The speed of jetstream.

Algebra Level 1

A passenger jet took three hours to fly 1800 miles in the direction of the jetstream. The return trip against the jetstream took four hours. What was the jet's speed in still air?

534 mph 525 mph 544 mph 548 mph

Saurav Sah by Saurav Sah , 20, India

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5 solutions

Let speed of jetstream be xmph and speed of stream be ymph. given (x+y) 3=1800 in direction of stream and (x-y) 4=1800 while returning i.e in opposite direction. After solving we get y=75 and x =450+75=525.

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i just find the ave. speed of the jet in different time gathered. then i add it. 600+450=1050, then i just divided it by two (because of what i have read, "in still air"- so maybe it is just the speed during the flight.) then the result is also 525. am i doing it wrong??

Christian Jude Salcedo - 7 years, 1 month ago

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yeah.. i also got the answer by finding the average of speed that is in jet stream and opposing the jet stream :)

edjoy rodelas - 7 years, 1 month ago

this problem is based on relative speed

Manmeswar Patnaik - 7 years, 1 month ago

525

Suleman Sahil - 7 years, 1 month ago
Muhammad Amir
Apr 28, 2014

1800/3= 600 mph --> With Jet-stream 1800/4= 450 mph --> Against Jet-stream

Speeds= (600+450)/2= 525

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v v = speed of the passenger jet in still air

y y = speed of the jetstream

v + y = 1800 3 = 600 v+y=\dfrac{1800}{3}=600 ( 1 ) \color{#D61F06}(1)

v − y = 1800 4 = 450 v-y=\dfrac{1800}{4}=450 ( 2 ) \color{#D61F06}(2)

Solving the two simultaneous system of equations, we get

v = 525 v=525 and y = 75 y=75

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Rifath Rahman
May 10, 2014

WITH JETSTREAM THE SPEED OF JET IS (1800/3)=600 MPH,AGAINST JETSTREAM THE SPEED IS (1800/4)=450 MPH SO (ACTUAL SPEED OF JET+JETSTREAM=600 MPH)+(ACTUAL SPEED OF JET-JETSTREAM=450 MPH)=2(ACTUAL SPEED OF JET)=1050 MPH OR ACTUAL SPEED OF JET=1050/2=525 MPH

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Gouthamreddy 525 because,here S=D÷T So 1800÷3=600 and again 1800÷4=450.so the avg of these two gives us the result

Goutham Reddy - 7 years ago
Saad Malik
Apr 27, 2014

An easy and a simple way to solve this is to find both of the speeds keeping the distance constant. You'll get 450 and 600. Now take the average of these two and there is your answer. However this solution works if you have all the data given (i.e: distance and time of both journeys)

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we ave the distance , and the distance is 1800miles , and we have the time that the plane took , so we just have to slove that , 1800/(4 24 60*60)

Bader Jegho - 7 years, 1 month ago

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