On a day with 4 mile an hour winds, Mr. Lee bikes three miles...(description)

Question

On a day with 4 mile an hour winds, Mr. Lee bikes three miles to the store and three miles back to his apartment. This trip takes him an hour. How long would it take Mr. Lee to bike a mile with no wind?

Answer 1

As the wind speed is given, we have to assume that the wind adds to his speed when blowing in the direction of his travel and subtracts from his speed when travelling against the wind.

If the wind is at his back when travelling to the store, the time it takes can be expressed as

3 miles / (X + 4) and when travelling back 3 miles / (X - 4) where X is his speed without any wind.

As the total trip takes one hour, so 3/(X+4) + 3/(X-4) = 1

After simplifying, you get X² -6X -16 = 0.

This leads to solutions of X = 8 and X = -2, Discard X =-2.

So, his speed without the wind is 8 mph. It will take him 1 hour/ 8 miles = 0.125 hours or 7.5 minutes to go one mile.

Answer 2

First find your given information:

D=6 miles (3 miles there + 3 miles back)

T=1 hour

V_wind= 3 mph

We're going to assume that the wind's velocity is always in opposite direction to Mr. Lee so:

V_Final = V_{Mr. Lee} - V_wind

We also know that V_Final=D/T

So D/T = V_{Mr. Lee} - V_wind

Rearrange to get V_{Mr. Lee} = D/T + V_wind

Plug in your given info to get V_{Mr. Lee} = (6 miles)/(1 hours) + 4 mph or 10 mph

T_{no wind}= D/V_{Mr. Lee} or (6 miles/10 mph)

T_{no wind}=0.6 hours

Answer 3

Assume Mr. Lee rides against the wind coming home.

Let x = the time without wind and y = the time against the wind.

Let r = Mr. Lee's rate.

d = rt or t = d/r

x = 3/r the time to the store

y = 3/(r-4) The time back home

x+y = 1

3/r + 3/(r-4) = 1

(3r-12 + 3r)/(r²-4r) = 1

r²-4r = 6r-12

r²-10r + 12 = 0

r = (10±√(100-48))/2 = 8.6, 1.4

But r-4 must be greater than zero, so choose 8.6.

x = 3/8.6 = .349 hours ≈ 21 minutes

Checking:

y = 3/(8.6-4) = 0.652 hours ≈ 39 minutes

21+39 = 60 minutes = 1 hour