A motorboat takes 5 hours to travel 100mi going upstream. The return trip takes 2 hours going downstream. What is the rate of the boat in still water and what is the rate of the current?
is this algebra ?
i think
Do you know if you are in a boat and the boat just "floats" , it will move downstream at the rate of the current ?
Yes I know how to calculate speed I don't know how to calculate the rate of the current
Do you know that rate * time = distance ?
yes
to do the problem, let x= rate of the boat y = rate of the current when the boat goes downstream, it's total rate is x+y any idea what the rate is going upstream ?
no
if you rowed upstream (instead of in still water), would you go faster or slower compared to still water ?
slower
if the current carried you backward at a rate of y, and you rowed forward at a rate of x how do you show how fast you are going (compared to still water) want to guess ?
the difference of the rates?
yes
how do you show a difference ?
dividing distance by time?
difference is shown by subtracting
i thought you meant something else, sorry
hopefully it make sense: the boat moves upstream at a rate of (x-y) and it goes downstream at a rate of (x+y)
A motorboat takes 5 hours to travel 100 miles going upstream. can you write a "rate * time = distance" equation using a rate (going upstream), an unknown time "t" and a distance 100 miles?
t x 5 = 100 t x 2 = 100
5 hours to travel 100 miles going upstream. time is 5 distance is 100 so it is rate * 5 = 100 but what rate should we use ? remember we are using x= rate of boat and y=rate of the current. how fast does the boat go upstream ?
20mph upstream 50mph downstream
the boat moves upstream at a rate of (x-y)
I forgot what x and y were
distance and time?
x is how fast the boat moves in still water y is how fast the current moves (x-y) is how fast the boat can go upstream
how do you find the current speed
we write two equations 5(x-y)= 100 and 2(x+y)= 100 and solve for x and y
20 and 50
@phi
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