A boat on the potomac river travels the same distnace downstream in 2/3 hour as it does going upstream in 1 hr. If the speed of the curent is 3 mph, find the speed of the boat in the still water.
you need to know a few things: 1) rate * time = distance 2) the speed going with the current is faster than going against the current if the current goes C and the boat goes B, going with the current, the speed is B+C going against the current, the speed is B-C
current is 3 mph, so the speed with the current is B+3 the speed against the current is B-3
downstream in 2/3 hour using rate * time = distance we write (B+3) * 2/3 = D we use the speed going with the current, and time = 2/3 hour
going upstream in 1 hr write (B-3) * 1 = D using the speed going against the current, and time = 1 hour
travels the same distance means we can say (B+3) * 2/3 = (B-3) * 1 can you solve for B ?
No, still have no idea what I'm doing
There are a few parts to this problem. One is finding the equations The second part is doing the algebra.
We can go over solving the equation (without understanding how we got the equation) start with (B+3) * 2/3 = (B-3) * 1 and solve for B. first, notice on the right side you have (B-3)*1 you can simplify that to (B-3) (because multiplying by 1 does not change (B-3) ) you have \[ \frac{2}{3}(B+3) = B-3 \] can you multiply both sides by 3?
Yeah and it will take out the 3 in the 2/3 and then multiply it by -3 on the other side so.. 2b+6=b-9
almost. to multiply by both sides by 3, write 3 times each side, like this \[ 3\cdot \frac{2}{3}(B+3) = 3\cdot (B-3) \] on the left side 3/3 is 1 so that simplifies to \[ 2(B+3) = 3(B-3) \] can you simplify both sides... by distributing the 2 on the left, and 3 on the right side?
2(B+3)=3(B-3) 2B+6=3b-9 then 2b-2b , and -2b from the 3b on the other side to get b 6=b-9 Then +9 and move it over to the 6 to get the b alone 15=B
B stands for the Boat's speed. so the answer to find the speed of the boat in the still water. is 15 miles per hour
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