A ship travels downstream for 36 miles and then makes the same return trip upstream. The trip downstream took 3/4 of an hour less then the trip upstream.
If the rate of the current is 4mph
find the rate of the ship in still water and the time it takes to complete each part of the trip.
speed down = s + 4
speed up = s - 4
time down = t
time up = t + .75
(s+4)(t) =36 so t = 36/(s+4)
(s-4)([36/(s+4)]+.75) = 36
36(s-4)/(s+4) +.75(s-4) = 36
36(s-4) +.75 (s^2-16) = 36(s+4)
.75 s^2 -12 -36(8) = 0
(3/4) s^2 = 4*25*3
s^2 = 4*4 *25
s = 4*5
s = 20
*****************************************
check
speed down = 24
time down = 36/24 = 6/4 = t
speed up = 16
time up = 36/16 = 9/4
9/4 - 6/4 = 3/4 sure enough
Thanks Damon, you just beat be to the correction of this earlier post.
https://www.jiskha.com/questions/1753073/A-ship-travels-downstream-for-36-miles-and-then-makes-the-same-return-trip-upstream
Whew :)
To find the rate of the ship in still water and the time it takes to complete each part of the trip, we can use the formula:
Distance = Speed * Time
Let's assume the rate of the ship in still water is S mph. Since the current is flowing downstream, it will help the ship's speed, so the effective speed is (S + 4) mph. Similarly, when the ship is traveling upstream, the current hinders the speed, making the effective speed (S - 4) mph.
Given that the downstream distance is 36 miles, we can set up the equation:
36 = (S + 4) * Time_downstream
Similarly, for the upstream trip, we know that the distance is 36 miles, and the time taken is 3/4 of an hour more than the downstream time. Hence, we have:
36 = (S - 4) * (Time_downstream + 3/4)
Now we can solve these two equations simultaneously to find the values of S and Time_downstream.
Let's start by solving the first equation for Time_downstream:
Time_downstream = 36 / (S + 4)
Substituting this into the second equation, we get:
36 = (S - 4) * (36 / (S + 4) + 3/4)
Simplifying, we get:
36 = (S - 4) * (36 + (3/4)(S + 4))
Now we can solve this equation to find the value of S.
After finding S, we can substitute it back into either of the original equations to find Time_downstream.