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Homework Help: Calculating Distance and Time Using Constant Velocity

  1. Aug 18, 2014 #1
    1. The problem statement, all variables and given/known data

    Another canoe question!!

    A canoe is headed directly across a river that is 200 m wide. Instead of moving with a constant velocity, the canoe moves with a constant acceleration of 4.0 x 10-2 m/s2. If the river is flowing with a constant velocity of 2.0 m/s, how long will it take for the canoe to reach the other side? How far down the river will it land? Sketch the shape of the path the canoe will follow.

    2. Relevant equations

    vf2 = vi2 + 2a X d

    sin a/a = sin b/b

    cos θ = Adjacent / Hypotenuse

    3. The attempt at a solution

    To calculate the velocity:

    Vf2 = 0 + 2(0.04) X 200
    Vf2 = 16m/s

    and determine the angle:

    sin Θ = (sin 90/16.12) x 2
    Θ = 7.12°


    I'm pretty sure I've messed something up at this point. My next step would be to try to figure out the distance of the hypotenuse in meters based on the angle that I calculated using CAH and the distance given (200m) which I think would be the adjacent side. I know something is wrong because the angle that I've calculated indicates the hypotenuse is shorter than the adjacent side of 200m.

    Did I start out with the wrong velocity formula?

    I appreciate any help you can provide!

    1. The problem statement, all variables and given/known data

    2. Relevant equations

    3. The attempt at a solution
  2. jcsd
  3. Aug 18, 2014 #2


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    Your intuition is correct. In doing the calculation above, you didn't handle the units correctly on the RHS of the equation.

    The correct calculation would have read:

    Vf2 = 16m[itex]^{2}[/itex]/s[itex]^{2}[/itex]

    Remember, you don't have Vf on the LHS, but Vf2.

    How do you calculate Vf?

    You've got to use the correct value of Vf for the triangle. But remember, Vf is constantly changing, because the canoe is accelerating as it travels across the river.
  4. Aug 21, 2014 #3
    I completely overlooked that. So I need to take the square root of 16 m2/s2 which gives me 4 m/s for vf. That means the hypotenuse is 20 m/s.

    the angle between the hypotenuse is now 5.79°. Using CAH, that would mean the hypotenuse is 200m which doesn't seem right.

    But for the sake of the calculation, that would mean to find time:

    t = (4-0)/0.04
    t = 100 s

    it would take 100s to reach the other side

    And to find the distance it has traveled down the river, we need to find the third side by using the Pythagorean Theorem:

    c2 = 200 + 2002 - 2(200)(200)cos 5.79
    c2 = 408
    c = 20.2 m

    This can't be right.

    Please bear with me, I'm not picking this up as quickly as I'd like.
  5. Aug 21, 2014 #4


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    That is the correct time.

    To find the distance it travelled upstream you don't need any triangles or the pythagorean theorem.

    You know how long the boat was in the water, and you know what the downstream velocity was for that time (2m/s) so how would you find the downstream displacement?
  6. Aug 23, 2014 #5
    I could add the vf of 4 m/s and downstream velocity of 2 m/s to get 6 m/s and multiply it by 100 s to get the downstream displacement of 600 m.

    Is it that easy?

    The book is also asking for me to sketch the shape of the path the canoe will follow. Would this suffice?

  7. Aug 23, 2014 #6


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    The 2m/s and the 4m/s are in different directions (perpendicular) so you can't just add them like that (you'd need to use pythagorean's theorem)

    But the thing is the question didn't ask for the total displacement, it's only asking for the downstream displacement (the unlabeled side of your picture in post#5) so you would just multiply 2m/s by 100 seconds (the 4m/s is directed across the stream, so it doesn't contribute do the downstream distance)

    About the sketch of the shape the path followed; yours would be correct IF the canoe had a constant velocity, but it doesn't have a constant velocity, so it will be a little trickier.
  8. Aug 23, 2014 #7
    I don't know whether it is correct or not, I am getting time as 100 s and shifted distance as 200 m.
    The path can not be a straight line (should be parabolic) as there is accelerated motion means there is a force acting on the canoe in the vertical direction where as there is a horizontal constant velocity (2 m/s) of river current. Assuming it starts from rest, it will have a final vertical velocity of 4 m/s and taking the average vertical velocity i.e. (0 + 4)/2 = 2 m/s, it will take 200/2 = 100 s to reach the opposite side, and in this time it will be shifted horizontally by 2 (river velocity) x 100 = 200 m
  9. Aug 23, 2014 #8
    I'm overlooking the easy stuff. Thanks for pointing that out! I really appreciate your help!!

    It will take 100s to get to the other side and the canoe will end up 200m downstream.

    Does the new sketch look better? It says to sketch the path so I'm assuming the shape just needs to be right but not exact. Would you think that's a safe assumption or should I plot the position/time coordinates? I'm sure that will only take me four days to figure out. :cry:

  10. Aug 23, 2014 #9


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    Hahah yes I think your sketch is good. If I were a teacher I would accept it; it shows that you understand what's happening.
    (You can't expect it to be perfect, you're not a computer!)

    If I submitted that and the teacher said "no, that's not exact enough" I would be very annoyed :tongue:

    (I think your sketch is a lot better looking than mine would be haha)
    Last edited: Aug 23, 2014
  11. Aug 23, 2014 #10
    x=x°+V°*t+1/2*a*t(square)---->x=1/2*0.004*10000=200m I think this is the right answer and it's a quarter of a circle. Still not so sure, but everything seems to be fine with it.
  12. Aug 23, 2014 #11


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    You mean the path is a quarter of a circle? That is not right.

    (When I say "x" I mean "the across-stream direction" and when I say "y" I mean "the down-stream direction")

    If the x-acceleration is constant, then the graph of the x-displacement vs. time is a parabola.

    Since the y-displacement is proportional to the time, (i.e. the downstream velocity is constant,) "the path" will also be parabolic.
    (Because "the path" is just "the graph of x-displacement vs. y-displacement")

    I wish I could say it more simply; I feel like I'm making it sound more complicated than it is.

    A simpler way to explain it may be like this:
    We know [itex]x(t)=0.02t^2[/itex] (a parabola) and we know [itex]y(t)=2t[/itex] (a straight line) and we want to find x(y) (the path it takes)
    We could just find [itex]t(y)=y/2[/itex] (also a straight line) and plug it into x(t) to get [itex]x(t(y))=x(y)=0.005y^2[/itex] which is a parabola

    (What I was trying to say in my original post is that, since y-position is proportional to time, replacing time with y-position will only have the effect of "stretching" the parabola in some way, but it will still be a parabola.)
    Last edited: Aug 23, 2014
  13. Aug 23, 2014 #12
    So what you are suggesting is that the graph of x-dispacement vs. y-displacement(or in your case porportional to time) is "the path". How do you think that such a thing is true ? I mean , if i stand up now and move forward with a constant accelerating velocity and then move backwards with a constant one , then i moved parabolicly then in a straight line ? No , sir i just moved back and forward. Plus if there is smth wrong with my calculation , I would like you to point it out , hence if nothing is wrong with it then "the path" is clearly a quarter of a circle.
  14. Aug 23, 2014 #13


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    This is not at all a similar situation. I think you misunderstood what I meant. Sorry, that is my fault, my orginial post was not very clear. I just edited it though.

    Your calculation is fine but your calcuation has nothing at all to do with the shape of the path. Also your equation has nothing to do with a quarter of a circle, so I do not understand why you say that "it is clearly a quarter of a circle"
  15. Aug 23, 2014 #14
    So in my calculation y=200 and we also have x=200 , imo that's a quarter of a circle.
    I understand your point, i also did at the beginning before you edited it , but the parabola may also ,when stretched, be a circle. Still, it might also undergo a parabolic motion and have x and y coordinates equal to 200m. So I think you might be right , unless someone calculates the curve.
  16. Aug 23, 2014 #15


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    That only describes the end point (and we also know the beginning point is x=0 y=0). There could be ANY path that takes you there (a straight line, a parabola, a quarter of a circle, some wacky curve, or anything else) there's no reason that it needs to be a quarter of a circle.

    Well when I say "stretched", I meant that only one axis is "stretched" (or "compressed")
    Sorry I'm not using very mathematically precise language, and that may be the problem.

    Suppose you have a parabola [itex]y=x^2[/itex], what I mean by "stretching the parabola" is changing the equation to [itex]y=kx^2[/itex]
    (visually, this "stretches" or "compresses" the parabola, (that's why I used that language,) but it still results in a parabola)

    Sorry for causing all this confusion. I'm a visual thinker, so for me that seems like the easiest way to explain things, but I should really be more precise (sorry).

    The equation of the curve is [itex]x=0.005y^2[/itex]
    You can see when y=0, x=0 and when y=200, x=200 (so this equation satisfies our beginning and end points)

    The way I derived this equation is at the end of post #11
  17. Aug 23, 2014 #16
    Well you are right.
  18. Aug 24, 2014 #17

    Thank you very much for your input!
  19. Nov 4, 2015 #18
    Hi everyone. So I have this same problem as a bonus question. Is the angle of the triangle even necessary to find for the final answer or is downstream an appropriate direction?
    Last edited: Nov 4, 2015
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