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Math Help - help (acceleration)

  1. #1
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    help (acceleration)

    A jumbo jet must reach a speed of 365 km/h on the runway for takeoff. What is the least constant acceleration needed for takeoff from a 2.05 km runway?
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  2. #2
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    use the kinematics equation ...

    v_f^2 = v_0^2 + 2a(\Delta x)

    rearranged ...

    a = \frac{v_f^2 - v_0^2}{2(\Delta x)}

    you might want to convert to meters and seconds unless a solution in \frac{km}{hr^2} is o.k.
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    Quote Originally Posted by skeeter View Post
    use the kinematics equation ...

    v_f^2 = v_0^2 + 2a(\Delta x)

    rearranged ...

    a = \frac{v_f^2 - v_0^2}{2(\Delta x)}

    you might want to convert to meters and seconds unless a solution in \frac{km}{hr^2} is o.k.

    how do you that you would use that equation??

    so it would go like this?

    360km^2 = 0^2 +2(a)1.91km

    360km^2 = 2(a) 1.91km

    (360km^2)/2(1.91km)= a


    that right?
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    Quote Originally Posted by Legendsn3verdie View Post
    A jumbo jet must reach a speed of 365 km/h on the runway for takeoff. What is the least constant acceleration needed for takeoff from a 2.05 km runway?
    The airplane starts from rest, so initial velocity Vo = 0

    V = Vo +at
    365 = 0 +at
    a*t = 365 -----------(i)

    We don't know a, we don't know t.
    But we know the distance to be travelled to get to V = 365 km/h.
    d = (average velocity)*t
    ave V = (1/2)(Vo +V) = (1/2)(0 +at) = at/2
    So,
    d = (at/2)*t = (a*t^2)/2
    We know d, and we can get t in terms of a from (i), so, substitutions,
    2.05 = (a*(365/a)^2)/2
    4.10 = (365^2)/a
    a = (365^2) / 4.10 = 32,494 km/hr/hr

    In meters/sec/sec, that would be
    (32,494 km/(hr)^2)*(1000m/1km)*[(1hr)^2 / (3600sec)^2]
    = 2.50725 m/sec/sec ------------------answer.
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    i lost you from here:

    Quote Originally Posted by ticbol View Post
    We don't know a, we don't know t.
    But we know the distance to be travelled to get to V = 365 km/h.
    d = (average velocity)*t
    ave V = (1/2)(Vo +V) = (1/2)(0 +at) = at/2
    So,
    d = (at/2)*t = (a*t^2)/2
    We know d, and we can get t in terms of a from (i), so, substitutions,
    2.05 = (a*(365/a)^2)/2
    4.10 = (365^2)/a
    a = (365^2) / 4.10 = 32,494 km/hr/hr

    In meters/sec/sec, that would be
    (32,494 km/(hr)^2)*(1000m/1km)*[(1hr)^2 / (3600sec)^2]
    = 2.50725 m/sec/sec ------------------answer.

    looks like your using another equation i m not aware of
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    Quote Originally Posted by Legendsn3verdie View Post
    i lost you from here:




    looks like your using another equation i m not aware of
    It is called, er, "building up".
    Kidding aside, it is basic Physics of uniform linear motion. It is starting from scratcth, building up.

    Oh, I lost you from the 365 = at?

    Or, was it from the distance = (ave velocity)*time?

    Uh, sorry, but I do not memorize too many formulas. I always rely on the basic ones only and then I build up or derive the other formulas from those basics only. I am very poor at memorization.
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  7. #7
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    Quote Originally Posted by Legendsn3verdie View Post
    how do you that you would use that equation??

    so it would go like this?

    360km^2 = 0^2 +2(a)1.91km

    360km^2 = 2(a) 1.91km

    (360km^2)/2(1.91km)= a


    that right?
    that would be almost correct ... the numerator has units km^2/hr^2, so your solution would be in units of km/hr^2.

    But, this is your original post ...

    A jumbo jet must reach a speed of 365 km/h on the runway for takeoff. What is the least constant acceleration needed for takeoff from a 2.05 km runway?
    why the change in values?
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    Quote Originally Posted by skeeter View Post
    that would be almost correct ... the numerator has units km^2/hr^2, so your solution would be in units of km/hr^2.

    But, this is your original post ...

    why the change in values?

    because i m doing the problems in the book.. i dont want to know the answer.. i kinda want to get the answer myself, but what does it matter if i see all the steps? .. so i almost had the answeR? all i had to do was convert? i m clueless on how to convert km^2/hr^2 to km/hr^2
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  9. #9
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    there was no conversion involved ... when you square the velocity in units of km/hr in the numerator, you get units of km^2/hr^2.

    km^2/hr^2 divided by km is km/hr^2, a valid (but not often used) unit for acceleration.
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    Quote Originally Posted by skeeter View Post
    there was no conversion involved ... when you square the velocity in units of km/hr in the numerator, you get units of km^2/hr^2.

    km^2/hr^2 divided by km is km/hr^2, a valid (but not often used) unit for acceleration.
    ok so did i do this right?

    (360km^2/hr^2 = 0^2 +2(a)1.91km

    360km^2/hr^2 = 2(a) 1.91km

    (360km^2hr^2)/2(1.91km)= a

    a = 94.24km/hr^2
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  11. #11
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    you forgot to square the 360.
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    (360km^2/hr^2 = 0^2 +2(a)1.91km

    12960km^2/hr^2 = 2(a) 1.91km

    (12960km^2hr^2)/2(1.91km)= a

    a = 33926.70 km/hr^2

    .. u sure u need to square the 360?
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