Jay and Diane walked to a dock at 3 miles per hour, jumped into a boat, and motored to Dillion at 7 miles per hour. If the total distance was 16 miles and the trip took 3 hours in all, how far did they go by boat?
2007-01-03 15:13:27 · 3 answers · asked by MIke 1 in Science & Mathematics ➔ Mathematics
The main equation for this is d = rt, distance = rate times time. In this case, t = d/r.
Let w be the distance they walked and b the distance they covered in the boat.
The total distance walked and boated is 16:
w + b = 16
Solving for w:
w = 16 - b
Since t = d/r, the distance walked divided by 3 mph plus the distance boated divided by 7mph will be 3 hours, the total time travelled:
w/3 + b/7 = 3
Multiply through by 21 to make this easier to work with:
7w + 3b = 63
Substitute in the value for w:
7(16 - b) + 3b = 63
Solve:
112 -7b + 3b = 63
49 = 4b
b = 12.25
Plug back in the equation for w:
w = 16 - 12.25 = 3.75
So they walked for 3 3/4 miles and boated for 12 1/4 miles.
2007-01-03 17:10:09 · answer #1 · answered by Jim Burnell 6 · 0⤊ 0⤋
Let
w = distance walked
b = distance traveled by boat
Given
16 miles = total distance traveled
3 hours = total time for trip
q = rate walking = 3 miles/hour
r = rate by boat = 7 miles/hour
Find b.
w + b = 16
w = 16 - b
time = distance/rate
3 = w/3 + b/7 = (16 - b)/3 + b/7
63 = 7(16 - b) + 3b
63 = 112 - 7b + 3b
4b = 49
b = 49/4 = 12 ¼ miles
They went 12 ¼ miles by boat.
2007-01-04 01:27:54 · answer #2 · answered by Northstar 7 · 0⤊ 0⤋
Let t1 (hr) be the time of walking.Then the time of motoring is (3-t1) (hr).
Since distance = velocity x time, and the total distance can be separated into two parts: walking distance (3t1) and boating distance 7(3-t1), we can write the following equation based on the total distance:
3t1 + 7(3-t1) = 16 miles
Solve for t1,
t1 = (21-16)/(7-3) = 1.25 hr
The distance they went by boat is
7(3-t1) = 12.25 miles
2007-01-04 01:11:30 · answer #3 · answered by sahsjing 7 · 0⤊ 0⤋