finding the speed of the current

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bailey07

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Feb 19, 2008
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A motorboat maintained a constant speed of 28 miles per hour relative to water in going 40 miles upstream and then returning. The total time for the trip was 3.5 hours. Find the speed of the current

ok so the equation that i set up is:
since its 40 miles up & 40 returning i split the 3.5 hours in half and got 105 minutes
28-v=105/60 & 28+v=105/60
(28-v)(105)=(28+v)(105)
-105v+2940=2940+105v
-105v=105v
v=-1 ...thats not rightt.. what am i missing?
 
bailey07 said:
A motorboat maintained a constant speed of 28 miles per hour relative to water in going 40 miles upstream and then returning. The total time for the trip was 3.5 hours. Find the speed of the current

ok so the equation that i set up is:
since its 40 miles up & 40 returning i split the 3.5 hours in half and got 105 minutes<<< You can only do that if the speeds are equal . In this case, they are not.
Let the speed of the current be 'v mph'

then

speed upstream = 28 - v

speed downstream = 28 + v

then solving for time -

40/(28+v) + 40/(28-v) = 3.5................now solve for 'v'


28-v=105/60 & 28+v=105/60
(28-v)(105)=(28+v)(105)
-105v+2940=2940+105v
-105v=105v
v=-1 ...thats not rightt.. what am i missing?
Click to expand...
 
bailey07 said:
ok thank you...why do you have to do 40/(28+v)...isnt it times?
Click to expand...

time = distance/speed

time required to go upstream = 40/(28 - v)

time required to go downstream = 40/(28 + v)

total time = (time required to go upstream) + (time required to go downstream)
 
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