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Stressed-out girl

  1. Sep 2, 2004 #1
    Pleeeeeeeeeeeeeeeeease Help!


    A test rocket is fired vertically upward from a well. Acatapult gives it an initial speed of 79.2 m/s at ground level. Its engines then fire and it accelerates upward at 4.10 m/s^2 until it reaches an altitude of 950m. At that point its engines fail, and the rocket goes into free fall, with an acceleration of -9.80 m/s^2. (You will need to consider the motion while the engine is operating separate from the free-fall motion.)

    a) How long is the rocket in motion above the ground?

    b) What is its maximum altitude?

    c) What is its veloctiy just before it collides with the Earth?
     
    Last edited: Sep 2, 2004
  2. jcsd
  3. Sep 2, 2004 #2
    Kathy Kool buys a sports car that can accelerate at the rate of 5.40 m/s^2. She decides to test the car by racing with another speedster, Stan Speedy. Both start from rest, but experienced Stan leaves the starting line 1.00 s before Kath. Stan moves with a constant acceleration of 4.00 m/s^2, and Kathy maintains an acceleration of 5.40 m/s^2.

    a) Find the time at which Kathy overtakes Stan.

    b) Find the distance she travels before she catches him.

    c) Find the speeds of both cars at the instant she overtakes him.
    Kathy_______ Stan___________
     
  4. Sep 3, 2004 #3

    Math Is Hard

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    Howdy, Elizabeth!

    Please show a little work that you have done on the problems or at the at the very least, how you were thinking about approaching the problems.

    If you do this, there are lots of people who will be eager to jump in and give you some help. :smile:
     
  5. Sep 3, 2004 #4

    Mk

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    What he said, and post a question like this next time in the homework forums :) Welcome!
     
  6. Sep 3, 2004 #5

    Chronos

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    Math is a notorious cross-poster. :smile:
    Elizabeth, try using this
    [tex]D = v_it+\frac{at^2}{2}[/tex]

    [tex]D =[/tex] distance, meters
    [tex]v_i =[/tex] initial velocity, meters/sec
    [tex]t =[/tex] time in seconds
    [tex]a =[/tex] acceleration, meters/sec
     
  7. Sep 3, 2004 #6
    Try This

    Elizabeth,

    You have to break this problem up into parts. In the first time interval, the rocket is being acted on by the engines acceleration. Thats Interval 1. Then the engines fail and the rockets descent begins to slow, more and more until it is falling towards the earth. This is the second part (the freefall). First, find the time interval 1, which is how long it will take the rocket to accelerate itself to a altitude of [itex] 950m [/itex].

    This is given by:

    [tex] \Delta y = V_0t + \frac{1}{2}at^2 [/tex]

    Now solve for [itex] t [/itex]. Tuck this answer away so that we can finish this part of the problem later.

    Part (b) asks you to find the maximum height attained by the rocket. Remeber, the rocket propels itself to an altitude of 950m, and after that is when the engines fail and the rocket begins to slow down. We need to know how fast the rocket is going at the time of the engine failure. [itex] V_f [/itex]. Use this equation:

    [tex] V_f = V_0 + at [/tex]

    We know the initial speed of the rocket (from the catapult), we know its acceleration from the engines, and we know the time from our first answer in part (a).

    Okay, so far so good. Now this equation comes to mind:

    [tex] V^2_f = V^2_0 - 2g \Delta y[/tex]

    Now lets plug in what we know so that we can get [itex] \Delta y [/itex], which is the height the rocket attains. (Note, this is the height from the 950m mark, not from the ground) Solve for [itex] \Delta y [/itex] and add the original 950 meters. This gives you the height from above the ground.

    The last part of the problem is to find the velocity just before the rocket hits the earth. Lets treat the 950 meters point as the origin or point of reference. The earth is 950 m BELOW this so we need to solve for when [itex] y = -950 [/itex]. Use:

    [tex] \Delta y = V_0t - \frac{1}{2}gt^2 [/tex]

    This will allow you to solve for [itex] t [/itex], which is the time for the second interval we talked about earlier. Add these times together and that should be part (a). Now, Last but not least, we need to find the final velocity.

    [tex] V_f = V_0-gt [/tex]

    Solve for [tex] V_f [/tex] where [tex] V_0 [/tex] is the initial velocity the rocket reached at 950 meters, and [itex] t [/itex] is the time we just calculated. Remember, since the rocket will be falling down, your answer should be negative as velocity is a vector quantity (Unless you are using a different frame of reference than I was for this problem).

    Hope this helps, I was working the problem quickly here at work. I will double check everything later.

    On a side note, If you learn these concepts and learn how to *think* about the problem, the second problem you listed will come to you with greater ease than this one. SO GO STUDY!!

    :smile:
     
    Last edited: Sep 3, 2004
  8. Sep 3, 2004 #7
    Yes just show some work so we can show you more
     
  9. Sep 3, 2004 #8
    ouch looking at the time of your post it sounds like we were too late anyway hopefully though i am wrong
     
  10. Sep 3, 2004 #9
    You need to do the work but Ill give you the answers in case you need to check your anser.


    [tex] t_{total} = 40.2s [/tex]

    [tex] d_{top} = 1673m [/tex]

    [tex] v_{final} = 181m/s [/tex]
     
  11. Sep 3, 2004 #10
    ok

    I'll second those answers,

    :surprised
     
  12. Sep 3, 2004 #11

    Math Is Hard

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    I think maybe my kitty avatar needs to have a little pink bow around her neck! :biggrin:
     
  13. Sep 4, 2004 #12
    lol i couldn't even tell taht was a cat until you said it
    then again it is 3 in the morning.
     
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