Applications of Linear Equations?

Question

The question:
A plane takes 4 hours to travel 3280 miles with the wind. The return trip, against the wind, takes 5 hours. Find the spped of the plane in still air and the speed of the wind.


If you have any ideas, please let me know. Thanks so much!!

Answer 1

Use distance = rate * time or d = r * t.
Let p = plane speed, w = wind speed. Then,
3280 = (p + w) * 4
3280 = (p - w) * 5

Multiplying top equation by 5 and bottom equation by 4 gives,
3280 * 5 = 20 * p + 20 * w
3280 * 4 = 20 * p - 20 * w

Adding the two equations together to eliminate w gives,
3280 * 9 = 40 * p, or p = 738.

Substituting this value of p into one of the top two equations and solving then gives w = 82.

Answer 2

Use distance = rate * time or d = r * t.
Let p = plane speed, w = wind speed. Then,
3280 = (p + w) * 4
3280 = (p - w) * 5

Multiplying top equation by 5 and bottom equation by 4 gives,
3280 * 5 = 20 * p + 20 * w
3280 * 4 = 20 * p - 20 * w

Adding the two equations together to eliminate w gives,
3280 * 9 = 40 * p, or p = 738.

Substituting this value of p into one of the top two equations and solving then gives w = 82.

Answer 3

Alright, it is an application of linear equations.

Let the speed of the plane is still air be P miles/hour
Let the speed of wind be W miles/hour.

Then in the first flight with the wind, the effecive speed of the plane will be P+W
ie., P+W = 3280/4 = 820 miles/hour

similary,
while coming bcak the plane is moving against the wind and hence the effective speed of the plane will be P-W
and P-W = 3280/5 = 656 miles/hour

Henxce we have (1) and (2)
P+W = 820-------(1)
P-W = 656--------(2)

Now, (1)-(2) will give,
P-P+W+W = 820-656
=>2W = 164
=> W = 82 miles/hour

Similarly. (1)+(2) will give,
P+P+W-W = 820+656
=> 2P = 1476
=> P = 738 miles/hour



Peace out.

Answer 4

Let us assume that the speeds of air and plane are constant during the entire trip.
Also let v = speed of plane and u = speed of wind.

When the plane is travelling with the wind then the net speed of the plane would be the sum of its original speed plus the speed of the wind.
Hence net speed of plane is (v+u).
Now distance travelled by plane with the wind = (speed of plane * time taken)
Hence 4*(u+v) = 3280
or (u+v)=3280/4 =820 -------------------- (i)

Now speed of plane against wind is (v-u) ( since speed of plane would be the sum of its speed with respect to ground and the negative of the speed of wind as the plane is travelling in a direction opposite to that of the wind).
Hence distance travelled by the plane against wind=5*(v-u)=3280
or (v-u) = 3280/5 = 656 --------------------- (ii)

Adding (i) and (ii) we get,
2*v = 1476
or v = 738.

Also from (ii) u+v = 820
or u= 820 - 738 = 82

Thus speed of plane is 738 miles per hour and that of wind is 82 miles per hour

Answer 5

If X is speed of the plane and Y is speed of the wind, then you can reduce your problem to two simple equations
X + Y = (3280)/4 - speed when you have plane going with the wind, and
X - Y = (3280/5) - speed when you have plane going agains the wind
X+Y = 820
X-Y = 656
solve for X and Y

Answer 6

let x mph and y mph be the respective speeds of the plane and the wind
since distance = vel x time
then,
3280 = (x+y).4
and 3280 = (x-y).5

solving for x and y, we get
x = 738 and y = 82
hence speed of the plane = 738mph
and speed of the wind = 82mph

Answer 7

Let w = speed of the wind
Let p = speed of plane in still air

4*(p+w) = 3280
5*(p-w) = 3280

4p + 4w = 3280
5p - 5w = 3280

20p + 20w = 16400
20p - 20w = 13120

40p = 29520

p = 738

4 * 738 + 4*w = 3280

2952 + 4w = 3280

4w = 328
w = 82

Answer 8

presuming the speed of air and plane are constant:
suppopse p = speed of plane in miles per hour
a = speed of air miles per hour

p+a = 3280/4 = 820-------------------i

p-a = 3280/5 = 656--------------------ii

solving i & ii

p= 738 miles per hour
a= 82 miles per hour