Question

Two planes left simultaneously from the same airport and headed in the same direction towards another airport 3600 km away. The speed of one of the planes was 200 km/hour slower than the speed of the other plane, and so it arrived at its destination 1.5 hours after the faster plane. Find the speeds of both planes.

Answer 1

Answer: The speed of the faster lane is 800 km/hour and the slower plane is 600 km/hour

Step-by-step explanation:

Answer 2

The speed of the faster plane is 800 km/h and the speed of the slower plane is 600 km/h.

The given parameters:

Distance traveled by both planes, d = 3600 km

• Let the speed of the faster plane = v
• The speed of the slower plane, u = v - 200 km/h
• Let the time taken for the faster plane = t hours
• Time taken for the slower plane to arrive = t + 1.5 hours

The distance traveled by both planes is the same;

d = u(t + 1.5) ---(1)

d = (v - 200) (t + 1.5)

3600 = vt + 1.5v - 200t  - 300  

d = vt  ----(2)

3600  =  vt

$v = \frac{3600}{t}$

$3600 = \frac{3600}{t} t \ +\ 1.5 (\frac{3600}{t} ) \ - \ 200 t - 300\\\\3600 = 3600 + \frac{5400}{t} - 200t - 300\\\\300 = \frac{5400}{t} - 200t\\\\300t = 5400 - 200t^2\\\\200t^2 +300t - 5400 = 0\\\\2t^2 + 3t- 54= 0\\\\(2t - 9)(t + 6)= 0\\\\t = 4.5 \ \ or \ \ -6 \ \\\\t = 4.5 \ hours$

The speed of the faster plane is calculated as;

$v = \frac{3600}{t} \\\\v = \frac{3600}{4.5} \\\\v = 800 \ km/h$

The speed of the slower plane is calculated as follows;

$u = v- 200\\\\u = 800 - 200\\\\u = 600 \ km/h$

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