An ultralight plane had been flying for 40 min when a change of wind direction doubled its ground speed. The entire trip of 160 mi took 2 h. How far did the plane travel during the first 40 min?
.
28 mi
B.
32 mi
C.
36 mi
D.
40 mi
2006-11-14 02:49:43 · 7 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics
The answer is B. To find it, all we have to do is use the equation for velocity: V = d/t and do some fancy footwork relating the two velocities and times traveled together.
Let v be the velocity during the first leg of the journey.
Let V = 2v be the velocity during the last leg.
D is the total distance traveled, and is equal to 160 miles.
t = 40 minutes for first leg
T= (120 - t) minutes for last leg of journey
Then we can make this assertion:
D = d(1) + d(2) = vt + VT.
Substituting in the assigned values, we get:
160 = v*40 + 2v*(120-40) = v*40 + v*[2*(80)] = v*200.
Dividing through by 200 gets us: 4/5 = v. Then V = 2*v = 8/5.
Since we are dealing with miles and minutes here, we interpret these velocities to be (4/5) mi/min and (8/5) mi/min respectively.
Since we know the velocity and travel time of the first leg, we simply multiply to find the distance traveled there:
d(1) = (4/5) mi/min*40 min = 32 mi.
Similarly, we know the velocity of the second leg and that its travel time was 80 minutes. Multiplying these, we get:
d(2) = (8/5) mi/min * 80 min = 128 mi.
We check by adding d(1) and d(2).
D = d(1) + d(2) = 32 mi. + 128 mi. = 160 mi.
We are done.
2006-11-14 04:14:04 · answer #1 · answered by MathBioMajor 7 · 0⤊ 0⤋
The plane was traveling twice as fast for 80 minutes out of 120. Two times as fast for two times the amount of time will make those 80 minutes account for four times as much distance as the first 40 minutes, or 80%. Knowing this, the first 40 minutes must account for 20% of the 160 miles, or 32 miles. The correct answer is B.
2006-11-14 11:00:39 · answer #2 · answered by Forward Kindness 3 · 0⤊ 0⤋
40 min = 2/3 hour
plane traveled at x mph for 2/3 hour
and at 2x for 1 1/3 hours
make formula based on D = r t
y x 2/3 + 2y x 1 1/3 = 160
2/3 y + 2 2/3 y = 160
3 1/3 y = 160
y = 48
so for 40 minutes plane flew at 48 mph
so d=rt = 2/3 x 48 = 32 miles
answer b.
2006-11-14 11:03:55 · answer #3 · answered by ignoramus 7 · 0⤊ 0⤋
Let x= rate for first 40 minutes, 2x= rate for remaining 80 minutes
rt=d
x(40/60) + 2x(80/60)=160
x(2/3)+2x(4/3)=160
multiply by 3
2x+8x=480
10x=48
x=48
this is the rate for the first 40 minutes so multiply it by 2/3 and get the distance
32miles
2006-11-14 10:58:43 · answer #4 · answered by mom 7 · 0⤊ 0⤋
B
2006-11-14 10:59:15 · answer #5 · answered by Halo 5 · 0⤊ 0⤋
B.
2006-11-14 11:00:11 · answer #6 · answered by openpsychy 6 · 0⤊ 0⤋
C. ?
2006-11-14 10:53:33 · answer #7 · answered by HEATHER V 1 · 0⤊ 0⤋