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Thread: velocity and acceleration

  1. #1
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    velocity and acceleration

    An airplane takes off after it travelled adistance of 1350 m on the runway. The airplane started from rest,moved with aconstant acceleration for 3o seconds before taking off .what was the speed at the time of take off ? ( I need to solve by integral with my greatfull)
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  2. #2
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    Hello, BAHADEEN!

    An airplane takes off after it travelled adistance of 1350 m on the runway.
    The airplane started from rest, moved with a constant acceleration
    for 30 seconds before taking off.
    What was the speed at the time of take-off?

    Let $\displaystyle \,a$ be the constant acceleration (in $\displaystyle \text{m/sec}^2$)

    Then the velocity is: .$\displaystyle v(t) \;=\;\int a\,dt \:=\:at + v_o$
    . . where $\displaystyle \,v_o$ is the plane's initial velocity.

    The plane started from rest, so $\displaystyle v_o = 0.$
    . . The velocity function is: .$\displaystyle v(t) \:=\:at$ .[1]


    Then the position function is: .$\displaystyle s \;=\;\int at\,dt \:=\:\frac{1}{2}at^2 + s_o$
    . . where $\displaystyle \,s_o$ is the plane's initial position.
    Assume that $\displaystyle s_o = 0$, we have: .$\displaystyle s(t) \:=\:\frac{1}{2}at^2$ .[2]


    We are told that $\displaystyle s = 1350$ when $\displaystyle t = 30.$

    Substitute into [2]: .$\displaystyle 1350 \:=\:\frac{1}{2}a(30^2) \quad\Rightarrow\quad a \:=\:3 $


    Substitute $\displaystyle t = 30,\;a = 3$ into [1]:

    . . $\displaystyle v(30) \:=\:(3)(30) \:=\:90\text{ m/sec.}$

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