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Grade 10 physics help

  1. Apr 9, 2007 #1
    1. What us the Average acceleration of a rocket that starts from a rest, rises vertically and attains a velocity of 1500km/hr [up] in 20 seconds?



    2. Relevant equations



    3.
    v(with arow thing on top)= 1500km/hr[up]= 41.67m/s
    (triangle th ing)t=20s
    a(with arowo n top)= ?

    forula needed: a(arow on top)= v(with arow thing on top)/(triangle)t

    a= 41.67m/s / 20s
    a= 2.0835 or 2.08m/s^2


    [/b
     
  2. jcsd
  3. Apr 9, 2007 #2

    mjsd

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    and your question is ?
     
  4. Apr 9, 2007 #3

    danago

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    Have another look at your conversion from km/h to ms-1
     
  5. Apr 9, 2007 #4

    GmL

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    The triangle is called "delta", which is defined as "the change in...". The arrow on top signifies that the quantity is a vector; for example, the v with the arrow shows that it is a velocity, which inherently contains the direction (as opposed to speed, which is only a magnitude without a direction).

    Now having said all that, your equations look fine. The average acceleration is the change in velocity divided by the change in time, or (delta v)/(delta t). You converted the velocity incorrectly. That is your problem. Might want to check that again.

    EDIT: Looks like someone (danago) asked the question first.
     
    Last edited: Apr 9, 2007
  6. Apr 9, 2007 #5
    I have and im realy confused... the answe should be 75.00 (km/h)/s [up] or 20.83 m/s^2
    i just cant get it.... I got 8/10 on a little test but i did not get the right answers its so confusing and she did not tell me what i did rong...
     
  7. Apr 9, 2007 #6
    Do you know the equation for average acceleration? You are told the intial velocity, the "final" velocity, and the time...if you know how to find the average velocity and understand that, hopefully you can correlate that with finding the average acceleration.
     
  8. Apr 9, 2007 #7

    hage567

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    If you are going to express it in km/h/s then it is straightforward division of the two numbers you were given (since v0 is zero). But as the others have said, your conversion of the velocity into m/s is wrong. Remember, there are 1000 meters in 1 kilometer.
     
  9. Apr 9, 2007 #8
    should it be 1.500 m/s?
     
  10. Apr 9, 2007 #9

    hage567

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    No. If you have 1500 km, how many meters is that?
     
  11. Apr 9, 2007 #10

    danago

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    1500 km/h is the same as 1500000 m/h right? Since there are 1000m in a kilometer. Dividing by 60, you get 25000 m/min, since there are 60 minutes in an hour. Doing the same again (this time, going from minutes->seconds), you get 416.67 m/s. Doing all that is the same as divinding by 3.6, so to convert km/hr to m/s, just divide by 3.6.
     
  12. Apr 9, 2007 #11
    opps i whent the oposite way.
    I should have 1500000m
     
  13. Apr 9, 2007 #12
    ok thanks Dango :D Im geting it now.
     
  14. Apr 9, 2007 #13
    can I ask what i did rong in this question?

    A dragster slows down from 28m/s[N] in a time of 12s using a parachute and brakes. Calculate the displacement drurring this acceleration(answer should be 245m[n])

    delta t=12s
    vector vi=25m/s[n]
    vector vf=13m/s[n]
    delta d=?
    delta d=(delta vector v) (delta t)
    delta d=(25-13m/s)(12s)
    delta d= 144m
     
    Last edited: Apr 9, 2007
  15. Apr 9, 2007 #14

    danago

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    Im not quite sure what it is youre asking. It slows from 28m/s^2? Should that be 28m/s? And youre asking how long did the acceleration last, yet the answer is 245 meters?? Shouldnt it be in seconds?
     
  16. Apr 9, 2007 #15
    Sory about that i coppied part of the question rong. I fixed it.
     
  17. May 13, 2009 #16
    A=??
    V1=0 A= v2-v1
    v2=1500 _________
    t1=0 t2-t1
    t2=20

    A= 1500-0
    ________ = 1500
    20-0 20 A= 75km/hr/s

    V= Velocity
    A= Accleration
    T= Time
     
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