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Thread: Another Kinematics problem

  1. #1
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    Another Kinematics problem

    A rocket is launched straight up with constant acceleration. Four seconds after liftoff, a bolt falls off the side of the rocket. The bolt hits the ground 6.60 later.

    What is the rocket's acceleration?



    I'm wanting to use the 3 kinematics equations but I feel like i'm missing information. I know what the answer is already but I want to understand how to get it. It seems I can't simply use the equations for 1D kinematics so if anyone can show me i'd be much obliged.
    Last edited by JonathanEyoon; Jan 19th 2009 at 08:00 AM.
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    A rocket is launched straight up with constant acceleration. Four seconds after liftoff, a bolt falls off the side of the rocket. The bolt hits the ground 6.60 later.

    What is the rocket's acceleration?


    Let accln. of the rocket=a

    velocity of Bolt= velocity of rocket after 4 sec.= 4a .............upwards.................first eq of kinematics...


    dist. covered by rocket in 4 sec.= 1/2 XaX 4X4= 8a ................................second eq of kinematics


    now the Bolt falls from this height(8a) with accln. due to gravity in 6.6sec.

    second eq of kinematics

    s=ut-1/2 g t^2 .............................free fall
    8a= 4a-1/2 X (9.8 X6.6X6.6)
    solve this......
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  3. #3
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    while the bolt rides with the rocket (assuming the rocket started from rest) ...

    $\displaystyle v = 4a$

    $\displaystyle \Delta y = \frac{(4a)^2 - 0^2}{2a} = 8a$

    when the bolt gets free, $\displaystyle a = -g$ ...

    $\displaystyle \Delta y = 4a \cdot t - \frac{1}{2}gt^2$

    $\displaystyle -8a = 4a(6.6) - \frac{1}{2}g(6.6)^2$

    $\displaystyle -34.4a = - \frac{1}{2}g(6.6)^2$

    $\displaystyle a = \frac{g(6.6)^2}{2(34.4)} \approx 6.2 \, m/s^2$
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  4. #4
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    Quote Originally Posted by JonathanEyoon View Post
    A rocket is launched straight up with constant acceleration. Four seconds after liftoff, a bolt falls off the side of the rocket. The bolt hits the ground 6.60 later.

    What is the rocket's acceleration?



    I'm wanting to use the 3 kinematics equations but I feel like i'm missing information. I know what the answer is already but I want to understand how to get it. It seems I can't simply use the equations for 1D kinematics so if anyone can show me i'd be much obliged.
    Split the problem into two parts! Part 1 and part 2!

    In part 1, the bolt is being accelerated from rest for 4 seconds.

    $\displaystyle v_1 = u_1 + a_1t_1 $

    $\displaystyle v_1 = a_1t_1 $

    $\displaystyle v_1 = 4a_1 $

    We don't know what value of the acceleration, that's what we're trying to find! But we do know one thing about it, it's negative, since the rocket flies upwards.

    Now let's have a look at part 2. In part 2 the bolt has a negative velocity $\displaystyle v_1 = 4a_1 $, and it accelerates towards the ground with $\displaystyle a_2 = 9.81 ms^{-2} $. And it then hits the ground and has velocity 0, and this process takes 6.6 seconds!

    $\displaystyle v_2 = u_2 + a_2t_2 $

    $\displaystyle 0 = 4a_1 + 9.81(6.6) $

    How's that?
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  5. #5
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    you guys are the best. I appreciate it!!!
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