Question 157616: At 3pm a plane leaves London to fly to Madrid which is 1260 miles away. At 4PM a plane leaves Madrid to fly to London. The second plane flies 40 MPH faster than the first plane. If the planes pass one another at 4:30pm, what is the speed of each plane?
Found 2 solutions by ptaylor, amalm06:
Answer by ptaylor(2198)   (Show Source):
You can put this solution on YOUR website!
Distance(d) equals Rate(r) times Time(t) or d=rt; r=d/t and t=d/r
Let r=speed of the London-to-Madrid plane (1st plane)
Then r+40=speed of the Madrid-to-London plane (2nd plane)
Distance 1st plane travels=r*1+r*0.5=r*1.5
Distance 2nd plane travels=(r+40)*0.5
Now we know that when the above two distances add up to 1260 mi, the two planes will be in the process of passing each other. So, our equation to solve is:
1.5r+0.5(r+40)=1260 get rid of parens
1.5r+0.5r+20=1260 subtract 20 from each side
1.5r+0.5r+20-20=1260-20 collect like terms
2r=1240 divide each side by 2
r=620 mi/hr-------------------------speed of 1st plane
r+40=620+40=660 mi/hr-------------------speed of 2nd plane
CK
in 1.5 hr, 1st plane travels (1.5)(620)=930 mi
in 0.5 hr, 2nd plane travels (0.5)(660)=330 mi
930+330=1260
1260=1260
Hope this helps-----ptaylor
Answer by amalm06(224)  (Show Source):
You can put this solution on YOUR website! Note that the total distance flown by both planes when they meet is 1260 mi. Therefore, one plane flies x mi, while the other plane flies 1260-x mi.
Using d=rt, where d refers to the distance traveled along one dimension, and r refers to the average speed, we have the following system of equations:
x=1.5r
1260-x=0.5(r+40)
Solve for r:
1.5r=1240-0.5r
3r=1240
r=620 mph (Answer), r=660 mph (Answer)
|
|
|
