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Math Help - Acceleration of a rocket ..

  1. #1
    Junior Member xterminal01's Avatar
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    Acceleration of a rocket ..

    A rocket is launched straight up with a constant acceleration of 200 feet per sec each sec.

    A. Find the rocket's velocity 20 sec after launch
    B. Find its altitude at the end of 20 sec
    The results are a. 550 ft per sec b. 40,550 ft high
    I am just having problems showing the steps !!!
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  2. #2
    No one in Particular VonNemo19's Avatar
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    Quote Originally Posted by xterminal01 View Post
    A rocket is launched straight up with a constant acceleration of 200 feet per sec each sec.

    A. Find the rocket's velocity 20 sec after launch
    B. Find its altitude at the end of 20 sec
    The results are a. 550 ft per sec b. 40,550 ft high
    I am just having problems showing the steps !!!
    Where'd you get those results?

    Start with the acceleration function and integrate your way to distance.
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  3. #3
    MHF Contributor
    skeeter's Avatar
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    Quote Originally Posted by xterminal01 View Post
    A rocket is launched straight up with a constant acceleration of 200 feet per sec each sec.

    A. Find the rocket's velocity 20 sec after launch
    B. Find its altitude at the end of 20 sec
    The results are a. 550 ft per sec b. 40,550 ft high
    I am just having problems showing the steps !!!
    I do not agree with your "answers".

    a)

    v = v_0 + at

    a rocket starts from rest, v_0 = 0

    v = 200(20) = 4000 ft/s

    b) d = v_0 t + \frac{1}{2}at^2

    d = \frac{1}{2}(200)(20^2) = 40000 ft
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  4. #4
    No one in Particular VonNemo19's Avatar
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    Quote Originally Posted by xterminal01 View Post
    A rocket is launched straight up with a constant acceleration of 200 feet per sec each sec.

    A. Find the rocket's velocity 20 sec after launch
    B. Find its altitude at the end of 20 sec
    The results are a. 550 ft per sec b. 40,550 ft high
    I am just having problems showing the steps !!!
    \int{200}dt=200t+c=V(t)

    at t=0, v(t)=0, therefore c=0


    \int{200t}=100t^2+c=s(t)

    again, at t=0, s(t)=0, therefore c=0

    Altitude= s(20)=100(20)^2=40,000
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