Math help please!!!!?

traveling down stream a boat can go 12 miles in 2 hours. Going up stream it makes only 2/3 this distance in twice the time. What is the rate of the boat in stil water, and what is the rate of the current?

Can you show me how to do this not just the answer?

3 Answers

  • 1 decade ago
    Favorite Answer

    Let the speed of the boat in still water be v, and let the speed of the current be c. Then the speed of the boat while traveling downstream is r = v + c, because the current pushes the boat along, which adds to its speed. The boat's speed traveling upstream is r' = v - c. The current subtracts from the boat's speed because it pushes against the direction of the boat's travel. Let t be the time it takes for the boat to travel downstream, and t' = 2t be the time it takes for the boat to travel upstream.

    Using the distance formula for the distance, d, traveled downstream, we get this:

    d = rt

    d = (v + c) t

    For the distance, d', traveled upstream, we get this:

    d' = r't'

    d' = (v - c) 2t.

    We would like to be able to equate the two distances so we can solve for one of the variables, but we can't just yet, because d' = (2/3)d, not d. But if we multiply d' by (3/2), then (3/2)d' = d. So let's do that, and see what we get:

    (3/2)d' = (3/2)(v - c) 2t

    (3/2)d' = (v - c)(3/2)(2t)

    (3/2)d' = (v - c) 3t.

    Now we can equate the two distances:

    d = (3/2)d'

    (v + c) t = (v - c) 3t

    vt + ct = 3vt - 3ct

    vt - 3vt = -3ct - ct

    -2vt = -4ct

    -2vt/-2t = -4ct/-2t

    v = 2c.

    What the result above tells us is that the speed of the boat in still water is equal to twice the speed of the current. Now we are getting somewhere, because we can substitute one or the other variable into one of our original equations to determine everything. Let's plug v = 2c into our first equation:

    d = (v + c) t

    12 = (2c + c) 2

    12/2 = [(2c + c) 2] / 2

    6 = (2c + c)

    6 = 3c

    6/3 = 3c/3

    2 = c

    The speed, or rate, of the current is 2 mph (miles per hour). So the rate of the boat in still water is v = 2c = 2 (2 mph) = 4 mph.

    Now we just have to confirm our answers:

    d = (v + c) t

    12 = (4 + 2) 2

    12 = (6) 2

    12 = 12 This checks.

    d' = (2/3)d = (v - c) t'

    d' = (2/3)(12) = (v - c) 2t

    8 = (4 - 2)(2)(2)

    8 = (2)(2)(2)

    8 = 8 This also checks.

    Our answers have produced correct results for each distance equation, so they are correct.

    The rate of the boat in still water is 4 mph.

    The rate of the current is 2 mph.

  • Anonymous
    1 decade ago

    These questions are always solved by the old standby equation, d=r*t, distance equals rate times time.

    But in this case, there's two different ''r's'' r, I'll call it rb, for the rate of the boat...another r, I'll call it rr, for the rate of the river...if I were writing this out, these would be r(subscript b) and r(subscript r)...

    So, d=(rr+rb) * t

    This equation is used for both the trip upstream, and the trip downstream...with one little detail. When we calculate upstream speed, the rr (rate of the river) is negative.

  • 1 decade ago


    4y_4x=8 x :still rate

    y:current rate

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