Word problem on finding speed based on time, distance

Click For Summary
The discussion revolves around solving a word problem involving the airspeed of an unladen swallow affected by wind. The problem states that the swallow flies three-quarters of a mile downwind and returns in 4 minutes, with the wind speed given as 6 miles per hour. The user attempts to set up equations for the speeds in both directions but struggles to find the correct airspeed of 24 miles per hour. Participants point out the need for proper bracket usage in the quadratic formula and confirm that the denominator should be 2a, not 4a. The conversation emphasizes the importance of careful calculations and correct formula application in solving such problems.
ducmod
Messages
86
Reaction score
0
Mod note: Moved the text under Relevant equations to the attempt section
1. Homework Statement

Hello!
Can't wrap my head around this easy problem. Would be grateful for help in spotting out my mistakes.
One day, Donnie observes that the wind is blowing at 6 miles per hour. An unladen swallow
nesting near Donnie's house flies three quarters of a mile down the road (in the direction of
the wind), turns around, and returns exactly 4 minutes later. What is the airspeed of the
unladen swallow? (Here, `airspeed' is the speed that the swallow can
y in still air.)

Homework Equations

The Attempt at a Solution


Here is how I approached the solution:
x - airspeed in miles/ hour
v1 - speed of the swallow in the direction of the wind, in miles/hour
v2 - speed on the way back in miles/ hour
v1 = x + 6
v2 = x - 6
time to fly 3/4 of a mile = 3/4 : v1
time to fly back = 3/4 : v2
overall time = 1/15 (4 minutes)

then, overall time: 1/15 = 3/4 : (x+6) + 3/4 : (x-6)

When I try to solve this, I do not get the result any way near the desired one, and I do not get correct roots for x.
 
Last edited by a moderator:
Physics news on Phys.org
ducmod said:
When I try to solve this, I do not get the result any way near the desired one, and I do not get correct roots for x.
Check that calculation once again .
 
Qwertywerty said:
Check that calculation once again .
I did quite a few times :)
the answer should be 24 miles/hour

15*3(x-6) + 15*3(x+6) - 4(x^2 - 36) = 0
-4x^2 + 90x + 144 = 0
doesn't give the 24m/h
 
ducmod said:
I did quite a few times :)
the answer should be 24 miles/hour

15*3(x-6) + 15*3(x+6) - 4(x^2 - 36) = 0
-4x^2 + 90x + 144 = 0
doesn't give the 24m/h

Okay - show your working of the quadratic equation's solution . It's correct till now .
 
Qwertywerty said:
Okay - show your working of the quadratic equation's solution . It's correct till now .
-90 +- √90*90 +4*4*144 / -16
-90 - 102 / - 16 = 12
 
ducmod said:
-90 +- √90*90 +4*4*144 / -16
-90 - 102 / - 16 = 12
Two things - First - You should be using brackets appropriately .
Second - What's the formula used to solve a quadratic ? Does the denominator use 4*a or 2*a ?
 
Qwertywerty said:
Two things - First - You should be using brackets appropriately .
Second - What's the formula used to solve a quadratic ? Does the denominator use 4*a or 2*a ?
 
exactly! thank you, and sorry for not posting correct brackets. of course, it is 2a in the denominator :) Thank you!
 

Similar threads

Replies
5
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 3 ·
Replies
3
Views
4K
Replies
5
Views
3K
Replies
1
Views
4K
  • · Replies 22 ·
Replies
22
Views
4K
Replies
38
Views
5K
Replies
2
Views
2K
  • · Replies 26 ·
Replies
26
Views
4K