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College phys neeed help

  1. Sep 5, 2009 #1
    college phys 1111... neeed help!!

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

    A space probe is desperately trying to escape the gravitational pull of a massive planet. After a final blast of its thrusters, the probe has a velocity of 2940 m/s away from the planet. Later, the probe is 176.4 km closer to the planet than it originally was, with a velocity of 8820 m/s toward the planet.

    You are to find (assume constant acceleration):
    a) the acceleration (magnitude and direction)
    b) How far beyond its starting point the probe got.
    c) How much time elapsed between initial and final positions.

    I don’t understand how to approach this problem…. How do I find acceleration when I don’t know the time?
     
  2. jcsd
  3. Sep 6, 2009 #2

    ideasrule

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    Re: college phys 1111... neeed help!!

    You know that the probe's velocity was originally 2940 m/s, but that after falling 176.4 km towards the planet (and losing a corresponding amount of potential energy), its velocity became 8820 m/s. You can use the conservation of energy to calculate g, the acceleration due to gravity.

    For b, you can use the conservation of energy again. c should be easy once you get the answer for a.
     
  4. Sep 6, 2009 #3
    Re: college phys 1111... neeed help!!

    i am sorry i still dont understand... =[
     
  5. Sep 6, 2009 #4
    Re: college phys 1111... neeed help!!

    i mean i know (Velocity final- Velocity initial)/time = acceleration but they dont tell the time and that is what is throwing me off...
     
  6. Sep 6, 2009 #5

    Redbelly98

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    Re: college phys 1111... neeed help!!

    Look at the kinematic equations for motion with constant acceleration. Look for the one equation that does not contain time t.
     
  7. Sep 6, 2009 #6
    Re: college phys 1111... neeed help!!

    v^2=vo^2+2a(x-xo) or
     
  8. Sep 6, 2009 #7
    Re: college phys 1111... neeed help!!

    v^2-vo^2=2a(x-xo)
     
  9. Sep 6, 2009 #8
    Re: college phys 1111... neeed help!!

    2940^2-8820^2=2a(x-176.4) and solve for a?
     
  10. Sep 6, 2009 #9

    Redbelly98

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    Re: college phys 1111... neeed help!!

    Yes. Um, you do know what the value of x is, right?
     
  11. Sep 6, 2009 #10
    Re: college phys 1111... neeed help!!

    ummm 0?
     
  12. Sep 6, 2009 #11

    ideasrule

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    Re: college phys 1111... neeed help!!

    Yup, exactly. But you used 176.4, which is in kilometers, whereas all your speeds are in m/s. You'll have to make the units agree.
     
  13. Sep 6, 2009 #12
    Re: college phys 1111... neeed help!!

    ok so 2940^2-8820^2=2a(0-109.6) i got a= 315459.85...... that doesnt sound right does it?
     
  14. Sep 6, 2009 #13
    Re: college phys 1111... neeed help!!

    thank you guys btw for ur help!!
     
  15. Sep 6, 2009 #14
    Re: college phys 1111... neeed help!!

    Well first convert x into meters, there are 1000 meters in 1km. Know you just plug and chug. And for some odd reason you have Vo as 8820 and V as 2940.
     
  16. Sep 6, 2009 #15
    Re: college phys 1111... neeed help!!

    ohhhh oops i think i converted the 176.4 km to miles..... soooo i should flip flop the v's? and 176.4 km to meters would be 176400?
     
  17. Sep 6, 2009 #16
    Re: college phys 1111... neeed help!!

    ok so i did it again and i got 196 for a now... that sounds kinda right. right?
     
  18. Sep 6, 2009 #17
    Re: college phys 1111... neeed help!!

    Looks good.
     
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