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Acceleration of a Boat Problem

  1. Dec 4, 2013 #1
    1. The problem statement, all variables and given/known data

    16. ) A row boat, starting from rest takes 11 seconds to travel 34 meters. If the boat maintains a uniform acceleration, what will be its instantaneous velocity at the end of the 11-second period??
    17.) At the point in the stream where the boat is crossing, the stream is 202 meters wide. If the boat in Question 16 maintains its final velocity for the remainder of the trip, how much time will it need to reach the opposite Shore?
    18.) What acceleration did the row boat initially experience in Question 16??

    So this is a Question of my Teacher
    Wherein i want to prove that I made the right answer. Can you please solve this ?? Because I am having difficulty in proving something without concrete evidence from other Physics geeks.... I know this is a really easy problem but I need a more concrete proof. I have already done this but I am trying to make sure. My Teacher tried doing the constant velocity where in the equation is v=d/t


    2. Relevant equations
    d= vit + at^2/2
    d= vt



    3. The attempt at a solution
    no.18)
    d=vit+at^2/2
    so since vi = 0 so i can remove vit
    d=at^2/2
    a=2d/t^2
    a=2(34)/11^2
    a= 68/121
    a= 0.56m/s^2

    no 16)
    vf = vit + at
    wherein vi = 0 so i can remove the vit in the equation
    vf= at
    vf= (0.56)(11)
    vf= 6.16m/s or if i use the not rounded off part of .56m/s^2 6.18m/s

    no 17)
    d= 168
    v= 6.16m/s
    since the part wanted was the distance from the boat to the other shore which will be 202-34= 168
    t=d/v
    t= 168m/6.16m/s
    t= 27.27s
    or
    t=168m/6.18/s
    t= 27.18s

    my Attempt
    no. 16 . ) 6.16m/s
    no. 17 . ) 27.27secs
    no. 18 . ) a= .56m/s^2

    My teacher's attempt
    no. 16 . ) 3.09m/s
    no. 17 . ) 56.38secs
    no. 18 . ) 0.02m/s^2

    I know my teacher is wrong already just looking at the Acceleration. I understand her because of the stress being put upon her though

    I know the Significant Figures is wrong the reason for that is because my teacher wanted to have a 2 decimal at the end. I tried to argue with her but she's the teacher and she's a stubborn one.
     
    Last edited: Dec 4, 2013
  2. jcsd
  3. Dec 4, 2013 #2

    NascentOxygen

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    Staff: Mentor

    You have to watch those significant figures. :smile:

    The data showed only 2 sig figs, so strictly speaking you should give answers only to 2 sig figs. But if you round acceleration to 2 sig figs, you can't then give time to 4 sig figs and expect that extra precision to make sense.
     
    Last edited: Dec 4, 2013
  4. Dec 4, 2013 #3

    PhanthomJay

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    Science Advisor
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    Gold Member

    Your answers more or less look correct, but you do not show how you arrived at the solution. Please show your work and we can check on the correctness of your attempt..
     
  5. Dec 4, 2013 #4

    PhanthomJay

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    Gold Member

    yes
    yes but it should round off to 6.2 but if teacher wants it to 2 decimal places then use the 6.18
    yeah.......you need a nice stop watch to check your result to the nearest one-one hundreth of a second....say OK

    I am sure she appreciates your understanding...I do...I've been wrong more than once...
    Did you know you can start with Q.16 and use d = [(vi + vf)/2]t to solve for vf? It seems to be the forgotten one of the kinematic equations for constant acceleration......:cry:
     
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