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Math Help - [SOLVED] Calculus Question

  1. #1
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    [SOLVED] Calculus Question

    Hello, I need to hand this in tomorrow. It's within a Calculus paper.

    A Rocket is modelled by a particle which moves along a vertical line. From launch, the rocket rises until its motor cuts out after 14 seconds. At this time it has reached a height of 480 metres above the launch pad and attained an upward velocity of  70ms^{-1}. From this time on, the rocket has constant upward acceleration -10ms^{-2} (due to the effect of gravity alone).

    Choose the s-axis (for the position of the particle that represents the rocket) to point upwards, with origin at the launch pad. Take  t=0 to be the time when the rocket motor cuts out.

    a) What is the maximum height (above the launch pad) reached by the rocket?

    b) How long (form launch) does the rocket take to reach this maximum height?

    c) After how long (from launch) does the rocket crash onto the launch pad?
    What I'm doing is wrong as I need to work out the answer to part b before I can answer part a. Plus this is a calculus question and I'm not sure how that fits in with my attempt at an answer. There must be an easier less complicated way.

    Formula for deceleration;

    b)

    deceleration = \frac{(Inital Velocity - final Velocity)} {time}

    We know at the point of it's maximum;

    -10ms^{-2} = \frac{(70ms^{-1} - 0)} t
     t = \frac{70ms^{-1}} {-10ms^{-2}}

    a)

    average velocity = \frac {distance} {time}

    from part (b) we know the know the time (t). Therefore;

     70ms^{-1}+10ms^{-2} = \frac {d} {t}

     d = \frac{70ms^{-1} +10ms^{-2}} {t}

    c)

     14 seconds + t + x

    acceleration = \frac {Final Velocity - Initial Velocity} {time}

     -10ms^{-2} = \frac {Final Velocity - 70ms^{-1}} {x}

     x = \frac {Final Velocity - 70ms^{-1}} {-10ms^{-2}}
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  2. #2
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    hi
    do you know the work-energy theorem?
    if yes you should use it to find the max height
    b) seems ok
    c) we'll see....
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  3. #3
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    Thanks for your reply, I'm not too familiar with the work-energy theoreom, I've been reading up on it, and from what I understand the formula that will be useful is;

    a=\frac{dv}{dt}=\frac{dv}{ds}\frac{ds}{dt}=v\frac{  dv}{ds}

    But I have no idea on how to apply that to this problem.

    Also could someone have a look at this, and tell me if it's useful?

    deceleration = \frac{(Inital Velocity - final Velocity)} {time}

    We know at the point of it's maximum;

    -10ms^{-2} = \frac{(70ms^{-1} - 0)} t

     t = \frac{70ms^{-1}} {-10ms^{-2}}

     t = \frac{70m}{s}\frac{-10m}{s^2}

     t = \frac{-700m^2}{s^3}

    I'm not too sure what to do here either?
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  4. #4
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    initial height when the motors cut off ... 480 m

    velocity at that time ... 70 m/s

    a = \frac{dv}{dt} = -10

    v = -10t + C

    v(0) = 70

    70 = C

    v = 70 - 10t

    \frac{dh}{dt} = 70 - 10t

    h = 70t - 5t^2 + C

    h(0) = 480

    C = 480

    h = 480 + 70t - 5t^2

    at max height ... v = 0

    h = 0 when it hits the ground

    solve for t in both situations.
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