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Please help me. I am dieing on these problems.

  1. Sep 27, 2006 #1
    Problem 1.
    A pelican has a fish in his mouth. He's traveling 3 m/s. He drops fish. What is fish's distance from pelican after 2 s?
    I did this
    Vi = 3m/s for both the fish and the pelican
    a = 9.8 (is this negative or pos?)
    Then i used
    x = Vi (T) 1/2 (a) (t) ^2
    I got for this
    25.6 m
    Did I do this right? If no, what i am doing wrong..
    Problem 2.
    I am so confused on this. I have tried this for 65 minutes now.
    Here it is.
    A speeding car is traveling 92.0 km/h toward a police car at rest. facing the same direction as the speeding car. If the police car begins accelerating when the speeding car is 250.0 m behind the police car, what must the police car's acceleration be in order for the police car to reach the speeding car's velocity at the moment the speeding car catches up? Assume that the speed car does not slow down.

    Ok i have no clue how to do this one. I tried everything
    Can someone please give me detailed instructions on this one and tell me if my number 1 was correct?
    Sorry if this is a burden
    THis is my first post!
  2. jcsd
  3. Sep 27, 2006 #2
    Sorry if I am sounding a bit irate but if I don't do good on this homework assignment this will be the break between a B and C. Plz help me :(.
  4. Sep 27, 2006 #3
    Please guys.. they're due tomorrow and its 10:25... i keep on trying and cant do it.. and really. 39 views and no answers.>?
  5. Sep 27, 2006 #4


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    Hello Gyros,

    Ok then....that means the pelican and fish are represented by the letters P and F instead of nice drawings :frown:

    Do these little sketches help you to get new ideas?

    http://img142.imageshack.us/img142/4568/pelicanbd7.jpg [Broken]

    You can try to separate the movement into x and y components. Does the initial velocity have any effect on the y component?
    The sign for the acceleration a (or g in the sketch) depends on where your y-axis would point to. In my sketch it is meant to point down to the ground, the acceleration g would be positive.

    http://img171.imageshack.us/img171/5224/policegh3.jpg [Broken]

    small s/p = speeding car/police car,
    small x = intersection


    Last edited by a moderator: May 2, 2017
  6. Sep 27, 2006 #5
    I am sorry but I am so confused... this is like our 4-5th day and my teacher gave us impossible problems.. could you like explain in non-picture form. I hope I am not asking too much >.>.
  7. Sep 27, 2006 #6


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    Hello Gyros,

    sorry, my sketches might have caused even more confusion :redface:

    Problem 1:

    Both the fish and the pelican have initial velocity of 3 m/s. There's some information missing:

    Does the pelican fly horizontally at all times? I assumed that this is the case and that's why the initial velocity doesn't have a y-component in my sketch. Both the pelican and the fish will have a velocity of 3 m/s in x-direction during the whole process. But it might be different in your problem. Can you look that up please? I also neglected any kind of air resistance. The distance of both will be represented by the height the fish "lost" in comparison to the pelican.

    Problem 2:

    I was trying to sketch a distance/time graph for both the speeding car and the police car. The speeding car has a known velocity [itex]v_s=92\,km/h[/itex] and the distance travelled will calculated by s=vt. The police car will accelerate (a=constant), so the distance will be calculated by [itex]s=\frac{1}{2}\cdot a\cdot t^2[/itex]. It also has a headstart of 250 m, that's why this is added to get [itex]s_p(t)[/itex].

    Both cars meet at the time [itex]t_x[/itex] in my sketch. The velocity of the police car [itex]a\cdot t_x[/itex] will be equal to the velocity of the speeding car[itex]v_s[/itex]. The travelled distances (including the headstart) will match as well [itex]s_s(t_x)=s_p(t_x)[/itex]. Using both equations, you'll be able to get the acceleration [itex]a_p[/itex] for the police car.
    Last edited: Sep 27, 2006
  8. Sep 27, 2006 #7
    I have no clue what you're talking about :(
    He has only told us 4 formulas so far
    VF = VI + at
    VF^2 = VI^2 + 2ax
    and i forget the other one like these 2
    and finally
    V = X/T
    can you explain the problem to me only using those problems? (btw i got 24.6 on the pelican one. I am pretty sure thats the answer too as long as gravity is 9.8 and not -9.8. Can you show me 2)?
  9. Sep 27, 2006 #8


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    I've edited post https://www.physicsforums.com/showpost.php?p=1097293&postcount=6" with more information regarding problem 2.

    The formulas I used are basically the ones you've provided.


    In problem 2 I've used this formula [itex]a\cdot t_x[/itex] to calculate the final velocity of the police car for the time [itex]t_x[/itex].


    Use this formula to calculate the distance travelled by the speeding car: [itex]s_s(t_x)=v_s\cdot t_x[/itex]

    One formula you might be missing is the distance travelled while accelerating. If a=const. you can use: [itex]s=\frac{1}{2}\cdot a\cdot t^2[/itex] (+s(initial)). I've used this equation to calculate the distance travelled by the police car [itex]s_p(t)[/itex].
    Last edited by a moderator: Apr 22, 2017
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