Answer:
Speed of the plane = 275 mph
Speed of wind = 25 mph
Step-by-step explanation:
When the plane travels against the wind, it moves slower due to the wind reducing its effective speed. If we identify the velocity of the plane with [tex]v_P[/tex] and the velocity of the wind with: [tex]v_W[/tex], we have that the actual velocity at which the plane travels is : [tex]v_P- v_W[/tex] .
Now we need to recall the relationship between speed (or velocity) , time (t), and distance traveled (d): [tex]v=\frac{d}{t} \\v*t=d[/tex]
We write then a first equation representing the travel of the plane against the wind, using 125 miles for the distance covered, and 30 minutes for the time employed flying with velocity [tex]v_P- v_W[/tex]:
[tex]d=v*t\\125=(v_P- v_W)*30\\125=30v_P-30v_W[/tex]
Now for the plane going with the wind in its second trip. Notice that in this case, the speed (or velocity) of the plane adds to the velocity of the wind, making it go at a faster resultant speed: [tex]v_P+ v_W[/tex]
Using now the info for the second trip: 125 miles taking it 25 minutes, and with a net speed of [tex]v_P+ v_W[/tex], we wirte the second equation:
[tex]d=v*t\\125=(v_P-+v_W)*25\\125=25v_P+25v_W[/tex]
since both equations we created equal 125 miles, we can say that the right hand side in both must equal each other:
[tex]125=30v_P-30v_W\\125=25v_P+25v_W\\30v_P-30v_W=125=25v_P+25v_W\\30v_P-30v_W=25v_P+25v_W\\30v_P-25v_P=25v_W+30v_W\\5v_P=55v_W\\v_P=11v_W[/tex]
Which tells us that the speed (or velocity) of the plane is 11 times that of the wind. We can use this info in the first equation we wrote, to find the speed of the wind, by replacing the speed of the plane with: "11 time the speed of the wind" ([tex]v_P=11v_W[/tex]):
[tex]125=(11v_W-v_W)*30=10v_W*30=300*v_W\\v_W=\frac{125}{300} \frac{miles}{minute}[/tex]
It is better to express the velocity in miles per hour (instead of miles per minute) so we use the fact that one hour equal 60 minutes to reduce our answer for the speed of the wind:
[tex]v_W=\frac{125}{300} \frac{miles}{minute}=\frac{125}{5} \frac{miles}{60minutes} =\frac{125}{5} \frac{miles}{hour} = 25 \frac{miles}{hour}[/tex]
Since we know that the speed of the plane is 11 times that of the wind, we find the speed of the plane by multiplying this result times 11:
[tex]v_P=11*v_W=11*25 \frac{miles}{hour}=275\frac{miles}{hour}[/tex]
we can therefore say that the speed of the wind is: 25 mph (abbreviation for [tex]\frac{miles}{hour}[/tex], and that of the plane is 275 mph.
