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Formula for a two-part constant-acceleration problem

  1. Sep 23, 2009 #1
    1. The problem statement, all variables and given/known data
    This is the problem: You have designed a rocket to be used to sample the local atmosphere for pollution. It is fired vertically with a constant upward acceleration of 16 m/s2. After 20 s, the engine shuts off and the rocket continues rising (in freefall) for a while. (Neglect any effects due to air resistance.) The rocket eventually stops rising and then falls back to the ground. You want to get a sample of air that is 11 km above the ground.

    Here is what I still need to find: (1) Determine the total time the rocket is in the air in seconds.

    (2) Find the speed of the rocket just before it hits the ground.



    2. Relevant equations



    3. The attempt at a solution

    I need formulas to figure these out. I have already done part of this problem by finding the highest point my rocket reaches. (8.4 km)

    I made a graph and found xf1 xf2 vf1 and vf2. Can somebody please give me the formulas to finding the total time and the speed before it hits the ground. Thanks.
     
  2. jcsd
  3. Sep 23, 2009 #2
    You already found the altitude and velocity of the rocket at the time the engine shuts off?

    For (1) You don't need anything more than this equation for 1 dimensional constant acceleration.

    [tex]
    x = x_0 + v_0 t + (1/2) a t^2
    [/tex]


    you know x_0 and v_0 at the time the engine shuts off. the rocket will reach the ground when x=0, so you can find out t. (t starts out with 0 again for the second part of the flight)
     
  4. Sep 23, 2009 #3
    Ok, thanks for your help.

    I got the answer for (2). I found the [tex] \sqrt{2gh} [/tex] which gave me 406.591 m/s.

    I'm confused about the first one though. It would be [tex] 0+0(t)+(1/2)at^2 [/tex] but that gives me [tex] 1/2(16)(20)^2 [/tex] which gives me 3200 meters. That's the height it reaches after the engine is released, I need to know the total time the rocket is in the air. From when it launches to after the engine shuts off and gravity brings it back to the ground, right?
     
  5. Sep 23, 2009 #4
    Vf=V0+a*t
    Vf2=V02+2*a*(y-y0)

    When you used the previous equation for part (1) for the first part of your rocket's motion (the accelerating upwards), you already knew the time (20s), so you didn't need to do any calculating.

    These are the kinematic equations for constant acceleration. You just plug and crank with these (often).

    Your textbook should list them.
     
  6. Sep 23, 2009 #5
    The 20 seconds was just the time when the engine was shut off, not the entire time the rocket is in the air.
     
  7. Sep 23, 2009 #6
    I have another problem to do similar to the other one. I don't know how to get this started either. Is it the same types of formulas again?

    Question: Two trains face each other on adjacent tracks. They are initially at rest, and their front ends are 40 m apart. The train on the left accelerates rightward at 1.15 m/s^2. The train on the right accelerates leftward at 1.13 m/s^2.

    (a) How far does the train on the left travel before the front ends of the trains pass in meters?

    (b) If the trains are each 150 m in length, how long after the start are they completely past one another, assuming their accelerations are constant in seconds?
     
  8. Sep 23, 2009 #7
    Yes, I'm aware. That is what I'm stating. When you used that equation previously, you ended up finding meters. You seemed to be confused, and I was just telling you that that part of the rocket's motion took 20 seconds, as the problem strictly stated.

    Now find the other parts and add it to the 20 seconds.

    You would use the same equations in that case, as well.
     
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