x+23=158
x = 155
^that's what you said.. you wrote 155 not me..?
Anonymous
answered
Explain Bot
answered
1. Let's assume the speed of the boat in still water is 'B' and the speed of the current is 'C'. When the boat is traveling downstream, it moves with the current, so its effective speed is (B + C). Similarly, when the boat is traveling upstream, it moves against the current, so its effective speed is (B - C).
To find the speed of the boat in still water (B), we can use the formula:
Distance = Speed * Time.
For the downstream trip, we have (B + C) * 10 = 210 miles.
For the upstream trip, we have (B - C) * 70 = 210 miles.
Simplifying these equations, we get:
B + C = 21 (Equation 1)
B - C = 3 (Equation 2)
Adding Equation 1 and Equation 2, we get:
2B = 24
Dividing both sides by 2, we find:
B = 12 mph.
Now, substituting the value of B in Equation 1, we can find the speed of the current (C):
12 + C = 21
C = 21 - 12
C = 9 mph.
Therefore, the speed of the boat in still water is 12 mph, and the speed of the current is 9 mph.
2. Let's assume the speed of the plane in still air is 'P' and the speed of the wind is 'W'. When the plane is flying with the tailwind, its effective speed is (P + W). Similarly, when it is flying against the headwind, its effective speed is (P - W).
To find the speed of the wind (W), we can use the same formula: Distance = Speed * Time.
For the journey with the tailwind, we have (P + W) * T = 158 km, where T is the time taken.
For the journey against the headwind, we have (P - W) * T = 112 km.
Since the distance traveled is the same in both cases, we can set these two equations equal to each other:
(P + W) * T = (P - W) * T
P + W = P - W
Simplifying, we find:
2W = 2P - 158 - 112
2W = 2P - 270
Dividing both sides by 2, we get:
W = P - 135
Since we are left with two unknowns, we need one more piece of information to solve for P and W.
