PurplePlug123
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I've been stuck on this for so long and I have an exam soon, anybody who can help me :'( ?

A light aircraft flew from Maseru to Nata and returned to Maseru.
The distance from Maseru to Nata is 1080km.
(i) On the outward flight, the average speed of the aircraft was x km/h. Write down an expression in terms of x, for the time taken in hours.
(ii) On the return flight the average speed was 30km/h greater than the average speed of the outward flight. Write down an expression in terms of x, for the time taken in hours for the return flight.

(a) The time taken on the return flight was half an hour less than the time on the outward flight. Form an equation in x and show that it reduces to x^2+30x-64800=0
(b) solution to the given quadratic equation ^^
(c) The time taken in hours on the outward flight.
(d) The average speed for the whole flight from Maseru to Nata and back to Maseru.

Respuesta :

(i)  speed = distance / time
so time =  distance / speed
here we have

time t = 1080/x  hours

(ii) return flight  time  = 1080 / (x + 30)  hours

(a)  1080/x - 1080/(x + 30) = 1/2

Multiplying  through by the LCD 2x(x + 30) we get:-

1080*2(x + 30) - 2x*1080 = x(x+30)
2160x + 64800 - 2160x = x^2 + 30x
x^2 + 30x - 64800  = 0

(b)  factoring;  -64800 = 270 * -240  ans 270-240 = 30 so we have

(x + 270)(x - 240) = 0   so x = 240  ( we ignore the negative -270)

So the speed for outward journey is 240 km/hr

(c) time ffor outward flight = 1080 / 240 =  4 1/2  hours

(d) average speed for whole flight = distance / time
   Time for outward journey = 4.5 hours and time for  return journey = d / v
= 1080 / (240+30) =  4 hours
 Therefore the average speed for whole journey =  2160 / 8.5 = 254.1 km/hr
aksnkj

The speed of outward flight and return flight is 240 km/hr and 270 km/hr, respectively. The average speed of the flight in the whole tour is 254.118 km/hr. The time of the outward flight and the return flight is 4.5 hours and 4 hours, respectively.

Given information:

The distance from Maseru to Nata is 1080 km.

Speed, s, is defined as,

[tex]s=\dfrac{d}{t}[/tex]

where d is the distance and t is the time.

(i).

On the outward flight, the average speed of the aircraft was x km/h.

So, the time t for the outward flight can be written as,

[tex]t=\dfrac{d}{s}\\t_1=\dfrac{1080}{x}[/tex]

(ii).

On the return flight, the average speed was 30km/h greater than the average speed of the outward flight.

So, the time t for the return flight can be written as,

[tex]t=\dfrac{d}{s}\\t_2=\dfrac{1080}{x+30}[/tex]

(a).

The time taken on the return flight was half an hour less than the time on the outward flight.

In the form of an equation, the above information can be written as,

[tex]t_1-t_2=\dfrac{1}{2}\\\dfrac{1080}{x}-\dfrac{1080}{x+30}=\dfrac{1}{2}\\1080(x+30-x)=x(x+30)\dfrac{1}{2}\\x^2+30x=2160\times30\\x^2+30x-64800=0[/tex]

(b).

Solve the above quadratic equation for x using middle term split as,

[tex]x^2+30x-64800=0\\x^2+270x-240x-64800=0\\x(x+270)-240(x-270)=0\\(x+270)(x-240)=0\\x=240,-270[/tex]

Speed can't be negative. So, the speed of the outward flight is 240 km/hr.

(c).

The time taken on the outward flight will be,

[tex]t_1=\dfrac{1080}{x}\\t_1=\dfrac{1080}{240}\\t_1=4.5\rm\;hr[/tex]

(d).

The time taken by the return will be 4.5-0.5=4 hours.

So, the total time of the tour from Maseru to Nata and back to Maseru will be 4.5+4=8.5 hours. And the total distance of the tour will be [tex]1080\times2=2160[/tex] km.

So, the average speed for the whole tour from Maseru to Nata and back to Maseru will be,

[tex]s=\dfrac{2160}{8.5}\\s=254.118\rm\;km/hr[/tex]

Therefore, the speed of outward flight and return flight is 240 km/hr and 270 km/hr, respectively. The average speed of the flight in the whole tour is 254.118 km/hr. The time of the outward flight and the return flight is 4.5 hours and 4 hours, respectively.

For more details, refer to the link:

https://brainly.com/question/9834403

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