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NEED PHYSICS HELP! am i doing this correctly?

  1. Nov 1, 2011 #1
    Once again the determined coyote is out to get the road runner. He has a new pair of power skates that provide a constant acceleration of 6.0 m/s^2 along the flat top of the mesa he starts off from rest 48.0 m from teh edge of a cliff just as the road runner zips past towards the cliff at constant speed. a) calculate how long it takes the coyote to reach the edge of the cliff. b) what is his speed as he gets there? c) calculate the minimum constant speed of the road runner in order to escape the clutches of the coyote. d) calculate how far from the cliffs edge the coyote lands in the canyon 78.4 m below. His power skates do not provide acceleration when he is air borne. e) what is his velocity on the cactus laden canyon base when he lands there?1. The problem statement, all variables and given/known data



    2. Relevant equations



    a) Calculate how long it takes the coyote to reach the edge of the cliff.
    V= V0 + at
    24.0 m/s= 0 + (6.0 m/s²)t
    t= 4.0sec

    b) What is his speed just as he gets there?
    V²= V0² + 2as
    V²= 0 + 2(6.0 m/s²)(48.0m)
    V²= 576 m²/s²
    V= √576 m/s
    V= 24.0 m/s

    c) Calculate the minimum constant speed of the roadrunner in order to escape the clutches of the coyote.
    Min Speed = 48.0 m/ 4.0 s = 12.0 m/s

    d) Calculate how far from the cliff’s edge the coyote lands in the canyon 78.4 m below. His power skates do not provide acceleration when he is air-borne.
    Y= y0 + Vy0(t) + (1/2)ay(t)²
    Y= y0 + Vy0(t) – (1/2)g(t)²
    0= 78.4 m + 0 – (1/2)(9.8 m/s²)t²
    0= 78.4 + 0 – 4.9t²
    4.9t²= 78.4
    t²= 16
    t= √16
    t= 4.0sec

    x= x0 + vx0(t) + (1/2)ax(t)²
    x= 0 +(24.0 m/s)(4.0s)+(1/2)(6.0 m/s²)(4.0s)²
    x= 144 m







    e) What is his velocity on the cactus laden canyon base when he lands there?
    Vx= Vx0 + ax(t)
    Vx= (24.0 m/s) + (6.0 m/s²)(4.0s)
    Vx= 48.0 m/s

    Vy= Vy0 + ay(t)
    Vy= Vy0 – g(t)
    Vy= 0 – (9.8 m/s²)(4.0s)
    Vy= -39.2 m/s

    V=(vx² +vy²)^1/2 = (48)² + (39.2)²)^1/2 = 61.972 m/s

    3. The attempt at a solution
     
  2. jcsd
  3. Nov 1, 2011 #2

    Delphi51

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    All correct except the (1/2)(6.0 m/s²)(4.0s)² in part (d) should be omitted because "His power skates do not provide acceleration when he is air borne."
    Jolly good work!
     
  4. Nov 1, 2011 #3

    SammyS

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    See comments above.
     
  5. Nov 1, 2011 #4

    Delphi51

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    It IS just a bit confusing unless you read (b) before (a).
    Maybe better to find the time in (a) using d = ½ a⋅t²
     
  6. Nov 1, 2011 #5
    Using the d=1.5a*t
    48.0=1/2 6.0* t^2
    48/(1/2*6.0)= 16
    =4.0 ??

    and for the Vx= Vx0 + ax(t)
    Vx= (24.0 m/s) + (6.0 m/s²)(4.0s)
    Vx= 48.0 m/s
    what did i do wrong?

    and for

    x= x0 + vx0(t) + (1/2)ax(t)²
    x= 0 +(24.0 m/s)(4.0s)+(1/2)(6.0 m/s²)(4.0s)²
    x= 144 m

    what should i use instead of the 6.0?
    Im pretty frustrated that i feel completely lost on my homeowork today thanks for the help!
     
  7. Nov 1, 2011 #6

    SammyS

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    What is the acceleration in the horizontal direction when he's airborne ?
     
  8. Nov 2, 2011 #7

    Delphi51

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    Wait, your first solution for (b) was correct!
    You can also get it with Vf = Vi + at = 0 + 6*4 = 24.
    For (d), the acceleration is zero when airborne:
    x= x0 + vx0(t) + (1/2)ax(t)²
    x= 0 +(24.0 m/s)(4.0s)+(1/2)(0)(4.0s)²
    x= 96 m

    You are very good at this, just a tiny bit of confusion about details in the question. Recommend you make a habit of writing down the numbers in the question in a way that makes sense to you at a glance.
    I wrote
    coyote.jpg
     
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