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Two navy planes leave an aircraft carrier at the same time and fly in opposite directions. After flying for 1 1/2 hour, the planes are 1,140 miles apart. If one plane flies 10 miles per hour faster than the other, what is the rate of the faster plane? Solution plz

Sagot :

Shinalcantaraidn
let 'x' be the velocity of one plane
    'y' be the velocity of the other
knowing that the formula for velocity is defined by the equation :
[tex]v = \frac{d}{t} [/tex]
where v is the velocity
         d is the distance traveled
         t is the time
modifying the equation you'll have:
[tex]d = vt[/tex]
since you are given the distance as 1,140mi and the time traveled as 1.5hours
and that the other plane is traveling 10mph faster than the other then you'll have:
xt + yt = 1,140mi   -----equation 1
x = y + 10mph  -----equation 2
substituting equation 2 to 1 and having the time as 1.5hrs,
xt +yt = 1,140
(y+10)(1.5) + y(1.5) = 1,140
1.5y +15 + 1.5y = 1,140
3y = 1,140 - 15
3y = 1125
y = 375mph
substitute to equation 2
x = y + 10
x = 375 + 10
x = 385mph
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